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Binomial distribution

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52341712
For each binomial probability expression, identify the values of \(n\), \(p\), and \(k\). a) \(P(X=11)=\binom{20}{11}(0.75)^{11}(0.25)^9\) b) \(P(X=2)=\binom{50}{2}(0.05)^2(0.95)^{48}\) c) \(P(X=5)=\binom{9}{5}\left(\frac14\right)^5\left(\frac34\right)^4\)

Hints

- Identify the total number of trials from the binomial coefficient. - The lower number in the binomial coefficient is the success count. - The base raised to the success-count exponent is \(p\). - The two exponents add to \(n\).

Solution

1. Compare each expression with \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\). 2. In part a), \(n=20\), \(p=0.75\), and \(k=11\). 3. In part b), \(n=50\), \(p=0.05\), and \(k=2\). 4. In part c), \(n=9\), \(p=\frac14\), and \(k=5\).

Answer

a) \(n=20\), \(p=0.75\), and \(k=11\) b) \(n=50\), \(p=0.05\), and \(k=2\) c) \(n=9\), \(p=\frac14\), and \(k=5\)
52341812
A binomial experiment has \(n=16\), \(p=\frac13\), and \(k=4\). a) Write the complete expression for \(P(X=4)\). b) State the exponent on the failure probability and explain how it is found from \(n\) and \(k\).

Hints

- Match the values with the standard binomial probability formula. - Find the failure probability from \(1-p\). - Subtract the number of successes from the number of trials. - The success and failure exponents sum to \(n\).

Solution

1. The binomial probability formula is \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\). 2. Substituting the values gives \(P(X=4)=\binom{16}{4}\left(\frac13\right)^4\left(\frac23\right)^{12}\). 3. The exponent on the failure probability is \(n-k=16-4=12\), the number of failures.

Answer

a) \(P(X=4)=\binom{16}{4}\left(\frac13\right)^4\left(\frac23\right)^{12}\) b) The exponent is \(12\), found from \(n-k\).
52340712
A binomial probability is written as \(P(X=6)=\binom{15}{k}p^6(0.75)^m\). Find the missing values \(k\), \(p\), and \(m\), and then evaluate the probability to four decimal places.

Hints

- Match each part of the expression with the standard binomial probability formula. - The success and failure exponents add to the total number of trials. - The two outcome probabilities add to \(1\).

Solution

1. Comparing the expression with \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\), the number of successes is \(k=6\). 2. The failure exponent is \(m=n-k=15-6=9\). 3. Since \(1-p=0.75\), \(p=0.25\). 4. Therefore, \(P(X=6)=\binom{15}{6}(0.25)^6(0.75)^9\approx0.0917\).

Answer

\(k=6\), \(p=0.25\), \(m=9\), and \(P(X=6)\approx0.0917\)
52340812
Complete each binomial probability expression by filling in the missing values. a) \(P(X=3)=\binom{10}{\square}(0.4)^{\triangle}(0.6)^{\bigcirc}\) b) \(P(X=\square)=\binom{n}{5}\left(\frac13\right)^5(\triangle)^7\) c) \(P(X=2)=\binom{\square}{2}p^2(1-p)^8\)

Hints

- The two exponents in a binomial probability add to \(n\). - The lower number in \(\binom{n}{k}\) is the number of successes. - The success and failure probabilities add to \(1\).

Solution

1. In part a), the success count is \(3\), so \(\square=3\) and \(\triangle=3\). The failure exponent is \(10-3=7\), so \(\bigcirc=7\). 2. In part b), the lower number in the binomial coefficient is the success count, so \(\square=5\). The total number of trials is \(5+7=12\), and the failure probability is \(1-\frac13=\frac23\), so \(\triangle=\frac23\). 3. In part c), the success and failure exponents add to \(n\). Thus, \(n=2+8=10\), so \(\square=10\).

Answer

a) \(\square=3\), \(\triangle=3\), and \(\bigcirc=7\) b) \(\square=5\) and \(\triangle=\frac23\), with \(n=12\) c) \(\square=10\)
52341212
Let \(X\) be binomial with \(n=12\) and \(p=\frac14\). Find \(P(X=4)\). Round to four decimal places.

Hints

- Find the probability of a failure on one trial. - Use the binomial coefficient to count the placements of the four successes. - Apply the exponents carefully. - Round only the final result.

Solution

1. Here, \(n=12\), \(k=4\), \(p=\frac14\), and \(1-p=\frac34\). 2. Use the binomial probability formula: \(P(X=4)=\binom{12}{4}\left(\frac14\right)^4\left(\frac34\right)^8\). 3. Since \(\binom{12}{4}=495\), the probability is \(495\cdot(0.25)^4\cdot(0.75)^8\approx0.1935777\). 4. Rounded to four decimal places, \(P(X=4)\approx0.1936\).

Answer

\(P(X=4)\approx0.1936\)
52341612
An archer hits the target independently with probability \(0.80\) on each shot. The archer takes \(12\) shots. Find the probability of each event. a) Exactly \(10\) hits b) Exactly \(8\) hits c) No hits

Hints

- Write the hit and miss probabilities as decimals. - Identify the total number of shots. - Use the binomial probability formula for each fixed number of hits. - Use scientific notation for a very small probability when helpful.

Solution

1. Let \(X\) be the number of hits. Then \(X\) is binomial with \(n=12\) and \(p=0.80\). 2. \(P(X=10)=\binom{12}{10}(0.8)^{10}(0.2)^2\approx0.2835\). 3. \(P(X=8)=\binom{12}{8}(0.8)^8(0.2)^4\approx0.1329\). 4. \(P(X=0)=\binom{12}{0}(0.8)^0(0.2)^{12}=4.096 \times 10^{-9}\).

Answer

a) \(P(X=10)\approx0.2835\), or about \(28.35\%\) b) \(P(X=8)\approx0.1329\), or about \(13.29\%\) c) \(P(X=0)=4.096 \times 10^{-9}\), or \(0.0000004096\%\)
52341912
Two factory machines produce components. Machine A produces a nondefective component with probability \(0.90\), while Machine B does so with probability \(0.70\). Assume component outcomes are independent for each machine. A sample of \(5\) components is taken from each machine. For which machine is it more likely that exactly \(4\) of the \(5\) components are nondefective? Support your answer with calculations.

Hints

- Exactly four nondefective components means exactly one defective component. - Use the binomial probability formula for each machine separately. - Compare the two probabilities after calculating them.

Solution

1. For Machine A, \(P(X=4)=\binom54(0.9)^4(0.1)=0.32805\). 2. For Machine B, \(P(X=4)=\binom54(0.7)^4(0.3)=0.36015\). 3. Since \(0.36015>0.32805\), exactly four nondefective components are more likely for Machine B.

Answer

Machine B. Its probability is \(0.36015\), compared with \(0.32805\) for Machine A.
52342012
Two archers, Jordan and Leah, each take \(3\) independent shots. Jordan hits the center with probability \(0.50\), while Leah hits the center with probability \(0.75\). Which event is more likely: Jordan gets exactly \(2\) center hits, or Leah gets center hits on all \(3\) shots?

Hints

- Identify the number of trials and success probability for each archer. - Count the arrangements that give Jordan exactly two hits. - There is only one outcome pattern in which Leah hits all three times. - Compare the resulting probabilities.

Solution

1. For Jordan, \(P(X=2)=\binom32(0.5)^2(0.5)=0.375\). 2. For Leah, \(P(Y=3)=(0.75)^3=0.421875\). 3. Since \(0.421875>0.375\), Leah getting three center hits is more likely.

Answer

Leah hitting the center on all \(3\) shots is more likely: \(0.421875\) compared with \(0.375\).
52342312
A basketball player makes each free throw independently with probability \(p=0.6\). During practice, the player takes \(5\) free throws. Use the binomial probability formula to find the probability of making exactly \(3\) shots.

Hints

- Identify \(n\), \(p\), and \(k\). - Find the probability of a missed free throw. - Use the binomial formula for an exact number of successes. - Include the binomial coefficient for the arrangements of the three made shots.

Solution

1. The parameters are \(n=5\), \(p=0.6\), and \(k=3\), with failure probability \(0.4\). 2. \(P(X=3)=\binom53(0.6)^3(0.4)^2\). 3. Since \(\binom53=10\), \((0.6)^3=0.216\), and \((0.4)^2=0.16\), the probability is \(10\cdot0.216\cdot0.16=0.3456\).

Answer

\(P(X=3)=0.3456\), or \(34.56\%\)
52342412
In a manufacturing process, \(20\%\) of components are defective. A sample of \(10\) components is selected, and the selections can be modeled as independent Bernoulli trials. Find the probability that exactly \(2\) components are defective.

Hints

- Define a success as selecting a defective component. - Identify the total number of trials and the success probability. - Find the failure probability. - Use the binomial formula for exactly \(2\) successes.

Solution

1. Let \(X\) be the number of defective components. Then \(X\) is binomial with \(n=10\), \(p=0.2\), and \(k=2\). 2. \(P(X=2)=\binom{10}{2}(0.2)^2(0.8)^8\). 3. Since \(\binom{10}{2}=45\), the probability is \(45\cdot0.04\cdot0.16777216=0.301989888\approx0.3020\).

Answer

\(P(X=2)\approx0.3020\), or about \(30.20\%\)
52343112
Each LED light bulb produced by a manufacturer independently has a \(5\%\) chance of having a slight color variation. A quality-control inspector selects \(30\) bulbs. a) Find the probability that none of the bulbs in the sample has a color variation. b) Find the probability that at most one bulb in the sample has a color variation.

Hints

- Decide whether the situation can be modeled as repeated independent trials with two outcomes. - Identify the values of \(n\) and \(p\). - Which values of \(X\) are included in “at most one”? - Use the binomial probability formula for exactly \(k\) successes.

Solution

1. Let \(X\) be the number of bulbs with a color variation. Then \(X\) has a binomial distribution with \(n=30\) and \(p=0.05\). 2. \(P(X=0)=\binom{30}{0}(0.05)^0(0.95)^{30}=(0.95)^{30}\approx 0.2146\). 3. For part b), \(P(X\le 1)=P(X=0)+P(X=1)\). 4. \(P(X=1)=\binom{30}{1}(0.05)(0.95)^{29}\approx 0.3389\). 5. Therefore, \(P(X\le 1)\approx 0.2146+0.3389=0.5535\).

Answer

a) The probability is approximately \(21.46\%\). b) The probability is approximately \(55.35\%\).
52343212
A biathlete hits the target on a standing shot with probability \(0.85\). Assume the outcomes of the shots are independent. During one shooting stage, she takes \(5\) shots. a) What is the probability that she makes exactly \(4\) hits? b) What is the probability that she makes at least \(4\) hits?

Hints

- What values of \(X\) are included when the problem asks for “at least \(4\)” hits? - Identify the values of \(n\) and \(p\). - Use the binomial probability formula for exactly \(k\) hits. - How many ways can \(4\) hits occur among \(5\) shots?

Solution

1. Let \(X\) be the number of hits. Then \(X\) has a binomial distribution with \(n=5\) and \(p=0.85\). 2. \(P(X=4)=\binom{5}{4}(0.85)^4(0.15)\approx 0.3915\). 3. For part b), \(P(X\ge 4)=P(X=4)+P(X=5)\). 4. \(P(X=5)=(0.85)^5\approx 0.4437\). 5. Therefore, \(P(X\ge 4)\approx 0.3915+0.4437=0.8352\).

Answer

a) The probability is approximately \(39.15\%\). b) The probability is approximately \(83.52\%\).
52343312
An archer independently takes \(4\) shots and hits the center of a target with probability \(0.60\) on each shot. Find the probability of exactly \(2\) center hits.

Hints

- Check that the hit probability stays constant from shot to shot. - Identify \(n\), \(p\), and \(k\). - Use the binomial probability formula. - Include the number of arrangements of the two hits.

Solution

1. The parameters are \(n=4\), \(p=0.6\), and \(k=2\). 2. The number of arrangements of two hits among four shots is \(\binom42=6\). 3. The probability is \(P(X=2)=\binom42(0.6)^2(0.4)^2=6\cdot0.36\cdot0.16=0.3456\).

Answer

\(P(X=2)=0.3456\), or \(34.56\%\)
52343412
A gardener knows that each seed of a certain flower variety germinates independently with probability \(0.80\). The gardener plants \(7\) seeds. Find the probability that exactly \(5\) seeds germinate.

Hints

- Model each seed as a success-or-failure trial. - Identify the binomial parameters. - Count the arrangements that contain exactly five germinations. - Include the probability of the two seeds that do not germinate.

Solution

1. The parameters are \(n=7\), \(p=0.8\), and \(k=5\). 2. The number of arrangements is \(\binom75=21\). 3. The probability of one arrangement with five germinations and two failures is \((0.8)^5(0.2)^2=0.0131072\). 4. Therefore, \(P(X=5)=21\cdot0.0131072=0.2752512\).

Answer

\(P(X=5)=0.2752512\), or about \(27.53\%\)
52344112
Use \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\) to calculate each binomial probability. Round to four decimal places when needed. a) \(P(X=3)\) for \(n=5\) and \(p=0.4\) b) \(P(X=2)\) for \(n=6\) and \(p=0.1\) c) \(P(X=0)\) for \(n=4\) and \(p=0.5\)

Hints

- Identify \(n\), \(p\), and \(k\) for each part. - Find the failure probability \(1-p\). - Evaluate the binomial coefficient before multiplying.

Solution

1. For part a), \(P(X=3)=\binom53(0.4)^3(0.6)^2=10\cdot0.064\cdot0.36=0.2304\). 2. For part b), \(P(X=2)=\binom62(0.1)^2(0.9)^4=15\cdot0.01\cdot0.6561=0.098415\approx0.0984\). 3. For part c), \(P(X=0)=\binom40(0.5)^0(0.5)^4=0.0625\).

Answer

a) \(0.2304\) b) \(0.0984\) c) \(0.0625\)
52344212
A binomial random variable has \(n=10\) and \(p=0.2\). Determine whether \(X=1\) or \(X=2\) is more likely.

Hints

- Calculate both probabilities separately. - Keep enough decimal places to compare them reliably. - Compare the final values directly.

Solution

1. \(P(X=1)=\binom{10}{1}(0.2)(0.8)^9=0.268435456\). 2. \(P(X=2)=\binom{10}{2}(0.2)^2(0.8)^8=0.301989888\). 3. Since \(0.301989888>0.268435456\), \(X=2\) is more likely.

Answer

\(X=2\) is more likely: about \(30.2\%\), compared with about \(26.8\%\) for \(X=1\).
52344312
The probability of an event \(A\) is given by \(P(A)=\binom{12}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{10}\). Describe a random experiment that this formula could model, and state event \(A\) in words.

Hints

- What do the exponents tell you about the total number of trials? - What does \(\binom{12}{2}\) tell you about the number of successes? - What familiar random device gives a success probability of \(\frac{1}{6}\)? - Decide which outcome will count as a success.

Solution

1. The expression has the binomial form \(\binom{n}{k}p^k(1-p)^{n-k}\), with \(n=12\), \(k=2\), and \(p=\frac{1}{6}\). 2. A suitable experiment is rolling a fair six-sided die \(12\) times and defining a success as rolling a \(6\). 3. Therefore, event \(A\) is rolling exactly two \(6\)s in the \(12\) rolls.

Answer

Random experiment: Roll a fair six-sided die \(12\) times. Event \(A\): Exactly two of the rolls show a \(6\).
52345312
Use a calculator to evaluate each binomial probability. Round each answer to four decimal places. a) \(P(X=3)\) when \(n=15\) and \(p=0.2\) b) \(P(X=10)\) when \(n=20\) and \(p=0.45\) c) \(P(X=2)\) when \(n=50\) and \(p=0.05\)

Hints

- Identify \(n\), \(p\), and \(k\) for each probability. - Many calculators use a command such as binompdf for an exact binomial probability. - Use the fifth decimal place to round to four decimal places.

Solution

1. For part a), \(P(X=3)=\binom{15}{3}(0.2)^3(0.8)^{12}\approx0.2501\). 2. For part b), \(P(X=10)=\binom{20}{10}(0.45)^{10}(0.55)^{10}\approx0.1593\). 3. For part c), \(P(X=2)=\binom{50}{2}(0.05)^2(0.95)^{48}\approx0.2611\).

Answer

a) \(0.2501\) b) \(0.1593\) c) \(0.2611\)
52345412
A fair six-sided die is rolled \(10\) times. Use a calculator to find each probability. Round to four decimal places. a) Exactly two rolls show a \(6\). b) Exactly five rolls show a \(6\). c) No rolls show a \(6\).

Hints

- Find the probability of rolling a \(6\) on one roll. - Identify the total number of trials. - Match each event to the required number of successes. - Use the binomial probability formula or your calculator's exact binomial probability command.

Solution

1. Let \(X\) be the number of rolls that show a \(6\). Then \(X\) is binomial with \(n=10\) and \(p=\frac{1}{6}\). 2. For part a), \(P(X=2)=\binom{10}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^8\approx0.2907\). 3. For part b), \(P(X=5)=\binom{10}{5}\left(\frac{1}{6}\right)^5\left(\frac{5}{6}\right)^5\approx0.0130\). 4. For part c), \(P(X=0)=\binom{10}{0}\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^{10}\approx0.1615\).

Answer

a) \(0.2907\) b) \(0.0130\) c) \(0.1615\)
52345512
A fair four-sided die, numbered \(1\) through \(4\), is rolled \(8\) times. Let \(X\) be the number of times a \(3\) is rolled. a) Find the probability \(p\) of a success, rolling a \(3\), on one roll. b) Use the binomial probability formula to find \(P(X=2)\). Show the binomial coefficient and the two powers before giving the final probability.

Hints

- What is the probability of one specified face on a fair four-sided die? - Identify \(n\), \(k\), and \(p\). - Calculate \(\binom{8}{2}\). - Include the probability of the six rolls that are not \(3\)s.

Solution

1. Since the four outcomes are equally likely, \(p=\frac{1}{4}=0.25\), and \(1-p=0.75\). 2. For \(n=8\) and \(k=2\), \(P(X=2)=\binom{8}{2}(0.25)^2(0.75)^6\). 3. The intermediate values are \(\binom{8}{2}=28\), \((0.25)^2=0.0625\), and \((0.75)^6\approx0.17798\). 4. Therefore, \(P(X=2)=28\cdot0.0625\cdot(0.75)^6=0.31146240234375\approx0.3115\).

Answer

a) \(p=0.25\) b) \(P(X=2)\approx0.3115\), or about \(31.15\%\)
52345612
A city survey reports that \(65\%\) of residents regularly ride a bicycle. A random sample of \(100\) residents is selected, and the count can be modeled with a binomial random variable \(X\). Use a calculator to find the probability that exactly \(60\) sampled residents regularly ride a bicycle. Round to four decimal places.

Hints

- Identify the total number of trials, the number of successes, and the success probability. - Decide whether the question asks for one exact value or a cumulative probability. - Use your calculator's exact binomial probability command.

Solution

1. The binomial parameters are \(n=100\), \(p=0.65\), and \(k=60\). 2. The required exact probability is \(P(X=60)=\binom{100}{60}(0.65)^{60}(0.35)^{40}\). 3. A calculator gives \(P(X=60)\approx0.0474\).

Answer

\(P(X=60)\approx0.0474\)
52346112
At a high school, \(80\%\) of students usually eat breakfast before school. A random sample of \(6\) students is selected. The table shows probabilities for the number of sampled students who ate breakfast before school. Use the table to find the probability of each event: - Event A: Exactly \(5\) or \(6\) students ate breakfast. - Event B: At least \(4\) students ate breakfast. - Event C: At most \(3\) students ate breakfast. <table> <thead> <tr> <th>Number of students \(k\)</th> <th>Probability \(P(X=k)\)</th> </tr> </thead> <tbody> <tr> <td>\(6\)</td> <td>\(0.2621\)</td> </tr> <tr> <td>\(5\)</td> <td>\(0.3932\)</td> </tr> <tr> <td>\(4\)</td> <td>\(0.2458\)</td> </tr> </tbody> </table>

Hints

- What operation corresponds to the word “or” when the outcomes cannot happen at the same time? - Which values of \(k\) are included in “at least \(4\)”? - How are the events “at least \(4\)” and “at most \(3\)” related? - How can you use a complement when the total probability is \(1\)?

Solution

1. For Event A, add the probabilities for \(k=5\) and \(k=6\): \(P(A)=0.3932+0.2621=0.6553\). 2. For Event B, add the probabilities for \(k=4\), \(k=5\), and \(k=6\): \(P(B)=0.2458+0.3932+0.2621=0.9011\). 3. Event C is the complement of Event B, so \(P(C)=1-P(B)=1-0.9011=0.0989\).

Answer

Event A: \(P(A)=0.6553\) Event B: \(P(B)=0.9011\) Event C: \(P(C)=0.0989\)
52358512
A random variable \(X\) has a binomial distribution with \(n=3\) and \(p=0.4\). Create a complete probability distribution table for \(X\).

Hints

- What values can \(X\) take in three trials? - Use the formula for the probability of exactly \(k\) successes in \(n\) independent trials. - Find the probability of a failure before substituting values. - A probability distribution table pairs each possible value of \(X\) with its probability.

Solution

1. The success probability is \(p=0.4\), the failure probability is \(1-p=0.6\), and the number of trials is \(n=3\). 2. Use the binomial probability formula \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\) for \(k\in\{0,1,2,3\}\). 3. \(P(X=0)=\binom{3}{0}(0.4)^0(0.6)^3=0.216\). 4. \(P(X=1)=\binom{3}{1}(0.4)^1(0.6)^2=0.432\). 5. \(P(X=2)=\binom{3}{2}(0.4)^2(0.6)^1=0.288\). 6. \(P(X=3)=\binom{3}{3}(0.4)^3(0.6)^0=0.064\).

Answer

<table> <tr> <td>\(k\)</td> <td>\(0\)</td> <td>\(1\)</td> <td>\(2\)</td> <td>\(3\)</td> </tr> <tr> <td>\(P(X=k)\)</td> <td>\(0.216\)</td> <td>\(0.432\)</td> <td>\(0.288\)</td> <td>\(0.064\)</td> </tr> </table>
52358612
A random variable \(X\) has a binomial distribution with \(n=5\) and \(p=0.2\). a) Calculate \(P(X=k)\) for every \(k\in\{0,1,2,3,4,5\}\). b) Check the distribution by adding all the individual probabilities.

Hints

- Use the binomial probability formula for each value of \(k\). - The binomial coefficient \(\binom{n}{k}\) counts the arrangements with exactly \(k\) successes. - Recall what the probabilities in any complete distribution must add to. - Track the exponent on the failure probability carefully.

Solution

1. The parameters are \(n=5\), \(p=0.2\), and \(1-p=0.8\). 2. Apply \(P(X=k)=\binom{5}{k}(0.2)^k(0.8)^{5-k}\): \(P(X=0)=\binom{5}{0}(0.2)^0(0.8)^5=0.32768\) \(P(X=1)=\binom{5}{1}(0.2)^1(0.8)^4=0.4096\) \(P(X=2)=\binom{5}{2}(0.2)^2(0.8)^3=0.2048\) \(P(X=3)=\binom{5}{3}(0.2)^3(0.8)^2=0.0512\) \(P(X=4)=\binom{5}{4}(0.2)^4(0.8)^1=0.0064\) \(P(X=5)=\binom{5}{5}(0.2)^5(0.8)^0=0.00032\) 3. The sum is \(0.32768+0.4096+0.2048+0.0512+0.0064+0.00032=1\), as required for a probability distribution.

Answer

a) \(P(X=0)=0.32768\); \(P(X=1)=0.4096\); \(P(X=2)=0.2048\); \(P(X=3)=0.0512\); \(P(X=4)=0.0064\); \(P(X=5)=0.00032\). b) \(0.32768+0.4096+0.2048+0.0512+0.0064+0.00032=1\).
52380612
An urn contains \(5\) red balls and \(3\) yellow balls. Two balls are drawn with replacement. The random variable \(X\) is the number of red balls drawn. a) Create the probability distribution of \(X\). b) Find the probability that at least one red ball is drawn.

Hints

- Identify the probability of drawing a red ball on one draw. - Because the draws are with replacement, use the same success probability for both trials. - Use the binomial probability formula for \(X=0\), \(X=1\), and \(X=2\). - For part b), consider using the complement of drawing at least one red ball.

Solution

1. On each draw, \(P(\text{red})=\frac{5}{8}\) and \(P(\text{yellow})=\frac{3}{8}\). Because the ball is replaced, the draws are independent. 2. Use the binomial probability formula with \(n=2\) and \(p=\frac{5}{8}\): \(P(X=0)=\binom{2}{0}\left(\frac{5}{8}\right)^0\left(\frac{3}{8}\right)^2=\frac{9}{64}\). \(P(X=1)=\binom{2}{1}\left(\frac{5}{8}\right)^1\left(\frac{3}{8}\right)^1=\frac{30}{64}=\frac{15}{32}\). \(P(X=2)=\binom{2}{2}\left(\frac{5}{8}\right)^2\left(\frac{3}{8}\right)^0=\frac{25}{64}\). 3. For at least one red ball, use the complement: \(P(X\ge1)=1-P(X=0)=1-\frac{9}{64}=\frac{55}{64}\).

Answer

a) <table border="1"> <tr> <td>\(x\)</td> <td>\(0\)</td> <td>\(1\)</td> <td>\(2\)</td> </tr> <tr> <td>\(P(X=x)\)</td> <td>\(\frac{9}{64}\)</td> <td>\(\frac{30}{64}\)</td> <td>\(\frac{25}{64}\)</td> </tr> </table> b) \(P(X\ge1)=\frac{55}{64}\approx85.9\%\)
52688012
A spinner has a red section that covers exactly one-fourth of the spinner. The spinner is spun independently \(6\) times. a) Find the probability that only the first two spins land on red. b) Find the probability that exactly two of the \(6\) spins land on red. c) Find the probability that none of the spins lands on red.

Hints

- Find the probability of red on one spin. - Distinguish between one specified order and any order with the same number of red outcomes. - Use a binomial coefficient to count arrangements when the positions are not specified. - For no red outcomes, multiply the probability of not red across all spins.

Solution

1. The probability of red is \(p=0.25\), and the probability of not red is \(0.75\). 2. For part a), the specified sequence is red, red, not red, not red, not red, not red. Its probability is \((0.25)^2(0.75)^4=\frac{81}{4096}\approx0.0198\). 3. For part b), the two red outcomes can occur in \(\binom62=15\) arrangements. Therefore, \(P(X=2)=\binom62(0.25)^2(0.75)^4=\frac{1215}{4096}\approx0.2966\). 4. For part c), \(P(X=0)=(0.75)^6=\frac{729}{4096}\approx0.1780\).

Answer

a) \(\frac{81}{4096}\approx0.0198\) b) \(\frac{1215}{4096}\approx0.2966\) c) \(\frac{729}{4096}\approx0.1780\)
52690412
A multiple-choice test has \(15\) questions. Each question has four answer choices, exactly one of which is correct. A student guesses independently and randomly on every question. Find the probability that the student answers exactly \(5\) questions correctly.

Hints

- Decide whether the questions can be modeled as independent trials with the same success probability. - Find the probability of guessing one question correctly. - Use the binomial probability formula for an exact number of correct answers.

Solution

1. This is a binomial setting with \(n=15\), \(p=0.25\), and \(k=5\). 2. \(P(X=5)=\binom{15}{5}(0.25)^5(0.75)^{10}\). 3. Since \(\binom{15}{5}=3003\), the probability is \(3003\cdot(0.25)^5(0.75)^{10}\approx0.1651\).

Answer

\(P(X=5)\approx0.1651\), or about \(16.51\%\)
52698112
A quiz has \(10\) multiple-choice questions. Each question has five answer choices, exactly one of which is correct. A student guesses independently and randomly on every question. a) How many correct-or-incorrect answer patterns contain exactly three correct answers? b) Find the probability that the student answers exactly three questions correctly. c) Find the probability that only the first three answers are correct and all remaining answers are incorrect.

Hints

- Choose which three of the ten positions contain correct answers. - Model the questions as independent trials with the same success probability. - Distinguish between all patterns with three correct answers and one specified pattern. - A fixed pattern does not need a binomial coefficient.

Solution

1. For part a), choose the three positions of the correct answers: \(\binom{10}{3}=120\). 2. For part b), the success probability is \(p=\frac15=0.2\). Thus, \(P(X=3)=\binom{10}{3}(0.2)^3(0.8)^7=0.201326592\approx0.2013\). 3. For part c), the order is fixed, so no binomial coefficient is used. The probability is \((0.2)^3(0.8)^7=0.0016777216\approx0.0017\).

Answer

a) \(120\) patterns b) \(P(X=3)\approx0.2013\), or about \(20.13\%\) c) \(0.0016777216\approx0.0017\), or about \(0.17\%\)
52698412
A spinner has three equal sections colored red, blue, and green. It is spun independently \(8\) times. a) Find the probability that exactly \(3\) spins land on red. b) Find the probability that exactly \(6\) spins land on either red or blue.

Hints

- Find the probability of one color on a spinner with three equal sections. - For red or blue, combine the probabilities of the two disjoint outcomes. - Use a binomial coefficient to count the arrangements of the successes. - Check that the success and failure probabilities add to \(1\).

Solution

1. For part a), define success as landing on red. Then \(p=\frac13\), so \(P(X=3)=\binom83\left(\frac13\right)^3\left(\frac23\right)^5=\frac{1792}{6561}\approx0.2731\). 2. For part b), define success as landing on red or blue. Then \(p=\frac23\), so \(P(Y=6)=\binom86\left(\frac23\right)^6\left(\frac13\right)^2=\frac{1792}{6561}\approx0.2731\).

Answer

a) \(\frac{1792}{6561}\approx0.2731\), or about \(27.31\%\) b) \(\frac{1792}{6561}\approx0.2731\), or about \(27.31\%\)
52699012
In a manufacturing process, the probability that a microchip is defective is \(5\%\), and chip statuses are independent. A nondefective chip is labeled \(N\), and a defective chip is labeled \(D\). Eight chips are inspected in order. Consider these three inspection records: (1) \((N, N, N, N, N, N, N, N)\) (2) \((D, N, N, N, N, N, N, N)\) (3) \((D, D, D, D, D, D, D, D)\) a) Which record has the smallest probability? Explain without calculating. b) Calculate the probabilities of records (1) and (2). c) Someone claims, “The probability that exactly one chip is defective equals the probability of record (2).” Evaluate this claim mathematically.

Hints

- Compare the probability of a defective chip with the probability of a nondefective chip. - For a specified record, multiply the probabilities in that order. - Count the possible positions of one defective chip among eight chips.

Solution

1. Record (3) is least likely because a defective chip has probability \(0.05\), much less than the probability \(0.95\) of a nondefective chip, and record (3) requires eight defects. 2. For record (1), \(P(1)=(0.95)^8\approx0.6634\). 3. For record (2), \(P(2)=(0.05)(0.95)^7\approx0.0349\). 4. Exactly one defective chip can occur in \(\binom81=8\) different positions. Therefore, \(P(X=1)=\binom81(0.05)(0.95)^7\approx0.2793\). 5. The claim is false because record (2) is only one of the eight arrangements with exactly one defective chip.

Answer

a) Record (3) b) \(P(1)\approx0.6634\); \(P(2)\approx0.0349\) c) The claim is false. \(P(X=1)\approx0.2793\), which is eight times the probability of record (2).
52699112
An archer hits a target with probability \(p=0.8\) on each shot. The archer takes \(4\) independent shots. a) Find the probability of exactly two hits. b) Find the probability that only the first and fourth shots are hits.

Hints

- Decide whether the order matters in each part. - Count the ways to place two hits among four shots. - Use the binomial probability formula when only the number of hits is specified. - For one specified sequence, multiply the probabilities along that sequence.

Solution

1. The parameters are \(n=4\), \(p=0.8\), and \(1-p=0.2\). 2. For part a), \(P(X=2)=\binom42(0.8)^2(0.2)^2=6\cdot0.64\cdot0.04=0.1536\). 3. For part b), the order is fixed as hit, miss, miss, hit. Its probability is \(0.8\cdot0.2\cdot0.2\cdot0.8=(0.8)^2\cdot(0.2)^2=0.0256\).

Answer

a) \(0.1536\), or \(15.36\%\) b) \(0.0256\), or \(2.56\%\)
52699712
Determine whether each situation can be modeled as a binomial experiment. Justify your decision. For each binomial experiment, give the number of trials \(n\) and the success probability \(p\). a) A container holds \(40\) screws, \(4\) of which are defective. Five screws are selected without replacement and checked for defects. b) A factory produces electronic components with a defect probability of \(0.5\%\). One hundred components are selected from ongoing production and tested, and their defect statuses are assumed independent. c) A fair six-sided die is rolled \(8\) times. A success is an even result.

Hints

- A binomial experiment requires independent trials and a constant success probability. - How does sampling without replacement affect the next trial? - When the population is effectively very large, can the probability be modeled as constant? - Does one die roll affect the next roll?

Solution

1. In a), the selections are without replacement from a small population. The probability of selecting a defective screw changes after each selection, and the trials are dependent. This is not a binomial experiment. 2. In b), the model assumes independent trials and a constant defect probability. This is a binomial experiment with \(n=100\) and \(p=0.005\). 3. In c), the die rolls are independent, and the probability of an even result is constant: \(p=\frac{3}{6}=0.5\). This is a binomial experiment with \(n=8\) and \(p=0.5\).

Answer

a) Not binomial; without replacement, the success probability changes and the trials are dependent. b) Binomial with \(n=100\) and \(p=0.005\). c) Binomial with \(n=8\) and \(p=0.5\).
52699812
Decide whether each situation is appropriately modeled as a binomial experiment. For each binomial experiment, give \(n\) and \(p\). a) A multiple-choice test has \(10\) questions, each with \(4\) answer choices and exactly one correct answer. A student guesses independently on every question. b) A committee of \(3\) students is selected without replacement from a class with \(12\) girls and \(13\) boys. A success is selecting a girl. c) A spinner has three equal sectors colored red, yellow, and blue. It is spun \(15\) times. A success is landing on yellow.

Hints

- Check whether there are exactly two relevant outcomes on each trial. - Does the success probability remain constant? - Can the trials be treated as independent?

Solution

1. In a), each question has two relevant outcomes, correct or incorrect. The guesses are independent and the success probability is constant at \(p=\frac{1}{4}=0.25\). This is binomial with \(n=10\). 2. In b), students are selected without replacement from a small group. The probability of selecting a girl changes after each selection, so this is not binomial. 3. In c), the spins are independent and the probability of yellow remains \(p=\frac{1}{3}\). This is binomial with \(n=15\).

Answer

a) Binomial with \(n=10\) and \(p=0.25\). b) Not binomial because selection is without replacement and the success probability changes. c) Binomial with \(n=15\) and \(p=\frac{1}{3}\).
52700112
A gardener plants \(15\) seeds of a particular flower variety. Each seed germinates independently with probability \(85\%\). Find the probability of each event. a) Exactly \(13\) of the \(15\) seeds germinate. b) The first \(13\) seeds germinate, and the last two do not.

Hints

- Decide whether the order matters in each part. - Count the possible arrangements when only the total number of germinating seeds is specified. - Compare an exact sequence with all sequences containing the same number of successes.

Solution

1. Let a germinating seed be a success. Then \(n=15\), \(p=0.85\), and \(1-p=0.15\). 2. For part a), the positions of the \(13\) germinating seeds are not specified, so \(P(X=13)=\binom{15}{13}(0.85)^{13}(0.15)^2\approx0.2856\). 3. For part b), the order is specified, so the probability is \((0.85)^{13}(0.15)^2\approx0.0027\).

Answer

a) \(0.2856\), or about \(28.56\%\) b) \(0.0027\), or about \(0.27\%\)
52700512
A basketball player makes a free throw with probability \(0.75\). The player takes \(12\) free throws. Briefly state the assumptions needed to model the series as a binomial setting, and find the probability that the player makes exactly \(9\) free throws.

Hints

- Recall the conditions required for a binomial setting. - Check whether the success probability stays the same for every trial. - Use the binomial formula for an exact number of successes. - Explain what independence means for successive shots.

Solution

1. A binomial model is appropriate if each shot has two outcomes, make or miss; the make probability remains \(0.75\); and the shots are independent. 2. The parameters are \(n=12\), \(k=9\), \(p=0.75\), and \(1-p=0.25\). 3. \(P(X=9)=\binom{12}{9}(0.75)^9(0.25)^3\). 4. Since \(\binom{12}{9}=220\), the probability is \(220\cdot(0.75)^9(0.25)^3\approx0.2581\).

Answer

The model requires two outcomes per shot, a constant make probability of \(0.75\), and independent shots. The probability of exactly \(9\) makes is \(P(X=9)\approx0.2581\), or about \(25.81\%\).
52701312
A multiple-choice test has \(6\) questions. Each question has three answer choices, exactly one of which is correct. A student guesses independently and randomly on every question. Explain why \(\left(\frac13\right)^2\left(\frac23\right)^4\) is not the probability that the student answers exactly two questions correctly.

Hints

- Are there several possible positions for the two correct answers? - What event does multiplying two success probabilities and four failure probabilities describe? - Think of all paths in a tree diagram that contain exactly two successes. - Does the order of the two correct answers matter to the event?

Solution

1. The test can be modeled as a binomial setting with \(n=6\) and \(p=\frac13\). 2. The expression \(\left(\frac13\right)^2\left(\frac23\right)^4\) gives the probability of one specific answer pattern containing two correct answers and four incorrect answers. 3. There are \(\binom62=15\) possible positions for the two correct answers. 4. Therefore, the full probability is \(P(X=2)=\binom62\left(\frac13\right)^2\left(\frac23\right)^4=\frac{80}{243}\approx0.3292\).

Answer

The expression represents only one specific arrangement of two correct and four incorrect answers. Because there are \(\binom62=15\) such arrangements, it must be multiplied by \(15\). Thus, \(P(X=2)=\frac{80}{243}\approx0.3292\).
52701512
In a manufacturing process, \(90\%\) of the components are nondefective. A sample of \(6\) components is selected, and the outcomes are modeled as independent Bernoulli trials. 1. Find the probability that only the last selected component is defective. 2. Find the probability that exactly one component in the sample is defective. 3. Find the probability that the first three selected components are nondefective.

Hints

- Distinguish a specified sequence from an event defined only by the total number of defects. - When only the first outcomes are specified, later outcomes do not affect the event. - Count the possible positions for a fixed number of defects.

Solution

1. The probability of a nondefective component is \(0.9\), and the probability of a defective component is \(0.1\). For the fixed sequence in part 1, \(P=(0.9)^5(0.1)=0.059049\). 2. Let \(X\) count defective components. Then \(X\) is binomial with \(n=6\) and \(p=0.1\). Thus, \(P(X=1)=\binom61(0.1)(0.9)^5=0.354294\). 3. Only the first three outcomes are restricted, so the remaining outcomes do not affect the event. The probability is \((0.9)^3=0.729\).

Answer

1. \(0.059049\), or about \(5.90\%\) 2. \(0.354294\), or about \(35.43\%\) 3. \(0.729\), or \(72.9\%\)
52702312
A fair six-sided die is rolled \(15\) times, and the rolls are independent. Let \(X\) be the number of \(6\)s rolled. Find each probability to four decimal places. 1. \(P(X=3)\) 2. \(P(X\le 4)\) 3. \(P(2\le X\le 5)\)

Hints

- Identify \(n\) and \(p\) for the experiment. - Distinguish between a probability for one exact value and a cumulative probability. - Express an interval probability as the difference of two cumulative probabilities. - Use the binomial probability functions on your calculator.

Solution

1. The random variable \(X\) has a binomial distribution with \(n=15\) and \(p=\frac{1}{6}\). 2. \(P(X=3)=\binom{15}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^{12}\approx 0.2363\). 3. A cumulative binomial calculation gives \(P(X\le 4)\approx 0.9102\). 4. \(P(2\le X\le 5)=P(X\le 5)-P(X\le 1)\). 5. Since \(P(X\le 5)\approx 0.9726\) and \(P(X\le 1)\approx 0.2596\), the probability is approximately \(0.7130\).

Answer

1. \(P(X=3)\approx 0.2363\) 2. \(P(X\le 4)\approx 0.9102\) 3. \(P(2\le X\le 5)\approx 0.7130\)
52702812
For a binomial random variable \(X\), define the cumulative distribution function by \(F(k)=P(X\le k)\). Use this definition to explain why \(F(k)-F(k-1)=P(X=k)\) for \(k\ge 1\).

Hints

- Write \(F(k)\) and \(F(k-1)\) as sums of individual probabilities. - Which outcome is included in \(F(k)\) but not in \(F(k-1)\)? - Identify the terms that cancel when the sums are subtracted.

Solution

1. By definition, \(F(k)=\sum_{i=0}^{k}P(X=i)\). 2. Similarly, \(F(k-1)=\sum_{i=0}^{k-1}P(X=i)\). 3. In the difference \(F(k)-F(k-1)\), every probability from \(P(X=0)\) through \(P(X=k-1)\) cancels. 4. The only remaining term is \(P(X=k)\). 5. Therefore, \(F(k)-F(k-1)=P(X=k)\).

Answer

Subtracting \(F(k-1)\) from \(F(k)\) removes all probabilities for values below \(k\), leaving exactly \(P(X=k)\).
52703912
For each expression, describe the probability it represents and calculate its value to four decimal places. a) \(\sum_{i=0}^{2}\binom{15}{i}(0.1)^i(0.9)^{15-i}\) b) \(1-\sum_{i=0}^{1}\binom{5}{i}(0.4)^i(0.6)^{5-i}\)

Hints

- Use the summation limits to identify the included values of the random variable. - For an expression beginning with \(1-\), identify the complement of the summed event. - Read \(n\), \(p\), and the possible values directly from each binomial term.

Solution

1. Part a) is the cumulative probability for at most \(2\) successes when \(n=15\) and \(p=0.1\). 2. \(P(X\le 2)\approx 0.8159\). 3. In part b), the sum is \(P(X\le 1)\) for \(n=5\) and \(p=0.4\). 4. Therefore, the full expression is \(1-P(X\le 1)=P(X\ge 2)\approx 0.6630\).

Answer

a) At most \(2\) successes; \(P\approx 0.8159\) b) At least \(2\) successes; \(P\approx 0.6630\)
52704512
In a manufacturing process, a microchip is defective with probability \(p=0.05\). A sample of \(8\) chips is selected, and chip statuses are independent. For each expression, describe an event in context whose probability it represents. a) \((0.05)^2(0.95)^6\) b) \(\binom82(0.05)^2(0.95)^6\) c) \(1-(0.95)^8\)

Hints

- Without a binomial coefficient, is the order fixed or unrestricted? - Interpret the exponents as counts of defective and nondefective chips. - Use the relationship between an event and its complement. - Interpret the binomial coefficient as the number of possible arrangements.

Solution

1. Let a defective chip be a success. Then \(n=8\), \(p=0.05\), and \(1-p=0.95\). 2. In part a), there is no binomial coefficient, so the expression represents one specified order containing two defective and six nondefective chips, such as only the first two chips being defective. 3. In part b), \(\binom82\) counts all possible positions of the two defective chips, so the expression represents exactly two defective chips in the sample. 4. In part c), \((0.95)^8\) is the probability that no chips are defective. Its complement represents at least one defective chip.

Answer

a) One specified sequence with two defective chips and six nondefective chips, such as only the first two chips being defective b) Exactly two of the eight chips are defective c) At least one of the eight chips is defective
52704612
A basketball player independently takes \(4\) free throws and makes each free throw with probability \(0.60\). Write an expression for the probability of each event. a) Only the first two shots are made. b) Exactly two shots are made. c) More than two shots are made.

Hints

- Does a specified order require counting several arrangements? - Use a binomial coefficient when the positions of the makes can vary. - Translate “more than two” into the possible numbers of makes. - Recall the formula for exactly \(k\) successes in a binomial setting.

Solution

1. The make probability is \(p=0.6\), and the miss probability is \(0.4\). 2. For part a), the sequence is make, make, miss, miss, so the probability is \((0.6)^2(0.4)^2=0.0576\). 3. For part b), the two makes can occur in any two positions, so \(P(X=2)=\binom42(0.6)^2(0.4)^2=0.3456\). 4. For part c), more than two makes means three or four makes. Therefore, \(P(X>2)=\binom43(0.6)^3(0.4)+\binom44(0.6)^4=0.4752\).

Answer

a) \((0.6)^2(0.4)^2=0.0576\) b) \(\binom42(0.6)^2(0.4)^2=0.3456\) c) \(\binom43(0.6)^3(0.4)+\binom44(0.6)^4=0.4752\)
52705512
An electronics manufacturer reports that \(4\%\) of the parts in a large shipment are defective. A customer randomly selects \(15\) parts. Assume the selections can be modeled as independent Bernoulli trials. Let \(X\) be the number of defective parts in the sample. Find the probability that the sample contains at least \(2\) defective parts.

Hints

- Identify the binomial parameters for the sample. - Would a complement be shorter than adding every case with at least \(2\) defective parts? - Which values are excluded by “at least \(2\)”?

Solution

1. Model \(X\) with a binomial distribution having \(n=15\) and \(p=0.04\). 2. Use the complement: \(P(X\ge 2)=1-P(X\le 1)\). 3. \(P(X=0)=(0.96)^{15}\approx 0.5421\). 4. \(P(X=1)=15(0.04)(0.96)^{14}\approx 0.3388\). 5. Therefore, \(P(X\ge 2)=1-(0.5421+0.3388)\approx 0.1191\).

Answer

The probability is approximately \(11.91\%\).
52706112
A manufacturer ships microchips in packages of \(20\). Each chip independently has a \(10\%\) probability of being defective. A package is rejected during quality control if it contains more than \(3\) defective chips. What percentage of packages should be expected to be rejected?

Hints

- Identify the probability distribution for the number of defective chips. - Translate “more than \(3\)” into an inequality for \(X\). - Consider using the complement of the rejection event. - Check which values belong in the cumulative probability.

Solution

1. Let \(X\) be the number of defective chips in a package. Then \(X\) has a binomial distribution with \(n=20\) and \(p=0.10\). 2. A package is rejected when \(X>3\), so the required probability is \(P(X\ge 4)\). 3. Use the complement: \(P(X\ge 4)=1-P(X\le 3)\). 4. A cumulative binomial calculation gives \(P(X\le 3)\approx 0.8670\). 5. Therefore, \(P(X\ge 4)\approx 1-0.8670=0.1330\), or about \(13.3\%\).

Answer

Approximately \(13.3\%\) of the packages are expected to be rejected.
52706212
A garden center sells packets containing \(12\) seeds of a certain flower variety. Each seed independently has a \(75\%\) probability of germinating. A customer considers a packet defective if fewer than \(8\) seeds germinate. Find the probability that a randomly selected packet is considered defective.

Hints

- Translate “fewer than \(8\)” into values of the random variable. - Decide whether to count germinating or non-germinating seeds. - Use either the cumulative probability directly or an appropriate complement. - Does the value \(8\) belong in the event?

Solution

1. Let \(X\) be the number of seeds that germinate. Then \(X\) has a binomial distribution with \(n=12\) and \(p=0.75\). 2. “Fewer than \(8\)” means \(X\le 7\). 3. \(P(X\le 7)=\sum_{k=0}^{7}\binom{12}{k}(0.75)^k(0.25)^{12-k}\). 4. Evaluating the cumulative probability gives \(P(X\le 7)\approx 0.1576\).

Answer

The probability is approximately \(0.1576\), or \(15.76\%\).
52706512
A biathlon athlete independently takes \(5\) shots and hits a target with probability \(p=0.8\) on each shot. Let \(X\) be the number of hits. A second athlete independently takes \(5\) shots and hits with probability \(0.2\) on each shot. Let \(Y\) be the number of hits. a) Calculate \(P(X=4)\) and \(P(Y=1)\). b) Without additional numerical calculations, explain why \(P(X=k)=P(Y=5-k)\) for every \(k\in\{0, 1, \ldots, 5\}\).

Hints

- Write both probabilities using the binomial formula. - Use the symmetry of \(\binom5k\) and \(\binom5{5-k}\). - Compare the powers in the two products.

Solution

1. For the first athlete, \(P(X=4)=\binom54(0.8)^4(0.2)=0.4096\). 2. For the second athlete, \(P(Y=1)=\binom51(0.2)(0.8)^4=0.4096\). 3. In general, \(P(X=k)=\binom5k(0.8)^k(0.2)^{5-k}\). 4. Also, \(P(Y=5-k)=\binom5{5-k}(0.2)^{5-k}(0.8)^k\). 5. Since \(\binom5{5-k}=\binom5k\), the two probabilities are equal.

Answer

a) \(P(X=4)=0.4096\) and \(P(Y=1)=0.4096\) b) The equality follows from \(\binom5k=\binom5{5-k}\) and the fact that the same factors \((0.8)^k\) and \((0.2)^{5-k}\) appear in both probabilities.
52708112
Determine whether the random variable \(X\) in each situation can be modeled with a binomial distribution. Justify your decision and state any needed assumptions. a) A bowl contains \(5\) red candies and \(15\) blue candies. A child takes \(4\) candies at once. The random variable \(X\) is the number of red candies selected. b) A spinner with three equal sectors—red, green, and blue—is spun \(10\) times. The random variable \(X\) is the number of spins that land on red. c) In a city with \(500{,}000\) households, about \(12\%\) own an electric vehicle. A random sample of \(50\) households is surveyed. The random variable \(X\) is the number in the sample that own an electric vehicle.

Hints

- Is the probability of success the same on every trial? - Does replacement matter in each situation? - Is the number of trials fixed in advance? - Compare the sample size with the population size.

Solution

1. In a), the candies are selected without replacement from a small population. The probability of selecting a red candy changes during the selection, so \(X\) is not binomial. 2. In b), each spin has two relevant outcomes, red or not red. The spins are independent with constant probability \(p=\frac{1}{3}\), so \(X\sim\operatorname{Bin}(10,\frac{1}{3})\). 3. In c), sampling is technically without replacement, but \(50\) is extremely small compared with \(500{,}000\). If households can be treated as approximately independent and the ownership rate is stable at \(0.12\), a binomial model with \(n=50\) and \(p=0.12\) is a reasonable approximation.

Answer

a) Not binomial; the success probability changes in a small sample without replacement. b) Binomial with \(n=10\) and \(p=\frac{1}{3}\). c) Approximately binomial with \(n=50\) and \(p=0.12\), assuming approximate independence and a stable population rate.
52708612
In a city survey, each person independently has a \(60\%\) chance of reporting that they regularly commute to work by bicycle. A group of \(15\) people is selected. Find the probability that a majority of the group regularly commutes by bicycle.

Hints

- How many people are needed for a majority in a group of \(15\)? - Define a random variable for the number of regular bicycle commuters. - Use a cumulative binomial probability or its complement.

Solution

1. Let \(X\) be the number of people in the group who regularly commute by bicycle. Then \(X\) has a binomial distribution with \(n=15\) and \(p=0.60\). 2. A majority of \(15\) people means at least \(8\) people. 3. \(P(X\ge 8)=1-P(X\le 7)\). 4. Since \(P(X\le 7)\approx 0.2131\), \(P(X\ge 8)\approx 0.7869\).

Answer

The probability that a majority, at least \(8\) people, regularly commutes by bicycle is approximately \(78.69\%\).
52708712
At a distribution center, \(5\%\) of packages are routed to the wrong facility. At the end of a day, an inspector randomly selects \(40\) packages. a) State the conditions under which the number of misrouted packages in the sample can be modeled with a binomial distribution. b) Define a suitable random variable \(X\), identify the parameters \(n\) and \(p\), and explain them in context. c) Find the probability that exactly two sampled packages were misrouted.

Hints

- Recall the conditions for a binomial setting. - In this context, define a success as a misrouted package. - Connect the sample size and individual probability to the binomial formula.

Solution

1. A binomial model is appropriate if each package has two classifications, misrouted or not misrouted; the misrouting probability remains \(0.05\); and the package outcomes are independent. 2. Let \(X\) be the number of misrouted packages among the \(40\) sampled packages. Then \(n=40\) is the sample size, and \(p=0.05\) is the probability that one package is misrouted. 3. \(P(X=2)=\binom{40}{2}(0.05)^2(0.95)^{38}\approx0.2777\).

Answer

a) Each package must have two outcomes, the misrouting probability must remain \(0.05\), and the outcomes must be independent. b) \(X\) is the number of misrouted packages; \(n=40\); \(p=0.05\). c) \(P(X=2)\approx0.2777\), or about \(27.77\%\)
52709012
In a large forest, \(80\%\) of the trees are pine trees. A forester inspects a sample of \(200\) trees. Assume the sample can be modeled as independent Bernoulli trials, and let \(X\) be the number of pine trees in the sample. a) Find the probability of each event. \(E\): Exactly \(165\) of the inspected trees are pine trees. \(F\): From \(155\) through \(170\) of the inspected trees, inclusive, are pine trees. b) Describe the meaning of \(1-P(X\ge 175)\) in context.

Hints

- Identify \(n\) and \(p\) from the context. - Express the inclusive interval in Event \(F\) as a difference of cumulative probabilities. - What event is the complement of at least \(175\) pine trees?

Solution

1. The random variable \(X\) has a binomial distribution with \(n=200\) and \(p=0.80\). 2. \(P(E)=P(X=165)=\binom{200}{165}(0.80)^{165}(0.20)^{35}\approx 0.0495\). 3. \(P(F)=P(155\le X\le 170)=P(X\le 170)-P(X\le 154)\approx 0.9717-0.1651=0.8066\). 4. The complement of \(X\ge 175\) is \(X\le 174\). Therefore, \(1-P(X\ge 175)\) is the probability that at most \(174\) of the \(200\) trees are pine trees.

Answer

a) \(E\): \(P(E)\approx 4.95\%\) \(F\): \(P(F)\approx 80.66\%\) b) The expression is the probability that at most \(174\) of the \(200\) inspected trees are pine trees.
52710812
A multiple-choice test has \(15\) questions. Each question has \(4\) answer choices, exactly one of which is correct. A test taker guesses independently and randomly on every question. Write an expression for the probability of each event. a) Exactly \(6\) questions are answered correctly. b) At most \(1\) question is answered correctly. c) At least \(3\) questions are answered correctly.

Hints

- What is the probability of a correct answer when guessing among \(4\) choices? - Identify the total number of trials. - “At most” includes all values from \(0\) through the stated number. - For “at least,” a complement can reduce the number of terms.

Solution

1. Let \(X\) be the number of correct answers. Then \(X\) has a binomial distribution with \(n=15\) and \(p=0.25\). 2. For part a), \(P(X=6)=\binom{15}{6}(0.25)^6(0.75)^9\). 3. For part b), \(P(X\le 1)=(0.75)^{15}+15(0.25)(0.75)^{14}\). 4. For part c), use the complement: \(P(X\ge 3)=1-\sum_{k=0}^{2}\binom{15}{k}(0.25)^k(0.75)^{15-k}\).

Answer

a) \(\binom{15}{6}(0.25)^6(0.75)^9\) b) \((0.75)^{15}+15(0.25)(0.75)^{14}\) c) \(1-\sum_{k=0}^{2}\binom{15}{k}(0.25)^k(0.75)^{15-k}\)
52711712
A manufacturer finds that each LED diode independently has a \(1\) in \(20\) chance of being defective. a) How many defective diodes should the manufacturer expect in a daily production run of \(1400\) diodes? b) Find the probability that exactly \(1\) diode is defective in a random sample of \(25\) diodes. c) Find the probability that at most \(2\) diodes are defective in a random sample of \(25\) diodes.

Hints

- First determine the probability that one diode is defective. - Which measure of a distribution describes its long-run average? - Use the binomial probability formula for a fixed number of trials. - “At most” means that several individual outcomes must be added.

Solution

1. The probability that one diode is defective is \(p=\frac{1}{20}=0.05\). 2. For part a), the expected number is \(E(X)=np=1400\cdot0.05=70\). 3. For part b), let \(X\sim\operatorname{Bin}(25,0.05)\). Then \(P(X=1)=\binom{25}{1}(0.05)(0.95)^{24}\approx 0.364986\). 4. For part c), \(P(X\le 2)=\sum_{k=0}^{2}\binom{25}{k}(0.05)^k(0.95)^{25-k}\approx 0.872894\).

Answer

a) The expected number is \(70\) defective diodes. b) The probability is approximately \(36.50\%\). c) The probability is approximately \(87.29\%\).
52713512
A basketball player makes free throws with probability \(0.75\). The player takes \(45\) free throws. a) State the conditions needed to model the number of made free throws with a binomial distribution. b) Find the most likely number of made free throws.

Hints

- Recall the conditions for a binomial setting. - Identify what must remain constant from shot to shot. - Relate the mode of a binomial distribution to \((n+1)p\). - Remember that the number of makes must be an integer.

Solution

1. A binomial model requires two outcomes per shot, a constant make probability of \(0.75\), and independent shots. 2. Here, \(n=45\) and \(p=0.75\). 3. Since \((n+1)p=46\cdot0.75=34.5\) is not an integer, the unique mode is \(\lfloor34.5\rfloor=34\).

Answer

a) Each shot must have two outcomes, the make probability must remain \(0.75\), and the shots must be independent. b) The most likely number of made free throws is \(34\).
52713912
A random variable \(X\) has a binomial distribution with \(n=80\) and \(p=0.45\). Without calculating the individual probabilities, justify the value of \(k\) for which \(P(X=k)\) is greatest.

Hints

- Which measure describes the center of a binomial distribution? - Near which value would you expect the tallest bar of the distribution? - Recall the relationship between the mode and the parameters of a binomial distribution.

Solution

1. For a binomial distribution, when \((n+1)p\) is not an integer, the unique mode is \(\lfloor(n+1)p\rfloor\). 2. Here, \((80+1)\cdot0.45=81\cdot0.45=36.45\), which is not an integer. 3. Therefore, the mode is \(\lfloor36.45\rfloor=36\), so \(P(X=36)\) is the greatest probability.

Answer

The greatest probability occurs at \(k=36\).
52719112
Determine whether each experiment can be modeled as binomial. Give \(n\) and \(p\) when appropriate. a) An archer takes \(12\) shots. The probability of hitting the bull’s-eye is constant at \(0.80\), and the shots are treated as independent. Each shot is recorded as a hit or miss. b) Five cards are drawn without replacement from a well-shuffled standard \(52\)-card deck. Each draw is recorded as an ace or not an ace.

Hints

- How many relevant outcomes are recorded on each trial? - Does the success probability stay the same? - Does one trial affect the following trials?

Solution

1. In a), there are two relevant outcomes, the success probability is constant, and the trials are independent. This is binomial with \(n=12\) and \(p=0.80\). 2. In b), drawing without replacement changes the probability of an ace after each draw. For example, after the first draw the next probability is either \(\frac{3}{51}\) or \(\frac{4}{51}\). The trials are dependent, so the experiment is not binomial.

Answer

a) Binomial with \(n=12\) and \(p=0.80\). b) Not binomial because the ace probability changes without replacement.
52719412
In a manufacturing process, each component independently has a \(4\%\) chance of being defective. A random sample of \(50\) components is selected for quality control. a) Find the probability of each event. (1) Exactly \(2\) components in the sample are defective. (2) At least \(4\) components in the sample are defective. (3) At most \(3\) components in the sample are defective. b) During one shift, \(100\) components have already been inspected and none were defective. Find the probability that the next component inspected is defective.

Hints

- Identify \(n\) and \(p\) for the binomial distribution. - For an “at least” event, consider its complement. - What does independence imply about the probability of the next outcome?

Solution

1. Let \(X\sim\operatorname{Bin}(50,0.04)\). 2. For part a)(1), \(P(X=2)=\binom{50}{2}(0.04)^2(0.96)^{48}\approx 0.276233\). 3. For part a)(2), \(P(X\ge 4)=1-P(X\le 3)\approx 0.139131\). 4. For part a)(3), \(P(X\le 3)=\sum_{k=0}^{3}\binom{50}{k}(0.04)^k(0.96)^{50-k}\approx 0.860869\). 5. For part b), independence means that the outcomes of the first \(100\) inspections do not change the defect probability for the next component. The probability remains \(0.04\).

Answer

a) (1) \(P(X=2)\approx 0.2762\) (2) \(P(X\ge 4)\approx 0.1391\) (3) \(P(X\le 3)\approx 0.8609\) b) The probability is \(0.04\), or \(4\%\).
52721612
At a distribution center, \(15\%\) of packages are set aside because their addresses cannot be read automatically. During one shift, \(120\) packages are processed, and outcomes are treated as independent. Find the probability that at least \(12\) and at most \(24\) packages are set aside.

Hints

- Identify the binomial parameters. - Write the stated range as an inequality for \(X\). - Use a difference of cumulative probabilities. - To include \(12\), subtract the cumulative probability through which value?

Solution

1. Let \(X\sim\operatorname{Bin}(120,0.15)\) be the number of packages set aside. 2. The required event is \(12\le X\le 24\). 3. Using the binomial cumulative distribution function, \(P(12\le X\le 24)=P(X\le 24)-P(X\le 11)\approx 0.9472-0.0420=0.9052\).

Answer

\(P(12\le X\le 24)\approx 0.9052\), or about \(90.52\%\)
52789912
Determine whether each experiment can be described by a binomial model. Justify your decision. When appropriate, give the success probability \(p\), the number of trials \(n\), and the meaning of \(X\). a) A container holds \(40\) balls, \(8\) of which are red. Five balls are drawn with replacement. Each draw is recorded as red or not red. b) A basketball player has a \(75\%\) free-throw success rate and takes \(10\) free throws. The number of made shots is recorded. c) A box contains \(20\) batteries, \(2\) of which are defective. Two batteries are drawn without replacement and tested.

Hints

- What conditions must a binomial experiment satisfy? - Does the probability of success change from trial to trial? - How does replacement affect independence? - Are there real-world factors that might challenge a constant success probability?

Solution

1. In a), replacement keeps the red-ball probability constant at \(p=\frac{8}{40}=0.20\), and the draws are independent. The model is binomial with \(n=5\), and \(X\) is the number of red balls drawn. 2. In b), under the modeling assumptions that the success probability remains constant and the shots are independent, the model is binomial with \(n=10\) and \(p=0.75\). The random variable \(X\) is the number of made free throws. 3. In c), drawing without replacement changes the defect probability on the second draw. The trials are dependent, so the model is not binomial.

Answer

a) Yes: \(n=5\), \(p=0.20\), and \(X\) is the number of red balls drawn. b) Yes, assuming independence and constant probability: \(n=10\), \(p=0.75\), and \(X\) is the number of made shots. c) No; without replacement, the success probability changes.
52871912
A gardener plants \(6\) seeds of a particular flower variety. Each seed has a germination probability of \(80\%\), and the germination outcomes are independent. a) Explain why the germination outcomes can be modeled as a sequence of \(6\) Bernoulli trials. b) Find the probability that exactly \(5\) of the \(6\) seeds germinate. c) Find the probability that at most \(1\) seed germinates.

Hints

- What conditions must hold for a sequence of Bernoulli trials? - Does each seed have the same germination probability? - Use the binomial probability formula for an exact number of successes. - “At most \(1\)” includes \(0\) and \(1\) success.

Solution

1. Each trial has two outcomes, germination or no germination. The probability of germination is constant at \(0.80\), and the outcomes are independent. Therefore, the conditions for Bernoulli trials are met. 2. Let \(X\sim\operatorname{Bin}(6,0.80)\). Then \(P(X=5)=\binom{6}{5}(0.80)^5(0.20)=0.393216\). 3. For part c), \(P(X\le 1)=P(X=0)+P(X=1)=(0.20)^6+6(0.80)(0.20)^5=0.0016\).

Answer

a) There are two outcomes per seed, the success probability is constant, and the outcomes are independent. b) The probability is \(0.393216\), or approximately \(39.32\%\). c) The probability is \(0.0016\), or \(0.16\%\).
52872012
A factory produces glass bottles. Historical data show that \(4\%\) of the bottles contain small air bubbles. A random sample of \(10\) bottles is selected for quality control. a) State the conditions under which this sample can be modeled using Bernoulli trials. For parts b) and c), assume those conditions hold. b) Find the probability that exactly \(1\) bottle in the sample contains an air bubble. c) Find the probability that at least \(2\) bottles in the sample contain air bubbles.

Hints

- Consider what must be true for independent trials with a constant probability. - Use the binomial formula for exactly \(1\) success. - What is the complement of “at least \(2\)”? - Write \(4\%\) as a decimal before calculating.

Solution

1. Each bottle must be classified into one of two outcomes, with a constant probability \(p=0.04\), and the outcomes must be independent. Independence is reasonable if the population is very large or sampling is effectively with replacement. 2. Let \(X\sim\operatorname{Bin}(10,0.04)\). Then \(P(X=1)=\binom{10}{1}(0.04)(0.96)^9\approx 0.277014\). 3. For part c), use the complement: \(P(X\ge 2)=1-[P(X=0)+P(X=1)]\approx 0.058154\).

Answer

a) Each trial has two outcomes, the probability \(0.04\) is constant, and the outcomes are independent. b) The probability is approximately \(0.2770\), or \(27.70\%\). c) The probability is approximately \(0.0582\), or \(5.82\%\).
52873112
A spinner has \(4\) equal sections, and \(1\) section is red. The spinner is spun independently \(4\) times. a) Find the probability that the spinner lands on red exactly once. b) A person claims, “Since the probability of red on each spin is \(25\%\), the probability of getting red at least once in \(4\) spins is greater than \(90\%\).” Use a calculation to evaluate the claim.

Hints

- What count of red outcomes is required by “exactly once”? - How are the probabilities of red and not red related? - For “at least once,” consider the complement event. - Compare your result with \(90\%\).

Solution

1. Let \(X\sim\operatorname{Bin}(4,0.25)\). 2. For part a), \(P(X=1)=\binom{4}{1}(0.25)(0.75)^3=0.421875\). 3. For part b), \(P(X\ge 1)=1-P(X=0)=1-(0.75)^4=0.68359375\). 4. Since \(68.36\%<90\%\), the claim is false.

Answer

a) The probability is \(0.421875\), or \(42.1875\%\). b) The claim is false. The probability of red at least once is approximately \(68.36\%\), not more than \(90\%\).
52873212
A multiple-choice test has \(5\) questions. Each question has \(3\) answer choices, exactly one of which is correct. A student guesses independently on every question. Create a probability distribution table for the number \(X\) of questions answered correctly. Round each probability to four decimal places.

Hints

- Find the probability of guessing one question correctly. - Identify the possible values of \(X\). - Use the formula for exactly \(k\) successes in a fixed number of independent trials. - Check that the unrounded probabilities add to \(1\).

Solution

1. The number of correct answers is binomial with \(n=5\), \(p=\frac{1}{3}\), and \(1-p=\frac{2}{3}\). 2. Apply \(P(X=k)=\binom{5}{k}\left(\frac{1}{3}\right)^k\left(\frac{2}{3}\right)^{5-k}\) for \(k\in\{0,1,2,3,4,5\}\). 3. The probabilities are \(P(X=0)=\frac{32}{243}\approx0.1317\), \(P(X=1)=\frac{80}{243}\approx0.3292\), \(P(X=2)=\frac{80}{243}\approx0.3292\), \(P(X=3)=\frac{40}{243}\approx0.1646\), \(P(X=4)=\frac{10}{243}\approx0.0412\), and \(P(X=5)=\frac{1}{243}\approx0.0041\).

Answer

<table> <tr> <td>\(k\)</td> <td>\(0\)</td> <td>\(1\)</td> <td>\(2\)</td> <td>\(3\)</td> <td>\(4\)</td> <td>\(5\)</td> </tr> <tr> <td>\(P(X=k)\)</td> <td>\(0.1317\)</td> <td>\(0.3292\)</td> <td>\(0.3292\)</td> <td>\(0.1646\)</td> <td>\(0.0412\)</td> <td>\(0.0041\)</td> </tr> </table>
52873412
A spinner has \(3\) equal sections colored red, blue, and yellow. The spinner is spun independently \(6\) times. 1. Find the probability that the spinner lands on blue exactly \(2\) times. 2. Find the probability that the spinner lands on red at least \(5\) times.

Hints

- First determine the probability of one specified color on a single spin. - Which possible counts are included in “at least \(5\)”? - Model each specified color as a success in a sequence of Bernoulli trials. - Add the probabilities of the required cases.

Solution

1. For either specified color, the success probability is \(p=\frac{1}{3}\). 2. For part 1, \(P(X=2)=\binom{6}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^4=\frac{80}{243}\approx 0.329218\). 3. For part 2, \(P(X\ge 5)=P(X=5)+P(X=6)\). 4. Therefore, \(P(X\ge 5)=\binom{6}{5}\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)+\left(\frac{1}{3}\right)^6=\frac{13}{729}\approx 0.017833\).

Answer

1. The probability is \(\frac{80}{243}\approx 32.92\%\). 2. The probability is \(\frac{13}{729}\approx 1.78\%\).
52873612
A factory produces LED light bulbs. Each bulb is independently defective with probability \(0.05\). An inspector selects \(8\) bulbs and tests them. Find the probability that a) all \(8\) bulbs are nondefective, b) exactly \(2\) bulbs are defective, c) at least \(1\) bulb is defective.

Hints

- What is the probability that one bulb is nondefective? - Use the binomial model for repeated trials. - For “at least \(1\),” consider the complement. - What is the complement of at least \(1\) defective bulb?

Solution

1. Let \(X\sim\operatorname{Bin}(8,0.05)\), where \(X\) is the number of defective bulbs. 2. For part a), \(P(X=0)=(0.95)^8\approx 0.663420\). 3. For part b), \(P(X=2)=\binom{8}{2}(0.05)^2(0.95)^6\approx 0.051456\). 4. For part c), \(P(X\ge 1)=1-P(X=0)\approx 0.336580\).

Answer

a) The probability is approximately \(66.34\%\). b) The probability is approximately \(5.15\%\). c) The probability is approximately \(33.66\%\).
52874212
An electronic component is defective with probability \(p=0.10\). For quality control, \(15\) components are randomly selected from a large shipment, with replacement. Find the probability that the sample contains no defective components or exactly \(1\) defective component.

Hints

- What does \(0\) defects contribute to the calculation? - Identify \(n\) and \(p\). - Calculate the cases of \(0\) and \(1\) defect separately. - Sampling with replacement supports a binomial model.

Solution

1. Let \(X\sim\operatorname{Bin}(15,0.10)\). 2. The required probability is \(P(X\le 1)=P(X=0)+P(X=1)\). 3. Thus, \(P(X\le 1)=(0.90)^{15}+15(0.10)(0.90)^{14}\approx 0.549043\).

Answer

The probability is approximately \(0.5490\), or \(54.90\%\).
52874712
A spinner has \(5\) equal sections. One section has a star and awards a prize. A player spins the spinner independently \(10\) times. Find the probability that the spinner lands on the prize section at least \(3\) times.

Hints

- Determine the probability of winning on one spin. - Consider whether the complement is easier to calculate. - Which counts belong to the complement of “at least \(3\)”? - Use the binomial distribution for repeated independent spins.

Solution

1. The success probability is \(p=\frac{1}{5}=0.20\). Let \(X\sim\operatorname{Bin}(10,0.20)\). 2. Use the complement: \(P(X\ge 3)=1-P(X\le 2)\). 3. Thus, \(P(X\ge 3)=1-\sum_{k=0}^{2}\binom{10}{k}(0.20)^k(0.80)^{10-k}\approx 0.322200\).

Answer

The probability is approximately \(32.22\%\).
52874812
A large bag of gummy bears contains several colors, and the probability of selecting a red gummy bear is \(20\%\). A person makes \(15\) selections with replacement. Find the probability that at most \(2\) of the selected gummy bears are red.

Hints

- Which counts are included in “at most \(2\)”? - Sampling with replacement keeps the probability constant. - Substitute the given values into the binomial formula. - Add the probabilities of all included outcomes.

Solution

1. Let \(X\sim\operatorname{Bin}(15,0.20)\). 2. At most \(2\) means \(X=0,1,\) or \(2\). 3. Therefore, \(P(X\le 2)=\sum_{k=0}^{2}\binom{15}{k}(0.20)^k(0.80)^{15-k}\approx 0.398023\).

Answer

The probability is approximately \(39.80\%\).
52874912
Sarah and Tim play \(7\) table-tennis games. The player who wins at least \(4\) games wins the match. Sarah wins each game with probability \(0.55\), independently of the other games. Find the probability that Sarah wins the match.

Hints

- How many games must Sarah win? - Model the independent games using a binomial distribution. - Which values of \(X\) result in a match win? - Add the probabilities for all winning counts.

Solution

1. Let \(X\sim\operatorname{Bin}(7,0.55)\), where \(X\) is the number of games Sarah wins. 2. Sarah wins the match when \(X\ge 4\). 3. Therefore, \(P(X\ge 4)=\sum_{k=4}^{7}\binom{7}{k}(0.55)^k(0.45)^{7-k}\approx 0.608288\).

Answer

The probability that Sarah wins the match is approximately \(60.83\%\).
52875012
Marc and Lucas play a series of \(10\) independent video games. Marc wins each game with probability \(0.40\). The overall winner is the player who wins more than half of the games. If each player wins \(5\) games, there is no overall winner. Find the probability that Marc is the overall winner.

Hints

- How many wins are needed to win more than half of \(10\) games? - Are the trials independent with a constant success probability? - List the win totals that make Marc the overall winner. - Add their binomial probabilities or use a suitable complement.

Solution

1. Let \(X\sim\operatorname{Bin}(10,0.40)\), where \(X\) is the number of games Marc wins. 2. Marc is the overall winner when \(X\ge 6\). 3. Therefore, \(P(X\ge 6)=\sum_{k=6}^{10}\binom{10}{k}(0.40)^k(0.60)^{10-k}\approx 0.166239\).

Answer

The probability that Marc is the overall winner is approximately \(16.62\%\).
52875412
A factory produces electronic components. Each component is independently defective with probability \(0.04\). A sample of \(15\) components is selected for quality control. Find the probability that a) exactly \(1\) component is defective, b) at least \(2\) components are defective, c) no components are defective.

Hints

- Identify the success probability for a defective component. - For “at least \(2\),” consider using the complement. - What value of \(X\) corresponds to no defective components? - The probability of a nondefective component is \(1-p\).

Solution

1. Let \(X\sim\operatorname{Bin}(15,0.04)\). 2. For part a), \(P(X=1)=\binom{15}{1}(0.04)(0.96)^{14}\approx 0.338804\). 3. For part b), \(P(X\ge 2)=1-[P(X=0)+P(X=1)]\approx 0.119110\). 4. For part c), \(P(X=0)=(0.96)^{15}\approx 0.542086\).

Answer

a) The probability is approximately \(33.88\%\). b) The probability is approximately \(11.91\%\). c) The probability is approximately \(54.21\%\).
52876112
An independent movie theater has \(13\) seats available for a special screening. Because each person with a reservation independently has a \(10\%\) probability of not attending, the theater accepts \(15\) reservations. Find the probability that more people attend than there are seats.

Hints

- What is the probability that one person with a reservation attends? - Which attendance counts exceed the number of seats? - Calculate those exact probabilities and add them.

Solution

1. Each person attends with probability \(0.90\). Let \(X\sim\operatorname{Bin}(15,0.90)\). 2. The theater is over capacity when \(X>13\), so \(X=14\) or \(X=15\). 3. Therefore, \(P(X>13)=\binom{15}{14}(0.90)^{14}(0.10)+(0.90)^{15}\approx 0.549043\).

Answer

The probability that the theater is over capacity is approximately \(54.90\%\).
52876212
A small restaurant has \(10\) outdoor tables. Each evening reservation is independently canceled at the last minute with probability \(0.25\). The owner accepts \(12\) reservations for one evening. Find the probability that there are enough tables for everyone who arrives.

Hints

- What is the probability that one reservation is used? - Which attendance counts create a shortage of tables? - Consider calculating the complement event.

Solution

1. Each reservation is used with probability \(0.75\). Let \(X\sim\operatorname{Bin}(12,0.75)\). 2. There are enough tables when \(X\le 10\). 3. Using the complement, \(P(X\le 10)=1-[P(X=11)+P(X=12)]\approx 0.841618\).

Answer

The probability is approximately \(84.16\%\).
52876312
An automaker finds that \(8\%\) of a certain vehicle model have a minor paint defect. A random sample of \(30\) vehicles is inspected. Assume the sampled vehicles' defect statuses are independent. What number of vehicles with a paint defect has the greatest probability?

Hints

- Model the number of affected vehicles with a binomial distribution. - The greatest probability occurs near the mean. - Use \((n+1)p\) to locate the mode. - Compare the nearby integer probabilities as a check.

Solution

1. Let \(X\) be the number of vehicles with a paint defect. Then \(X\) is binomial with \(n=30\) and \(p=0.08\). 2. Since \((n+1)p=31\cdot0.08=2.48\) is not an integer, the unique mode is \(\lfloor2.48\rfloor=2\). 3. As a check, \(P(X=2)\approx0.2696\), while \(P(X=3)\approx0.2188\).

Answer

\(2\) vehicles with a paint defect
52876412
At a greenhouse, each seed of a certain flower variety germinates independently with probability \(0.75\). Fifteen seeds are planted. Which number or numbers of germinating seeds have the greatest probability?

Hints

- Model germination as independent success-or-failure trials. - Calculate \((n+1)p\). - Recall what happens when \((n+1)p\) is an integer. - Check the two adjacent values near the mean.

Solution

1. Let \(X\) be the number of seeds that germinate. Then \(X\) is binomial with \(n=15\) and \(p=0.75\). 2. \((n+1)p=16\cdot0.75=12\), which is an integer. 3. When \((n+1)p\) is an integer, the two modes are that integer and one less. Therefore, the modes are \(12\) and \(11\). 4. Indeed, \(P(X=11)=P(X=12)\approx0.2252\).

Answer

\(11\) and \(12\) germinating seeds
53086612
A bag contains five balls labeled \(1\), \(2\), \(3\), \(4\), and \(5\). A ball is drawn at random, its number is recorded, and the ball is replaced. This process is repeated for a total of \(3\) draws. Find the probability that the ball labeled \(2\) is drawn exactly once.

Hints

- Define success for one draw. - Explain why replacement keeps the success probability unchanged. - Count the possible positions of the one successful draw.

Solution

1. Define success as drawing the ball labeled \(2\). Because the ball is replaced after each draw, the trials are independent with \(n=3\) and \(p=\frac15=0.2\). 2. The probability of exactly one success is \(P(X=1)=\binom31(0.2)(0.8)^2\). 3. Therefore, \(P(X=1)=3\cdot0.2\cdot0.64=0.384\).

Answer

\(P(X=1)=0.384\), or \(38.4\%\)
53088912
A gardener plants \(12\) bulbs of a rare tulip variety. Each bulb sprouts with probability \(0.95\), independently of the others. Find the probability that exactly \(11\) of the \(12\) bulbs sprout.

Hints

- Identify the number of trials and the success probability. - Check that each bulb has two possible outcomes. - Count the possible positions of the one bulb that does not sprout. - Include the failure probability.

Solution

1. Let \(X\) be the number of bulbs that sprout. Then \(X\) is binomial with \(n=12\), \(p=0.95\), and \(1-p=0.05\). 2. \(P(X=11)=\binom{12}{11}(0.95)^{11}(0.05)\). 3. Since \(\binom{12}{11}=12\), the probability is \(12\cdot(0.95)^{11}(0.05)\approx0.3413\).

Answer

\(P(X=11)\approx0.3413\), or about \(34.13\%\)
53089012
A basketball player independently takes \(10\) free throws during practice and makes each free throw with probability \(0.70\). Find the probability of making exactly \(7\) free throws.

Hints

- Define success and failure. - Identify the total number of shots. - Use the binomial formula for an exact number of makes. - Count the arrangements of seven makes among ten shots.

Solution

1. The binomial parameters are \(n=10\), \(p=0.7\), and \(k=7\). 2. The miss probability is \(0.3\). 3. \(P(X=7)=\binom{10}{7}(0.7)^7(0.3)^3\). 4. Since \(\binom{10}{7}=120\), the probability is \(120\cdot(0.7)^7(0.3)^3\approx0.2668\).

Answer

\(P(X=7)\approx0.2668\), or about \(26.68\%\)
53089312
A multiple-choice test has \(15\) questions. Each question has four answer choices, exactly one of which is correct. A student guesses independently and randomly on every question. Find the probability that the student answers no questions correctly.

Hints

- Find the probability of guessing one question incorrectly. - Use independence to multiply the probabilities across all questions. - Count how many consecutive incorrect answers are required.

Solution

1. The probability of an incorrect answer is \(\frac34=0.75\). 2. All \(15\) answers must be incorrect, so \(P(X=0)=(0.75)^{15}\). 3. Therefore, \(P(X=0)\approx0.01336\).

Answer

\((0.75)^{15}\approx0.01336\), or about \(1.34\%\)
53089412
A shipping company reports that \(95\%\) of its packages arrive at their destination within \(24\) hours. A customer tracks \(20\) independent shipments. Find the probability that all \(20\) packages arrive within \(24\) hours.

Hints

- Check that every shipment has the same success probability. - Interpret what it means for the event to occur on all \(20\) trials. - Identify \(n\), \(p\), and \(k\).

Solution

1. Let success mean that a package arrives within \(24\) hours. Then \(n=20\) and \(p=0.95\). 2. The event requires \(20\) successes, so \(P(X=20)=\binom{20}{20}(0.95)^{20}(0.05)^0=(0.95)^{20}\). 3. Therefore, \(P(X=20)\approx0.3585\).

Answer

\((0.95)^{20}\approx0.3585\), or about \(35.85\%\)
53089612
A multiple-choice test has \(10\) questions. Each question has four answer choices, exactly one of which is correct. A student guesses independently and randomly on every question. a) Define a random variable \(X\), and identify its distribution and parameters. b) Find the probability that the student answers exactly half of the questions correctly. c) Find the expected number of correct answers.

Hints

- Find the probability of guessing one question correctly. - Identify the two binomial parameters in context. - Use the binomial formula for an exact number of correct answers. - Recall the expected-value formula for a binomial random variable.

Solution

1. Let \(X\) be the number of correct answers. Then \(X\) is binomial with \(n=10\) and \(p=\frac14=0.25\). 2. Exactly half of the questions means \(X=5\). Thus, \(P(X=5)=\binom{10}{5}(0.25)^5(0.75)^5\approx0.0584\). 3. The expected value is \(E(X)=np=10\cdot0.25=2.5\).

Answer

a) \(X\) is the number of correct answers, and \(X\) is binomial with \(n=10\) and \(p=0.25\). b) \(P(X=5)\approx0.0584\), or about \(5.84\%\) c) \(E(X)=2.5\) correct answers
53091012
An online store includes a coupon in each promotional package with probability \(0.2\), independently of the other packages. A customer receives \(12\) packages. Let \(X\) be the number that contain a coupon. a) Find the probability that exactly \(3\) packages contain a coupon. b) Find the probability that at least \(1\) package contains a coupon.

Hints

- Does the coupon probability change from package to package? - Which distribution models a fixed number of independent trials with two outcome categories? - Use the binomial formula for exactly \(3\) coupons. - For “at least \(1\),” consider the complement of no coupons.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(12,0.2)\). 2. \(P(X=3)=\binom{12}{3}\cdot (0.2)^3\cdot (0.8)^9\approx 0.2362\). 3. Use the complement for part b): \(P(X\ge 1)=1-P(X=0)=1-(0.8)^{12}\approx 0.9313\).

Answer

a) \(P(X=3)\approx 0.2362\), or about \(23.62\%\) b) \(P(X\ge 1)\approx 0.9313\), or about \(93.13\%\)
53099112
A spinner has \(8\) equal sectors numbered \(1\) through \(8\). a) Describe one way to classify a single spin as a Bernoulli trial. Give the success probability \(p\) and failure probability \(q\). b) Give another definition of success that produces a different probability \(p'\).

Hints

- What distinguishes a Bernoulli trial from a more general random experiment? - How many relevant outcome categories does a Bernoulli trial use? - How can the eight numbers be separated into success and failure sets? - Is there only one valid way to define success?

Solution

1. One possible definition is “success means an even number.” The success set is \(\{2,4,6,8\}\), so \(p=\frac{4}{8}=0.5\) and \(q=1-p=0.5\). 2. Another possible definition is “success means a number greater than \(6\).” The success set is \(\{7,8\}\), so \(p'=\frac{2}{8}=0.25\).

Answer

a) Example: success is an even number. Then \(p=0.5\) and \(q=0.5\). b) Example: success is a number greater than \(6\). Then \(p'=0.25\).
53099312
An electronics manufacturer reports that \(12\%\) of its microchips are defective. A random sample of \(4\) chips is selected from ongoing production. The random variable \(X\) is the number of defective chips in the sample. Determine the probability distribution of \(X\), rounding probabilities to four decimal places.

Hints

- Decide whether the setting has repeated trials with two outcomes. - Identify the possible values of the number of defective chips. - Record the number of trials and the defect probability. - Apply the binomial probability formula to each possible value. - Check that the unrounded probabilities add to \(1\).

Solution

1. Model \(X\) as binomial with \(n=4\), \(p=0.12\), and \(1-p=0.88\). 2. Use \(P(X=k)=\binom{4}{k}(0.12)^k(0.88)^{4-k}\) for \(k\in\{0,1,2,3,4\}\). 3. The probabilities are \(P(X=0)=0.59969536\approx0.5997\), \(P(X=1)=0.32710656\approx0.3271\), \(P(X=2)=0.06690816\approx0.0669\), \(P(X=3)=0.00608256\approx0.0061\), and \(P(X=4)=0.00020736\approx0.0002\).

Answer

<table> <tr> <td>\(k\)</td> <td>\(0\)</td> <td>\(1\)</td> <td>\(2\)</td> <td>\(3\)</td> <td>\(4\)</td> </tr> <tr> <td>\(P(X=k)\)</td> <td>\(0.5997\)</td> <td>\(0.3271\)</td> <td>\(0.0669\)</td> <td>\(0.0061\)</td> <td>\(0.0002\)</td> </tr> </table>
53099412
In a region, \(28\%\) of households have fiber-optic internet service. Five households are selected independently at random. The random variable \(Y\) is the number of sampled households with fiber-optic service. Determine the complete probability distribution of \(Y\), rounding to four decimal places.

Hints

- Define what counts as a success and identify its probability. - Identify the number of trials. - Pair each possible value of \(Y\) with its binomial probability. - Use the binomial coefficient to count arrangements with exactly \(k\) successes.

Solution

1. Model \(Y\) as binomial with \(n=5\), \(p=0.28\), and \(1-p=0.72\). 2. Use \(P(Y=k)=\binom{5}{k}(0.28)^k(0.72)^{5-k}\) for \(k\in\{0,1,2,3,4,5\}\). 3. The probabilities are \(P(Y=0)\approx0.1935\), \(P(Y=1)\approx0.3762\), \(P(Y=2)\approx0.2926\), \(P(Y=3)\approx0.1138\), \(P(Y=4)\approx0.0221\), and \(P(Y=5)\approx0.0017\).

Answer

<table> <tr> <td>\(k\)</td> <td>\(0\)</td> <td>\(1\)</td> <td>\(2\)</td> <td>\(3\)</td> <td>\(4\)</td> <td>\(5\)</td> </tr> <tr> <td>\(P(Y=k)\)</td> <td>\(0.1935\)</td> <td>\(0.3762\)</td> <td>\(0.2926\)</td> <td>\(0.1138\)</td> <td>\(0.0221\)</td> <td>\(0.0017\)</td> </tr> </table>
53099712
An archer hits the center of a target with probability \(0.70\) on each shot and takes \(4\) shots. a) State the conditions under which this experiment can be modeled with a binomial distribution. b) Find the probabilities of exactly \(0\), \(1\), \(2\), \(3\), and \(4\) center hits.

Hints

- Identify the success and failure outcomes. - What are \(n\), \(p\), and \(1-p\)? - Use the binomial probability formula for each value of \(k\). - The five probabilities should add to \(1\).

Solution

1. Each shot must have two relevant outcomes, hit or miss. The hit probability must remain \(0.70\), and the shots must be independent. 2. Let \(X\sim\operatorname{Bin}(4,0.70)\). Using \(P(X=k)=\binom{4}{k}(0.70)^k(0.30)^{4-k}\): 3. \(P(X=0)=(0.30)^4=0.0081\). 4. \(P(X=1)=\binom{4}{1}(0.70)(0.30)^3=0.0756\). 5. \(P(X=2)=\binom{4}{2}(0.70)^2(0.30)^2=0.2646\). 6. \(P(X=3)=\binom{4}{3}(0.70)^3(0.30)=0.4116\). 7. \(P(X=4)=(0.70)^4=0.2401\).

Answer

a) Two relevant outcomes, constant probability \(p=0.70\), and independent shots. b) \(P(X=0)=0.0081\) \(P(X=1)=0.0756\) \(P(X=2)=0.2646\) \(P(X=3)=0.4116\) \(P(X=4)=0.2401\)
53101312
Let \(X\sim\operatorname{Bin}(30,0.2)\). Find each probability. a) \(P(X\le 4)\) b) \(P(X>8)\) c) \(P(5\le X\le 10)\)

Hints

- Express “more than” using a complement. - For an interval, determine which lower cumulative probability must be subtracted. - Use \(F(k)=P(X\le k)\).

Solution

1. Using the binomial cumulative distribution function, \(P(X\le 4)\approx 0.2552\). 2. Use the complement for part b): \(P(X>8)=1-P(X\le 8)\approx 0.1287\). 3. For part c), \(P(5\le X\le 10)=P(X\le 10)-P(X\le 4)\approx 0.7192\).

Answer

a) \(P(X\le 4)\approx 0.2552\) b) \(P(X>8)\approx 0.1287\) c) \(P(5\le X\le 10)\approx 0.7192\)
53101412
Let \(X\sim\operatorname{Bin}(80,0.35)\). Find each probability. a) \(P(X\le 20)\) b) \(P(22<X<32)\)

Hints

- Pay attention to strict versus inclusive inequalities. - List the integer values inside the open interval. - Express the interval probability as a difference of cumulative probabilities.

Solution

1. Using the binomial cumulative distribution function, \(P(X\le 20)\approx 0.0368\). 2. Since \(X\) is integer-valued, \(22<X<32\) means \(23\le X\le 31\). Therefore, \(P(22<X<32)=P(X\le 31)-P(X\le 22)\approx 0.6980\).

Answer

a) \(P(X\le 20)\approx 0.0368\) b) \(P(22<X<32)\approx 0.6980\)
53103112
A tour-bus company has \(50\) seats available for a day trip. Each customer with a reservation independently has a \(10\%\) probability of not arriving, so the company accepts \(54\) reservations. Find the probability that every customer who arrives will have a seat.

Hints

- What attendance count can the bus accommodate? - What is the probability that one customer arrives? - Model the arrivals as independent Bernoulli trials. - Use a cumulative probability for all acceptable attendance counts.

Solution

1. Each customer arrives with probability \(0.90\). Let \(X\sim\operatorname{Bin}(54,0.90)\). 2. Everyone has a seat when \(X\le 50\). 3. Therefore, \(P(X\le 50)\approx 0.801542\).

Answer

The probability is approximately \(80.15\%\).
53112912
A catering company estimates that each guest independently orders a vegetarian meal with probability \(0.15\). For an event with \(450\) guests, let \(X\) be the number of vegetarian orders. Find the probability that at least \(80\) guests order a vegetarian meal.

Hints

- Translate “at least \(80\)” into an inequality. - Use the complement if your technology returns \(P(X\le k)\). - Check the endpoint used in the complement carefully.

Solution

1. Let \(X\sim\operatorname{Bin}(450,0.15)\). 2. The required probability is \(P(X\ge 80)=1-P(X\le 79)\). 3. Using the cumulative binomial distribution gives \(P(X\ge 80)\approx 0.059003\).

Answer

The probability is approximately \(5.90\%\).
53113412
A binomial random variable \(X\) has parameters \(n=800\) and \(p=0.12\). Find \(P(80\le X\le 110)\).

Hints

- Recall what a cumulative binomial probability represents. - Express the interval probability as a difference of two cumulative probabilities. - To include the lower endpoint, subtract the cumulative probability through the integer immediately below it. - Use a calculator or statistical software to evaluate the cumulative probabilities.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(800,0.12)\). 2. Write the interval probability as \(P(80\le X\le 110)=P(X\le 110)-P(X\le 79)\). 3. Using the binomial cumulative distribution function, \(P(X\le 110)\approx 0.9405\) and \(P(X\le 79)\approx 0.0337\). 4. Therefore, \(P(80\le X\le 110)\approx 0.9405-0.0337=0.9068\).

Answer

\(P(80\le X\le 110)\approx 0.9068\), or about \(90.68\%\)
53113612
A factory produces components that meet quality standards with probability \(p=0.85\), independently of one another. A random sample of \(n=150\) components is inspected. Let \(X\) be the number that meet the standards. Find the probability that between \(122\) and \(132\) components, inclusive, meet the standards.

Hints

- Identify the binomial parameters and the inclusive interval. - Express the interval probability as a difference of cumulative probabilities. - Make sure the lower endpoint remains included when choosing the value to subtract.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(150,0.85)\). Its mean is \(\mu=np=150\cdot 0.85=127.5\). 2. Write the interval probability as \(P(122\le X\le 132)=P(X\le 132)-P(X\le 121)\). 3. Using the binomial cumulative distribution function, \(P(X\le 132)\approx 0.8757\) and \(P(X\le 121)\approx 0.0881\). 4. Using unrounded values, \(P(122\le X\le 132)\approx 0.7875\).

Answer

\(P(122\le X\le 132)\approx 0.7875\), or about \(78.75\%\)
53114212
A spinner has four equal sections labeled red, blue, green, and yellow. The spinner is spun \(40\) times independently. Let \(X\) be the number of times it lands on red. a) Find the probability that the spinner lands on red more than \(12\) times. b) Find the probability that it lands on red at least \(8\) times but no more than \(12\) times. c) Find the probability that it lands on red fewer than \(5\) times.

Hints

- Determine the probability of landing on one specified section. - Translate “more than \(12\)” into the corresponding cumulative-probability statement. - For “fewer than \(5\),” identify the greatest integer count that is included. - Express the inclusive interval as a difference of cumulative probabilities.

Solution

1. Each spin lands on red with probability \(p=\frac{1}{4}=0.25\), so \(X\sim\operatorname{Bin}(40,0.25)\). 2. For part a), use the complement: \(P(X>12)=1-P(X\le 12)\approx 1-0.8209=0.1791\). 3. For part b), \(P(8\le X\le 12)=P(X\le 12)-P(X\le 7)\approx 0.8209-0.1820=0.6389\). 4. For part c), because \(X\) is an integer count, \(P(X<5)=P(X\le 4)\approx 0.0160\).

Answer

a) \(P(X>12)\approx 0.1791\), or about \(17.91\%\) b) \(P(8\le X\le 12)\approx 0.6389\), or about \(63.89\%\) c) \(P(X<5)\approx 0.0160\), or about \(1.60\%\)
53114512
A sports club has a stadium section with \(1200\) seats. Because an average of \(15\%\) of season-ticket holders do not attend a game, the club sells \(1400\) season tickets. Assume that individual attendance is independent. Find the probability that more people attend than there are seats.

Hints

- Define success as one ticket holder attending. - Determine the first attendance count that exceeds capacity. - Use a binomial model for independent two-outcome trials. - Express the upper-tail probability using a cumulative probability.

Solution

1. Each season-ticket holder attends with probability \(0.85\). Let \(X\sim\operatorname{Bin}(1400,0.85)\). 2. The section is over capacity when \(X>1200\). 3. Therefore, \(P(X>1200)=1-P(X\le 1200)\approx 0.216938\).

Answer

The probability that the section is over capacity is approximately \(21.69\%\).
53116112
A spinner has two outcomes: win with probability \(p=0.3\), and lose with probability \(0.7\). The spinner is spun \(6\) times. a) Find the probability of exactly \(2\) wins. b) Briefly state the three main conditions needed to model this process as a binomial setting.

Hints

- Match the information in the problem to \(n\), \(p\), and \(k\). - Count the arrangements containing exactly two wins. - Identify what must remain unchanged across the spins.

Solution

1. The binomial parameters are \(n=6\), \(p=0.3\), and \(k=2\). 2. \(P(X=2)=\binom62(0.3)^2(0.7)^4\). 3. Since \(\binom62=15\), \((0.3)^2=0.09\), and \((0.7)^4=0.2401\), the probability is \(15\cdot0.09\cdot0.2401=0.324135\). 4. A binomial model requires two outcomes per trial, a constant success probability, and independent trials.

Answer

a) \(P(X=2)=0.324135\), or about \(32.41\%\) b) Each trial has two outcomes, the success probability remains constant, and the trials are independent.
53205912
A bag contains \(4\) red balls, \(3\) green balls, and \(3\) yellow balls, as shown. A ball is drawn at random and replaced after each draw, for a total of six draws. Find the probability that a green ball is drawn exactly twice. Give the result as a percentage rounded to one decimal place.
Figure for problem 532059

Hints

- Check whether the ball is replaced after each draw. - Count the total number of balls and the number of green balls. - Model green versus not green as a binomial setting. - Use the binomial formula for exactly two successes. - Convert the decimal to a percentage and round as directed.

Solution

1. There are \(10\) balls total, of which \(3\) are green, so the success probability is \(p=\frac{3}{10}=0.3\). 2. Replacement makes the six draws independent with the same probability, so \(X\) is binomial with \(n=6\) and \(p=0.3\). 3. \(P(X=2)=\binom62(0.3)^2(0.7)^4=0.324135\). 4. As a percentage rounded to one decimal place, the probability is \(32.4\%\).

Answer

\(32.4\%\)
53208712
The spinner shown has \(6\) equal sections: \(3\) yellow, \(2\) blue, and \(1\) red. It is spun independently \(6\) times. Find each probability. Round each percentage to two decimal places. a) The spinner lands on yellow exactly four times. b) The spinner lands on blue exactly two times.
Figure for problem 532087

Hints

- Determine each color probability from the equal sections. - Model one selected color versus all other colors as a binomial setting. - Use the binomial formula for exactly \(k\) successes. - Convert each result to a percentage and round to two decimal places.

Solution

1. The probability of yellow is \(\frac36=\frac12\), and the probability of blue is \(\frac26=\frac13\). 2. For part a), \(P(X=4)=\binom64\left(\frac12\right)^4\left(\frac12\right)^2=\frac{15}{64}=0.234375\). 3. Therefore, the probability in part a) is \(23.44\%\). 4. For part b), \(P(Y=2)=\binom62\left(\frac13\right)^2\left(\frac23\right)^4=\frac{80}{243}\approx0.329218\). 5. Therefore, the probability in part b) is \(32.92\%\).

Answer

a) \(\frac{15}{64}=23.44\%\) b) \(\frac{80}{243}\approx32.92\%\)
53210512
The spinner shown has \(10\) equal sections and is spun independently \(8\) times. Find the probability that the spinner lands on red exactly \(3\) times. Give the answer as a decimal rounded to three decimal places.
Figure for problem 532105

Hints

- Count the total sections and the red sections. - Identify \(n\), \(p\), and \(k\). - Use the binomial formula for exactly three red outcomes. - Round the final decimal to three places.

Solution

1. Four of the \(10\) sections are red, so \(p=\frac{4}{10}=0.4\), and \(1-p=0.6\). 2. Let \(X\) be the number of red outcomes in \(8\) spins. Then \(X\) is binomial with \(n=8\) and \(p=0.4\). 3. \(P(X=3)=\binom83(0.4)^3(0.6)^5\). 4. Since \(\binom83=56\), the probability is \(56\cdot(0.4)^3(0.6)^5=0.27869184\approx0.279\).

Answer

\(0.279\)
53215812
The spinner shown has \(8\) equal sections and is spun independently \(10\) times. Find each probability, expressed as a percentage rounded to two decimal places. a) The spinner lands on red exactly four times. b) The spinner lands on yellow exactly two times.
Figure for problem 532158

Hints

- Count the sections of each requested color. - Divide by the total number of sections to find the success probabilities. - Use the binomial formula for each exact number of outcomes. - Use \(n=10\) and the specified value of \(k\) in each part.

Solution

1. Three of the eight sections are red, so \(p_R=\frac38=0.375\). One section is yellow, so \(p_Y=\frac18=0.125\). 2. For part a), \(P(X=4)=\binom{10}{4}(0.375)^4(0.625)^6\approx0.247528\), or \(24.75\%\). 3. For part b), \(P(Y=2)=\binom{10}{2}(0.125)^2(0.875)^8\approx0.241600\), or \(24.16\%\).

Answer

a) \(24.75\%\) b) \(24.16\%\)
53598412
The spinner shown has \(8\) equal sections colored yellow, blue, and red. Each spin is independent. Find the probability of each event. a) In \(5\) spins, the spinner lands on blue exactly \(2\) times. b) In \(8\) spins, the spinner lands on yellow exactly \(4\) times. c) In \(4\) spins, the spinner lands on red exactly once.
Figure for problem 535984

Hints

- Count the sections of each color to find its probability on one spin. - Model the selected color versus all other colors as a binomial setting. - Identify \(n\), \(p\), and \(k\) for each part. - Use the binomial probability formula.

Solution

1. From the spinner, \(p_{\text{yellow}}=\frac48=0.5\), \(p_{\text{blue}}=\frac38=0.375\), and \(p_{\text{red}}=\frac18=0.125\). 2. For part a), \(P(X=2)=\binom52(0.375)^2(0.625)^3\approx0.3433\). 3. For part b), \(P(Y=4)=\binom84(0.5)^4(0.5)^4=\frac{70}{256}\approx0.2734\). 4. For part c), \(P(Z=1)=\binom41(0.125)(0.875)^3\approx0.3350\).

Answer

a) \(0.3433\), or about \(34.33\%\) b) \(0.2734\), or about \(27.34\%\) c) \(0.3350\), or about \(33.50\%\)
53604712
The spinner shown has \(5\) equal sections and is spun independently \(6\) times. Let \(X\) be the number of times the spinner lands on blue. Use the binomial probability formula to find \(P(X=3)\).
Figure for problem 536047

Hints

- Find the probability of blue on one spin. - Identify the total number of spins and the required number of blue outcomes. - Use the binomial formula for an exact number of successes.

Solution

1. One of the five equal sections is blue, so \(p=\frac15=0.2\). The parameters are \(n=6\) and \(k=3\). 2. \(P(X=3)=\binom63(0.2)^3(0.8)^3\). 3. Since \(\binom63=20\), the probability is \(20\cdot0.008\cdot0.512=0.08192\).

Answer

\(P(X=3)=0.08192\), or \(8.192\%\)
53745812
A fair six-sided number cube is rolled \(3\) times. Rolling a \(6\) is considered a success. The random variable \(X\) is the number of sixes rolled. a) Find \(P(X=k)\) for \(k\in\{0,1,2,3\}\). b) Verify that the probabilities add to \(1\).
Figure for problem 537458

Hints

- The probability of rolling a \(6\) is the same on every roll. - Find the probability of not rolling a \(6\). - Use the binomial probability formula or organize the outcomes with a tree diagram.

Solution

1. This is a binomial setting with \(n=3\), \(p=\frac{1}{6}\), and \(q=\frac{5}{6}\). 2. Use \(P(X=k)=\binom{3}{k}p^kq^{3-k}\): \(P(X=0)=\left(\frac{5}{6}\right)^3=\frac{125}{216}\approx0.5787\). \(P(X=1)=3\cdot\frac{1}{6}\cdot\left(\frac{5}{6}\right)^2=\frac{75}{216}\approx0.3472\). \(P(X=2)=3\cdot\left(\frac{1}{6}\right)^2\cdot\frac{5}{6}=\frac{15}{216}\approx0.0694\). \(P(X=3)=\left(\frac{1}{6}\right)^3=\frac{1}{216}\approx0.0046\). 3. The sum is \(\frac{125+75+15+1}{216}=\frac{216}{216}=1\).

Answer

a) \(P(X=0)=\frac{125}{216}\approx0.5787\), \(P(X=1)=\frac{75}{216}\approx0.3472\), \(P(X=2)=\frac{15}{216}\approx0.0694\), and \(P(X=3)=\frac{1}{216}\approx0.0046\). b) \(\frac{125+75+15+1}{216}=1\).
52343012
At a glass factory, each drinking glass independently has a \(3\%\) chance of having small air bubbles. The glasses are packed in boxes of \(12\). a) Find the probability that a randomly selected box contains no glasses with air bubbles. b) An inspector checks \(6\) such boxes. What is the probability that at least \(5\) of the boxes contain only glasses without air bubbles?

Hints

- First find the probability that one glass has no air bubbles. - How can you find the probability that all \(12\) glasses in a box have no air bubbles? - The phrase “at least” means that you must include more than one possible number of acceptable boxes. - Use the probability from part a) as the success probability for the \(6\) boxes.

Solution

1. The probability that one glass has no air bubbles is \(1-0.03=0.97\). 2. The probability that all \(12\) glasses in one box have no air bubbles is \(p_B=0.97^{12}\approx 0.6938\). 3. Let \(Y\) be the number of acceptable boxes among the \(6\) checked. Then \(Y\) has a binomial distribution with \(n=6\) and \(p=p_B\). 4. \(P(Y\ge 5)=P(Y=5)+P(Y=6)\). 5. \(P(Y=5)=\binom{6}{5}p_B^5(1-p_B)\approx 0.2954\). 6. \(P(Y=6)=p_B^6\approx 0.1116\). 7. Therefore, \(P(Y\ge 5)\approx 0.2954+0.1116=0.4070\).

Answer

a) The probability is approximately \(69.38\%\). b) The probability is approximately \(40.70\%\).
52345112
Two basketball players compete in a free-throw challenge. Assume all shots are independent. Player A makes each free throw with probability \(0.90\) and takes \(4\) shots. Player B makes each free throw with probability \(0.60\) and takes \(6\) shots. Which event is more likely? Event 1: Player A makes at least \(3\) shots. Event 2: Player B makes exactly \(4\) shots.

Hints

- Model each player’s shots with a binomial distribution. - Identify \(n\), \(p\), and the relevant values of the random variable for each player. - For Player A, “at least \(3\)” includes more than one case. - Use the binomial probability formula for each required case.

Solution

1. For Player A, let \(X\) be the number of made shots. Then \(X\) has a binomial distribution with \(n=4\) and \(p=0.90\). 2. \(P(X\ge 3)=P(X=3)+P(X=4)\). 3. \(P(X=3)=\binom{4}{3}(0.90)^3(0.10)=0.2916\), and \(P(X=4)=(0.90)^4=0.6561\). 4. Therefore, \(P(X\ge 3)=0.2916+0.6561=0.9477\). 5. For Player B, let \(Y\) be the number of made shots. Then \(Y\) has a binomial distribution with \(n=6\) and \(p=0.60\). 6. \(P(Y=4)=\binom{6}{4}(0.60)^4(0.40)^2=0.31104\). 7. Since \(0.9477>0.31104\), Event 1 is more likely.

Answer

Event 1 is more likely. Its probability is \(94.77\%\), compared with approximately \(31.10\%\) for Event 2.
52345212
In a game show, contestants choose between two strategies and guess every answer independently and randomly. Strategy A: Answer \(5\) multiple-choice questions, each with \(4\) choices and exactly one correct answer. To win, the contestant must answer at least \(3\) questions correctly. Strategy B: Answer \(10\) true-or-false questions. To win, the contestant must answer exactly \(5\) questions correctly. Which strategy has the greater probability of success?

Hints

- Determine the success probability for one question in each strategy. - Identify the number of questions in each strategy. - “At least \(3\)” includes \(3\), \(4\), and \(5\). - Calculate the required binomial probabilities and compare them.

Solution

1. For Strategy A, let \(X\sim\operatorname{Bin}(5,0.25)\). Then \(P(X\ge 3)=\sum_{k=3}^{5}\binom{5}{k}(0.25)^k(0.75)^{5-k}=\frac{106}{1024}\approx 0.103516\). 2. For Strategy B, let \(Y\sim\operatorname{Bin}(10,0.50)\). Then \(P(Y=5)=\binom{10}{5}(0.50)^{10}=\frac{252}{1024}\approx 0.246094\). 3. Since \(0.246094>0.103516\), Strategy B has the greater success probability.

Answer

Strategy B has the greater success probability: approximately \(24.61\%\), compared with approximately \(10.35\%\) for Strategy A.
52353712
A pharmaceutical company packages tablets in blister packs of \(20\). Suppose each tablet has the same probability \(p\) of being defective, independently of the others. The probability that a randomly selected blister pack contains at least one defective tablet is exactly \(5\%\). Find \(p\). Round your answer to four decimal places.

Hints

- Express “at least one defective tablet” using a complement. - What is the probability that all \(20\) tablets are not defective? - Set up an equation involving the probability of zero defective tablets and solve for \(p\). - Use a twentieth root when isolating \(1-p\).

Solution

1. Let \(X\) be the number of defective tablets in a pack. Then \(X\) has a binomial distribution with \(n=20\) and defect probability \(p\). 2. The condition is \(P(X\ge 1)=1-P(X=0)=0.05\). 3. Therefore, \(1-(1-p)^{20}=0.05\), so \((1-p)^{20}=0.95\). 4. Solve for \(p\): \(p=1-(0.95)^{1/20}\). 5. Numerically, \(p\approx 0.0025614\), which rounds to \(0.0026\).

Answer

The probability that one tablet is defective is approximately \(0.0026\), or \(0.26\%\).
52354012
A spinner is divided into \(5\) equal sections, and exactly \(1\) section is red. The spinner is spun \(8\) times, and the spins are independent. a) Find the probability that the spinner lands on red exactly \(3\) times. b) Find the probability that the spinner lands on red at least \(2\) times. c) What is the minimum number of spins needed so that the probability of landing on red at least once is greater than \(90\%\)?

Hints

- First find the probability of landing on red in one spin. - For part b), use the complement of getting \(0\) or \(1\) red result. - For part c), write the complement event “no red results.” - Consider how increasing the number of spins changes the probability of at least one red result.

Solution

1. Each spin has red probability \(p=\frac{1}{5}=0.2\). 2. For \(8\) spins, let \(X\) be the number of times the spinner lands on red. Then \(X\) has a binomial distribution with \(n=8\) and \(p=0.2\). 3. \(P(X=3)=\binom{8}{3}(0.2)^3(0.8)^5\approx 0.1468\). 4. \(P(X\ge 2)=1-[P(X=0)+P(X=1)]\). 5. Thus, \(P(X\ge 2)=1-[(0.8)^8+8(0.2)(0.8)^7]\approx 0.4967\). 6. For part c), require \(1-(0.8)^n>0.90\), so \((0.8)^n<0.10\). 7. Since \(n>\frac{\ln(0.10)}{\ln(0.8)}\approx 10.319\), the minimum number of spins is \(11\).

Answer

a) The probability is approximately \(14.68\%\). b) The probability is approximately \(49.67\%\). c) The spinner must be spun at least \(11\) times.
52354312
A multiple-choice vocabulary quiz has \(10\) questions. Each question has \(4\) answer choices, exactly one of which is correct. A student guesses randomly on every question. a) Find the probability that the student answers exactly \(3\) questions correctly. b) Find the probability that the student answers at least \(1\) question correctly. c) What should the minimum passing score be so that the probability of passing by guessing alone is less than \(15\%\)?

Hints

- What is the probability of guessing one question correctly? - For part b), use the complement of answering every question incorrectly. - For part c), compare upper-tail probabilities for possible passing scores until the probability falls below \(0.15\).

Solution

1. Let \(X\) be the number of correct answers. Then \(X\) has a binomial distribution with \(n=10\) and \(p=0.25\). 2. \(P(X=3)=\binom{10}{3}(0.25)^3(0.75)^7\approx 0.2503\). 3. \(P(X\ge 1)=1-P(X=0)=1-(0.75)^{10}\approx 0.9437\). 4. For a passing score of \(4\), \(P(X\ge 4)\approx 0.2241\), which is greater than \(0.15\). 5. For a passing score of \(5\), \(P(X\ge 5)\approx 0.0781\), which is less than \(0.15\). 6. Therefore, the minimum passing score should be \(5\) correct answers.

Answer

a) The probability is approximately \(25.03\%\). b) The probability is approximately \(94.37\%\). c) The quiz should require at least \(5\) correct answers to pass.
52354912
A hostel has \(45\) beds. For one weekend, it accepts \(52\) reservations because historical data show that each guest with a reservation arrives with probability \(0.80\). Assume the guests’ arrival decisions are independent. a) What is the probability that more guests arrive than there are beds? b) The hostel wants the probability of overbooking—more than \(45\) guests arriving—to be at most \(10\%\). What is the maximum number of reservations it may accept?

Hints

- Model the number of guests who arrive with a binomial random variable. - Which values of the random variable represent an overbooking? - Use a complement to calculate the probability of more than \(45\) arrivals. - For part b), test consecutive possible reservation totals and compare their overbooking probabilities with \(0.10\).

Solution

1. For part a), let \(X\) be the number of arriving guests. Then \(X\) has a binomial distribution with \(n=52\) and \(p=0.80\). 2. An overbooking occurs when \(X>45\), so \(P(X>45)=1-P(X\le 45)\). 3. A cumulative binomial calculation gives \(P(X\le 45)\approx 0.9177\). Therefore, \(P(X>45)\approx 0.0823\). 4. For part b), find the greatest integer \(n\) for which \(P(X>45)\le 0.10\), where \(X\) has a binomial distribution with parameters \(n\) and \(p=0.80\). 5. For \(n=52\), \(P(X>45)\approx 0.0823\), which satisfies the requirement. 6. For \(n=53\), \(P(X>45)\approx 0.1420\), which does not satisfy the requirement. 7. Because the overbooking probability increases as \(n\) increases, the maximum number of reservations is \(52\).

Answer

a) The probability is approximately \(8.23\%\). b) The hostel may accept at most \(52\) reservations.
52357712
An electronics manufacturer knows that \(5\%\) of its components are defective. A random sample of \(20\) components is selected. Let \(X\) be the number of defective components in the sample, and assume the situation can be modeled with independent Bernoulli trials. Find the probability of each event: a) Exactly \(1\) component is defective. b) At most \(2\) components are defective. c) At least \(1\) component is defective.

Hints

- Identify the success probability and the number of trials. - For “at most,” list every included value of the random variable. - Can you use a complement for “at least \(1\)”? - Use the binomial probability formula for exactly \(k\) successes.

Solution

1. The binomial parameters are \(n=20\) and \(p=0.05\). 2. \(P(X=1)=\binom{20}{1}(0.05)(0.95)^{19}\approx 0.3774\). 3. \(P(X\le 2)=\sum_{k=0}^{2}\binom{20}{k}(0.05)^k(0.95)^{20-k}\approx 0.9245\). 4. \(P(X\ge 1)=1-P(X=0)=1-(0.95)^{20}\approx 0.6415\).

Answer

a) \(P(X=1)\approx 0.3774\), or \(37.74\%\) b) \(P(X\le 2)\approx 0.9245\), or \(92.45\%\) c) \(P(X\ge 1)\approx 0.6415\), or \(64.15\%\)
52357812
An archer hits the target on each shot with probability \(p=0.7\). The archer takes \(10\) independent shots, and \(X\) is the number of hits. Find each probability. a) At least \(8\) hits b) More than \(5\) but fewer than \(9\) hits

Hints

- Translate each verbal condition into a set of possible values of \(X\). - List the hit counts included in each event. - Add the corresponding binomial probabilities. - Check carefully whether the boundary values are included.

Solution

1. Since \(X\sim\operatorname{Bin}(10,0.7)\), use \(P(X=k)=\binom{10}{k}(0.7)^k(0.3)^{10-k}\). 2. “At least \(8\)” means \(X=8,9,10\). Therefore, \(P(X\ge 8)=\sum_{k=8}^{10}\binom{10}{k}(0.7)^k(0.3)^{10-k}\approx 0.3828\). 3. “More than \(5\) but fewer than \(9\)” means \(X=6,7,8\). Therefore, \(P(5<X<9)=\sum_{k=6}^{8}\binom{10}{k}(0.7)^k(0.3)^{10-k}\approx 0.7004\).

Answer

a) \(P(X\ge 8)\approx 0.3828\), or about \(38.28\%\) b) \(P(5<X<9)\approx 0.7004\), or about \(70.04\%\)
52359312
A spinner is divided into \(4\) equal sections colored red, blue, green, and yellow. The spinner is spun \(10\) times, and the spins are independent. Find the probability of each event: a) Red appears exactly \(2\) times. b) Red appears at most \(1\) time. c) Red appears at least \(2\) times. d) The spinner lands on a color other than red at least \(9\) times.

Hints

- First find the probability of landing on red in one spin. - Which distribution models repeated spins under the same conditions? - Can a complement make any of the calculations shorter? - For part d), translate the number of non-red results into a condition on the number of red results.

Solution

1. Let \(X\) be the number of times red appears. Then \(X\) has a binomial distribution with \(n=10\) and \(p=0.25\). 2. \(P(X=2)=\binom{10}{2}(0.25)^2(0.75)^8\approx 0.2816\). 3. \(P(X\le 1)=(0.75)^{10}+10(0.25)(0.75)^9\approx 0.2440\). 4. \(P(X\ge 2)=1-P(X\le 1)\approx 1-0.2440=0.7560\). 5. At least \(9\) non-red results means at most \(1\) red result, so part d) has the same probability as part b): approximately \(0.2440\).

Answer

a) \(P\approx 0.2816\), or \(28.16\%\) b) \(P\approx 0.2440\), or \(24.40\%\) c) \(P\approx 0.7560\), or \(75.60\%\) d) \(P\approx 0.2440\), or \(24.40\%\)
52359412
A fair six-sided die is rolled \(6\) times, and the rolls are independent. Find the probability of each event: a) Exactly one \(6\) is rolled. b) No \(6\) is rolled. c) At least two \(6\)s are rolled. d) A number other than \(6\) is rolled at most \(5\) times.

Hints

- What are the probabilities of rolling a \(6\) and not rolling a \(6\) on one roll? - For “at least two,” consider subtracting the cases \(0\) and \(1\) from \(1\). - For part d), what is the complement of rolling a non-\(6\) all \(6\) times?

Solution

1. Let \(X\) be the number of \(6\)s rolled. Then \(X\) has a binomial distribution with \(n=6\) and \(p=\frac{1}{6}\). 2. \(P(X=1)=\binom{6}{1}\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^5\approx 0.4019\). 3. \(P(X=0)=\left(\frac{5}{6}\right)^6\approx 0.3349\). 4. \(P(X\ge 2)=1-[P(X=0)+P(X=1)]\approx 0.2632\). 5. A non-\(6\) is rolled at most \(5\) times exactly when at least one \(6\) is rolled. Therefore, the probability is \(1-P(X=0)\approx 0.6651\).

Answer

a) \(P\approx 0.4019\) b) \(P\approx 0.3349\) c) \(P\approx 0.2632\) d) \(P\approx 0.6651\)
52360712
A spinner has \(5\) equal sections, exactly \(1\) of which is blue. The spinner is spun \(6\) times, the spins are independent, and landing on blue counts as a success. a) Find the probability of getting at least \(2\) successes. b) Find the probability that the first success occurs on the fourth spin. c) An event \(A\) has probability \(P(A)=\binom{6}{3}(0.2)^3(0.8)^3\). State the values of \(n\), \(k\), and \(p\), and describe event \(A\) in context.

Hints

- For part a), compare a direct sum with using the complement of \(0\) or \(1\) success. - What must happen on the first \(3\) spins for the first success to occur on spin \(4\)? - Compare the expression in part c) with the binomial probability formula.

Solution

1. The success probability is \(p=\frac{1}{5}=0.2\). 2. For part a), let \(X\) be the number of successes in \(6\) spins. Then \(P(X\ge 2)=1-[P(X=0)+P(X=1)]\). 3. \(P(X\ge 2)=1-[(0.8)^6+6(0.2)(0.8)^5]=0.34464\). 4. For the first success to occur on the fourth spin, the first \(3\) spins must be failures and the fourth must be a success. The last \(2\) spins can have any outcomes. 5. Therefore, the probability is \((0.8)^3(0.2)=0.1024\). 6. In part c), \(n=6\), \(k=3\), and \(p=0.2\). Event \(A\) is getting exactly \(3\) blue results in \(6\) spins.

Answer

a) \(P(X\ge 2)=0.34464\), or approximately \(34.46\%\) b) The probability is \(0.1024\), or \(10.24\%\). c) \(n=6\), \(k=3\), and \(p=0.2\). Event \(A\) is getting exactly \(3\) blue results in \(6\) spins.
52361312
Examine each probability expression. Determine whether it is a correct use of the binomial probability formula. For each valid expression, identify \(n\), \(p\), and \(k\), and describe a possible experiment and event. a) \(P(X=3)=\binom{10}{3}(0.15)^3(0.85)^7\) b) \(P(X=4)=\binom{12}{4}(0.20)^4(0.80)^6\) c) \(P(X=8)=\binom{10}{8}(0.20)^2(0.80)^8\)

Hints

- Compare each expression with the general binomial probability formula. - Check the relationship between the upper number in the binomial coefficient and the exponents. - Verify that the success and failure probabilities add to \(1\). - For a valid model, describe repeated independent trials with two relevant outcomes.

Solution

1. In a), the coefficient and exponents match \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\). Here \(n=10\), \(k=3\), and \(p=0.15\). For example, a process with a \(15\%\) success probability is repeated \(10\) independent times, and the event is exactly \(3\) successes. 2. In b), the binomial coefficient indicates \(n=12\), but the exponents add to \(4+6=10\), not \(12\). Therefore, the expression is not a valid binomial probability formula. 3. In c), the expression is valid with \(n=10\), \(k=8\), and \(p=0.80\). The factors are written in the opposite order, but multiplication is commutative. For example, a basketball player with an \(80\%\) free-throw success rate takes \(10\) independent shots, and the event is exactly \(8\) made shots.

Answer

a) Yes: \(n=10\), \(p=0.15\), \(k=3\). One example is exactly \(3\) successes in \(10\) independent trials with success probability \(0.15\). b) No. The exponents add to \(10\), but the binomial coefficient indicates \(n=12\). c) Yes: \(n=10\), \(p=0.80\), \(k=8\). One example is exactly \(8\) made free throws in \(10\) independent attempts with success probability \(0.80\).
52361412
A candy maker produces chocolates, and \(4\%\) of them are misshapen. A random sample of \(15\) chocolates is selected. Let \(X\) be the number of misshapen chocolates, and assume the outcomes are independent. a) Explain why \(P(X=2)=\binom{15}{2}(0.04)^2(0.96)^{13}\) correctly represents the probability of selecting exactly \(2\) misshapen chocolates. b) An employee claims that \((0.96)^{15}+15(0.04)(0.96)^{14}\) gives the probability that at most \(1\) chocolate is misshapen. Evaluate the claim using the binomial probability formula.

Hints

- Check the conditions required for a binomial model. - What values of \(X\) are included in “at most \(1\)”? - Recall the values of \(\binom{n}{0}\) and \(\binom{n}{1}\). - Substitute \(k=0\) and \(k=1\) separately into the binomial probability formula.

Solution

1. For part a), there are \(15\) independent trials, each with two outcomes and a constant misshapen probability of \(0.04\). 2. The factor \(\binom{15}{2}\) counts the ways to choose the \(2\) misshapen chocolates, \((0.04)^2\) is the probability for those \(2\) outcomes, and \((0.96)^{13}\) is the probability that the other \(13\) chocolates are not misshapen. Therefore, the expression is correct. 3. For part b), “at most \(1\)” means \(X=0\) or \(X=1\). 4. \(P(X=0)=\binom{15}{0}(0.04)^0(0.96)^{15}=(0.96)^{15}\). 5. \(P(X=1)=\binom{15}{1}(0.04)(0.96)^{14}=15(0.04)(0.96)^{14}\). 6. Their sum is exactly the employee’s expression, so the claim is correct.

Answer

a) The expression correctly applies the binomial probability formula with \(n=15\), \(k=2\), and \(p=0.04\). b) The claim is correct because the expression is \(P(X=0)+P(X=1)=P(X\le 1)\).
52361512
A multiple-choice test has \(12\) questions. Each question has four answer choices, exactly one of which is correct. A student guesses randomly and independently on every question. Let \(X\) be the number of correct answers. a) Find the probability that the student answers at least \(5\) questions correctly. b) Find the smallest integer \(k\) for which the probability of getting at least \(k\) correct answers by guessing is less than \(10\%\).

Hints

- Determine the probability of answering one question correctly by guessing. - Identify the distribution for a fixed number of independent trials with a constant success probability. - Relate “at least \(k\)” to “at most \(k-1\)” using a complement. - For part b), test consecutive integer values of \(k\) with cumulative probabilities.

Solution

1. The probability of a correct guess is \(p=\frac{1}{4}=0.25\), so \(X\sim\operatorname{Bin}(12,0.25)\). 2. For part a), use the complement: \(P(X\ge 5)=1-P(X\le 4)\approx 1-0.8424=0.1576\). 3. For part b), seek the smallest \(k\) such that \(P(X\ge k)<0.10\), or equivalently \(P(X\le k-1)>0.90\). 4. For \(k=5\), \(P(X\ge 5)\approx 0.1576>0.10\). For \(k=6\), \(P(X\ge 6)=1-P(X\le 5)\approx 1-0.9456=0.0544<0.10\). 5. Therefore, the smallest qualifying value is \(k=6\).

Answer

a) \(P(X\ge 5)\approx 0.1576\), or about \(15.76\%\) b) \(k=6\)
52361712
A container holds \(20\) marbles. Some are blue, and the rest are yellow. A marble is drawn \(8\) times with replacement. The probability of drawing at most \(2\) blue marbles is approximately \(67.85\%\). How many blue marbles are in the container?

Hints

- Which probability distribution applies when the draws have two outcomes and are made with replacement? - How is the probability of drawing blue related to the number of blue marbles? - Write a cumulative probability for drawing at most \(2\) blue marbles. - Test possible whole-number counts of blue marbles and compare the result with the given probability.

Solution

1. Let \(k\) be the number of blue marbles. Then the probability of drawing blue on one draw is \(p=\frac{k}{20}\). 2. Let \(X\) be the number of blue marbles drawn in \(8\) draws. Because the draws are made with replacement, \(X\) has a binomial distribution with \(n=8\) and \(p=\frac{k}{20}\). 3. The condition is \(P(X\le 2)\approx 0.6785\). 4. Test possible integer values of \(k\). For \(k=4\), \(p=0.20\) and \(P(X\le 2)\approx 0.7969\). 5. For \(k=5\), \(p=0.25\) and \(P(X\le 2)\approx 0.6785\). 6. As \(k\) increases, \(P(X\le 2)\) decreases, so the matching value is unique. Therefore, the container has \(5\) blue marbles.

Answer

There are \(5\) blue marbles in the container.
52361812
A shipment contains \(25\) electronic components. For quality control, a component is selected, checked, and returned to the shipment before the next selection. This is repeated \(10\) times. The probability of selecting at least \(1\) defective component is approximately \(89.26\%\). How many defective components are in the shipment?

Hints

- Use the complement of selecting at least \(1\) defective component. - What is the probability of selecting no defective components in all \(10\) draws? - Express the one-draw defect probability as a fraction of the \(25\) components. - Solve the resulting equation for the unknown probability.

Solution

1. Let \(k\) be the number of defective components. Then the probability of selecting a defective component on one draw is \(p=\frac{k}{25}\). 2. Let \(X\) be the number of defective selections in \(10\) draws. Because each component is returned, \(X\) has a binomial distribution with \(n=10\) and \(p=\frac{k}{25}\). 3. Using the complement, \(P(X\ge 1)=1-P(X=0)=1-(1-p)^{10}\). 4. Set \(1-(1-p)^{10}\approx 0.8926\). Then \((1-p)^{10}\approx 0.1074\). 5. Taking the tenth root gives \(1-p\approx 0.8\), so \(p\approx 0.2\). 6. Therefore, \(k=25\cdot0.2=5\).

Answer

The shipment contains \(5\) defective components.
52362312
A binomial random variable \(X\) has parameters \(n=20\) and \(p=0.30\). Find each probability. a) \(P(X=6)\) b) \(P(X\le 4)\) c) \(P(X>7)\) d) \(P(5\le X\le 9)\)

Hints

- Decide whether each part asks for one exact value or a range of values. - “At most” probabilities are cumulative sums. - A complement can simplify an upper-tail probability. - For an interval, subtract the cumulative probability below the interval from the cumulative probability at its upper endpoint.

Solution

1. Use the binomial probability formula for part a): \(P(X=6)=\binom{20}{6}(0.30)^6(0.70)^{14}\approx 0.1916\). 2. A cumulative binomial calculation gives \(P(X\le 4)\approx 0.2375\). 3. Use a complement for part c): \(P(X>7)=1-P(X\le 7)\approx 1-0.7723=0.2277\). 4. For part d), subtract cumulative probabilities: \(P(5\le X\le 9)=P(X\le 9)-P(X\le 4)\approx 0.9520-0.2375=0.7145\).

Answer

a) \(P(X=6)\approx 0.1916\) b) \(P(X\le 4)\approx 0.2375\) c) \(P(X>7)\approx 0.2277\) d) \(P(5\le X\le 9)\approx 0.7145\)
52520712
The lifetime \(X\), in hours, of an LED bulb is normally distributed with mean \(\mu=15{,}000\) and standard deviation \(\sigma=1200\). 1. Find the probability that a randomly selected bulb lasts less than \(13{,}500\) hours. 2. A wholesaler samples \(200\) bulbs. Find the probability that at most \(30\) bulbs in the sample last less than \(13{,}500\) hours. Model the count with a binomial distribution.

Hints

- Standardize the lifetime cutoff. - Use the probability from part 1 as the binomial success probability. - Identify the sample size and the maximum allowed number of successes. - Use a cumulative binomial probability.

Solution

1. The standard score is \(z=\frac{13{,}500-15{,}000}{1200}=-1.25\). Therefore, \(P(X<13{,}500)=\Phi(-1.25)\approx0.10565\). 2. Let \(Y\sim\operatorname{Bin}(200,p)\), where \(p=\Phi(-1.25)\approx0.10565\). 3. Then \(P(Y\le30)=\sum_{k=0}^{30}\binom{200}{k}p^k(1-p)^{200-k}\approx0.9805\).

Answer

1. Approximately \(0.1056\), or \(10.56\%\) 2. Approximately \(0.9805\), or \(98.05\%\)
52524112
A hot-drink machine dispenses coffee with volume \(X\), in milliliters, normally distributed with mean \(\mu=200\) and standard deviation \(\sigma=5\). a) Find the probability that a cup contains less than \(195\,\text{mL}\). b) For a meeting, \(50\) cups are dispensed independently. Find the probability that at most \(10\) cups contain less than \(195\,\text{mL}\).

Hints

- Model the fill volume of one cup with a normal distribution. - Standardize the cutoff. - Use a binomial model for the number of cups below the cutoff. - Evaluate a cumulative binomial probability.

Solution

1. a) The standard score is \(z=\frac{195-200}{5}=-1\). Therefore, \(P(X<195)=\Phi(-1)\approx0.1587\). 2. b) Let \(Y\sim\operatorname{Bin}(50,p)\), where \(p=\Phi(-1)\approx0.15866\). 3. Then \(P(Y\le10)=\sum_{k=0}^{10}\binom{50}{k}p^k(1-p)^{50-k}\approx0.8406\).

Answer

a) Approximately \(0.1587\), or \(15.87\%\) b) Approximately \(0.8406\), or \(84.06\%\)
52524212
A factory produces precision bolts. Their diameters \(X\), in millimeters, are normally distributed with mean \(\mu=10.00\) and standard deviation \(\sigma=0.05\). A bolt fits correctly if its diameter is between \(9.95\,\text{mm}\) and \(10.05\,\text{mm}\). a) Find the proportion of bolts that fit correctly. b) A shipment contains \(200\) bolts. Find the probability that at least \(140\) fit correctly.

Hints

- Standardize the two diameter limits. - Relate the interval to a familiar central normal probability. - Use a binomial model for the number of correctly fitting bolts. - Use a complement for “at least” a specified count.

Solution

1. a) The endpoints have standard scores \(-1\) and \(1\). Thus, \(p=P(9.95\le X\le10.05)=\Phi(1)-\Phi(-1)\approx0.68269\). 2. b) Let \(Y\sim\operatorname{Bin}(200,p)\). Then \(P(Y\ge140)=1-P(Y\le139)\approx0.3291\).

Answer

a) Approximately \(0.6827\), or \(68.27\%\) b) Approximately \(0.3291\), or \(32.91\%\)
52525412
A manufacturer states that the lifetime \(X\), in hours, of an LED model is normally distributed with mean \(\mu=1500\) and standard deviation \(\sigma=60\). a) Find the probability that a randomly selected LED lasts at least \(1410\) hours. b) An industrial facility installs \(40\) LEDs. Find the probability that at least \(38\) of them last at least \(1410\) hours.

Hints

- Standardize the minimum lifetime. - Use symmetry of the normal distribution when helpful. - Use the result from part a) as a binomial success probability. - Add the probabilities for exactly \(38\), \(39\), and \(40\) successes.

Solution

1. a) The standard score is \(z=\frac{1410-1500}{60}=-1.5\). Thus, \(p=P(X\ge1410)=1-\Phi(-1.5)=\Phi(1.5)\approx0.93319\). 2. b) Let \(Y\sim\operatorname{Bin}(40,p)\). Then \(P(Y\ge38)=P(Y=38)+P(Y=39)+P(Y=40)\approx0.4947\).

Answer

a) Approximately \(0.9332\), or \(93.32\%\) b) Approximately \(0.4947\), or \(49.47\%\)
52529912
A manufacturer states that the lengths of its steel pins are normally distributed with mean \(\mu=120\,\text{mm}\) and standard deviation \(\sigma=0.5\,\text{mm}\). a) Describe how the manufacturer could have estimated \(\mu\) and \(\sigma\) from a sample. b) A customer buys a package of \(150\) pins. A pin meets specifications when its length is between \(119.5\,\text{mm}\) and \(120.5\,\text{mm}\). Find the probability that at least \(65\%\) of the pins in the package meet specifications.

Hints

- Think about which sample statistics estimate a population mean and standard deviation. - Compare the specification interval with the given mean and standard deviation. - After finding the probability for one pin, model the number of acceptable pins with a binomial distribution. - Convert \(65\%\) of \(150\) into a whole-number minimum.

Solution

1. a) The manufacturer could select a sufficiently large random sample of pins and measure their lengths. The sample mean estimates \(\mu\), and the sample standard deviation estimates \(\sigma\). A histogram or normal probability plot could be used to assess whether a normal model is reasonable. 2. b) For one pin, \(p=P(119.5\le X\le120.5)\). These endpoints are one standard deviation below and above the mean, so \(p=\Phi(1)-\Phi(-1)\approx0.6827\). 3. Let \(Y\) be the number of pins that meet specifications. Assuming independent pin lengths, \(Y\sim\operatorname{Binomial}(150,0.6827)\). 4. At least \(65\%\) of \(150\) means at least \(98\) pins because \(0.65\cdot150=97.5\). 5. Therefore, \(P(Y\ge98)=\sum_{k=98}^{150}\binom{150}{k}p^k(1-p)^{150-k}\approx0.8059\).

Answer

a) Use a sufficiently large random sample, estimate \(\mu\) with the sample mean, and estimate \(\sigma\) with the sample standard deviation. b) Approximately \(0.8059\), or \(80.59\%\)
52530012
A coffee company fills bags with ground coffee. The fill weight \(W\), in grams, is normally distributed with \(\mu=500\) and \(\sigma=4\). a) Find the probability that a randomly selected bag contains more than \(505\,\text{g}\). b) A grocery store receives \(400\) bags. Find the probability that at most \(10\%\) of the bags contain more than \(505\,\text{g}\). Assume the bag weights are independent.

Hints

- First find the probability for one bag using a standard score. - Model the number of bags above the cutoff with a binomial distribution. - Convert “at most \(10\%\)” into a maximum number of bags.

Solution

1. a) The standard score for \(505\,\text{g}\) is \(z=\frac{505-500}{4}=1.25\). 2. Therefore, \(p=P(W>505)=1-\Phi(1.25)\approx0.1056\). 3. b) Let \(X\) be the number of bags containing more than \(505\,\text{g}\). Then \(X\sim\operatorname{Binomial}(400,p)\). 4. At most \(10\%\) of \(400\) bags means \(X\le40\). 5. Thus, \(P(X\le40)=\sum_{k=0}^{40}\binom{400}{k}p^k(1-p)^{400-k}\approx0.3950\).

Answer

a) Approximately \(0.1056\), or \(10.56\%\) b) Approximately \(0.3950\), or \(39.50\%\)
52530112
At a bottling plant, the fill amount \(X\) in a \(750\,\text{mL}\) orange juice bottle is normally distributed with mean \(\mu=754\,\text{mL}\) and standard deviation \(\sigma=4\,\text{mL}\). a) Find the probability that a randomly selected bottle contains less than \(750\,\text{mL}\). b) A case contains \(12\) bottles. Find the probability that at most one bottle contains less than \(750\,\text{mL}\). Assume the fill amounts are independent. c) What is the minimum number of bottles that must be inspected so that the probability of finding at least one bottle containing less than \(750\,\text{mL}\) is at least \(95\%\)?

Hints

- Express the cutoff as a standard score. - Use a binomial model for the number of underfilled bottles in a fixed-size sample. - For “at least one,” consider the complement event of finding none. - Use logarithms to solve an inequality with the sample size in an exponent.

Solution

1. a) The standard score is \(z=\frac{750-754}{4}=-1\). Thus, \(p=P(X<750)=\Phi(-1)\approx0.1587\). 2. b) Let \(Y\) be the number of underfilled bottles in a case. Then \(Y\sim\operatorname{Binomial}(12,p)\). 3. Therefore, \(P(Y\le1)=(1-p)^{12}+12p(1-p)^{11}\approx0.4105\). 4. c) For \(n\) inspected bottles, \(P(\text{at least one underfilled})=1-(1-p)^n\). We need \(1-(1-p)^n\ge0.95\). 5. This is equivalent to \((1-p)^n\le0.05\), so \(n\ge\frac{\ln(0.05)}{\ln(1-p)}\approx17.34\). The smallest whole number that works is \(18\).

Answer

a) Approximately \(0.1587\), or \(15.87\%\) b) Approximately \(0.4105\), or \(41.05\%\) c) \(18\) bottles
52530212
The diameter \(D\) of a steel pin is normally distributed with mean \(\mu=20.00\,\text{mm}\) and standard deviation \(\sigma=0.05\,\text{mm}\). A pin is rejected if its diameter differs from the mean by more than \(0.10\,\text{mm}\). a) Use the Empirical Rule, without a calculator, to explain why about \(5\%\) of the pins are rejected. b) A box contains \(50\) independently produced pins. Find the probability that exactly two pins are rejected. c) A normal approximation may be used for a binomial distribution when the large-counts conditions \(np\ge10\) and \(n(1-p)\ge10\) hold. Find the minimum value of \(n\) that meets both conditions, and verify the conditions for \(n=200\).

Hints

- Compare the rejection cutoff with the standard deviation. - Use the Empirical Rule to estimate the probability of rejection for one pin. - Model the number of rejected pins with a binomial distribution. - Apply each large-counts inequality separately.

Solution

1. a) The rejection cutoff is \(0.10\,\text{mm}=2\sigma\) from the mean. By the Empirical Rule, about \(95\%\) of values lie within \(2\sigma\), so about \(5\%\) lie outside and are rejected. 2. b) Let \(X\) be the number of rejected pins. Using \(p\approx0.05\), \(X\sim\operatorname{Binomial}(50,0.05)\). 3. Therefore, \(P(X=2)=\binom{50}{2}(0.05)^2(0.95)^{48}\approx0.2611\). 4. c) The conditions are \(0.05n\ge10\) and \(0.95n\ge10\). These give \(n\ge200\) and \(n\ge10.53\), respectively. 5. Thus, the minimum sample size is \(n=200\). For \(n=200\), \(np=10\) and \(n(1-p)=190\), so both conditions hold.

Answer

a) About \(5\%\) b) Approximately \(0.2611\), or \(26.11\%\) c) The minimum is \(n=200\); at \(n=200\), the large-counts values are \(10\) and \(190\).
52531112
The fill weight \(X\) of a coffee bag is approximately normally distributed with mean \(\mu=500\,\text{g}\) and standard deviation \(\sigma=5\,\text{g}\). The bags are classified as follows: <table> <tr> <td>Category</td> <td>Underweight</td> <td>Standard</td> <td>Overweight</td> </tr> <tr> <td>Weight \(m\) in \(\text{g}\)</td> <td>\(m<490\)</td> <td>\(490\le m<510\)</td> <td>\(m\ge510\)</td> </tr> </table> a) Use the Empirical Rule to estimate the probability that a randomly selected bag is classified as Standard and the probability that it is classified as Overweight. b) Find the minimum number of independently selected bags needed so that the probability of including at least one Underweight bag is at least \(95\%\). Use the Empirical Rule estimate from part a).

Hints

- Express each boundary as a number of standard deviations from the mean. - Use symmetry to split the probability outside the central interval. - For “at least one,” use the complement event of selecting no Underweight bags. - Use logarithms to solve for a variable in an exponent.

Solution

1. a) The endpoints are \(490=500-2\cdot5=\mu-2\sigma\) and \(510=500+2\cdot5=\mu+2\sigma\). 2. By the Empirical Rule, \(P(490\le X<510)\approx0.95\). 3. The remaining probability is about \(0.05\), split equally between the tails. Thus, \(P(X\ge510)\approx0.025\), and the Underweight probability is also \(p\approx0.025\). 4. b) For \(n\) independent bags, the probability of at least one Underweight bag is \(1-(1-p)^n\). We need \(1-(0.975)^n\ge0.95\). 5. Therefore, \((0.975)^n\le0.05\), so \(n\ge\frac{\ln(0.05)}{\ln(0.975)}\approx118.33\). 6. The smallest whole number that works is \(119\).

Answer

a) Standard: approximately \(95\%\); Overweight: approximately \(2.5\%\) b) \(119\) bags
52535712
An orange juice filling machine produces bottles whose fill amount \(X\), in milliliters, is normally distributed with mean \(\mu=500\) and standard deviation \(\sigma=4\). a) Find the probability that a randomly selected bottle contains between \(495\,\text{mL}\) and \(505\,\text{mL}\). b) Find the proportion of bottles containing less than \(492\,\text{mL}\). c) Twelve bottles are selected independently for a case. Find the probability that at most one bottle in the case contains less than \(492\,\text{mL}\).

Hints

- Standardize each endpoint before using the normal distribution. - Use the probability from part b) as the success probability in part c). - Model the number of bottles below the cutoff with a binomial distribution. - “At most one” includes zero and one.

Solution

1. a) The endpoint standard scores are \(z_1=\frac{495-500}{4}=-1.25\) and \(z_2=\frac{505-500}{4}=1.25\). 2. Therefore, \(P(495\le X\le505)=\Phi(1.25)-\Phi(-1.25)\approx0.7887\). 3. b) The standard score for \(492\) is \(z=\frac{492-500}{4}=-2\). Thus, \(p=P(X<492)=\Phi(-2)\approx0.0228\). 4. c) Let \(Y\) be the number of bottles below \(492\,\text{mL}\). Then \(Y\sim\operatorname{Binomial}(12,p)\). 5. Thus, \(P(Y\le1)=(1-p)^{12}+12p(1-p)^{11}\approx0.9706\).

Answer

a) Approximately \(0.7887\), or \(78.87\%\) b) Approximately \(0.0228\), or \(2.28\%\) c) Approximately \(0.9706\), or \(97.06\%\)
52535812
The battery life \(X\), in hours, of a smartphone under heavy use is normally distributed with mean \(\mu=14\) and standard deviation \(\sigma=1.5\). a) Find the probability that a randomly selected phone has a battery life between \(12\) and \(15\) hours. b) Find the percentage of phones with a battery life longer than \(16\) hours. c) A retailer has \(50\) independently selected phones. Find the probability that at least \(5\) have a battery life longer than \(16\) hours.

Hints

- Identify the mean and standard deviation. - Use the complement for a probability above a cutoff. - Use the probability from part b) in a binomial model for part c). - “At least \(5\)” can be found using the complement of “at most \(4\).”

Solution

1. a) The endpoint standard scores are \(z_1=\frac{12-14}{1.5}=-1.3333\) and \(z_2=\frac{15-14}{1.5}=0.6667\). 2. Thus, \(P(12\le X\le15)=\Phi(0.6667)-\Phi(-1.3333)\approx0.6563\). 3. b) The standard score for \(16\) is \(z=\frac{16-14}{1.5}=1.3333\). Therefore, \(p=P(X>16)\approx0.0912\). 4. c) Let \(Y\) be the number of phones with battery life longer than \(16\) hours. Then \(Y\sim\operatorname{Binomial}(50,p)\). 5. Therefore, \(P(Y\ge5)=1-P(Y\le4)=1-\sum_{k=0}^{4}\binom{50}{k}p^k(1-p)^{50-k}\approx0.4844\).

Answer

a) Approximately \(0.6563\), or \(65.63\%\) b) Approximately \(9.12\%\) c) Approximately \(0.4844\), or \(48.44\%\)
52652212
A multiple-choice test has \(6\) questions. Each question has \(5\) answer choices, exactly one of which is correct. A student guesses randomly on every question. Describe an event whose probability is represented by each expression. a) \(\left(\frac{4}{5}\right)^6\) b) \(1-\left(\frac{4}{5}\right)^6\) c) \(6\cdot\frac{1}{5}\cdot\left(\frac{4}{5}\right)^5\) d) \(1-\left[\left(\frac{4}{5}\right)^6+6\cdot\frac{1}{5}\cdot\left(\frac{4}{5}\right)^5\right]\)

Hints

- First identify the probabilities of a correct and an incorrect answer. - Use the exponents to determine how many questions are represented. - What event is the complement of getting no correct answers? - In part d), identify the two probabilities being subtracted from \(1\).

Solution

1. Let \(X\) be the number of correct answers. Then \(X\) has a binomial distribution with \(n=6\) and \(p=\frac{1}{5}\). 2. The expression in part a) is \(P(X=0)\), the probability of answering no questions correctly. 3. The expression in part b) is \(1-P(X=0)=P(X\ge 1)\), the probability of answering at least \(1\) question correctly. 4. The expression in part c) is \(\binom{6}{1}\frac{1}{5}\left(\frac{4}{5}\right)^5=P(X=1)\), the probability of answering exactly \(1\) question correctly. 5. The expression in part d) is \(1-[P(X=0)+P(X=1)]=P(X\ge 2)\), the probability of answering at least \(2\) questions correctly.

Answer

a) The student answers none of the \(6\) questions correctly. b) The student answers at least \(1\) question correctly. c) The student answers exactly \(1\) of the \(6\) questions correctly. d) The student answers at least \(2\) questions correctly.
52685512
For a group of \(15\) people, assume that each person’s day of birth is independently and equally likely to be any of the \(7\) days of the week. a) Find the probability that exactly \(2\) people were born on a Sunday. b) Find the probability that at least \(1\) person was born on a Saturday. c) Find the probability that more than \(13\) people were born on a weekday from Monday through Friday.

Hints

- If all days are equally likely, what is the probability of one specified day? - Which probability model applies to repeated independent trials with two outcomes? - For “at least \(1\),” use the complement of getting none. - For part c), count how many days are included from Monday through Friday and identify the values greater than \(13\).

Solution

1. For a specified day of the week, the success probability is \(p=\frac{1}{7}\). 2. For part a), \(P(X=2)=\binom{15}{2}\left(\frac{1}{7}\right)^2\left(\frac{6}{7}\right)^{13}\approx 0.2889\). 3. For part b), use the complement: \(P(X\ge 1)=1-\left(\frac{6}{7}\right)^{15}\approx 0.9010\). 4. For part c), the probability of a Monday-through-Friday birth is \(p=\frac{5}{7}\). More than \(13\) means \(14\) or \(15\). 5. \(P(X>13)=\binom{15}{14}\left(\frac{5}{7}\right)^{14}\left(\frac{2}{7}\right)+\left(\frac{5}{7}\right)^{15}\approx 0.0450\).

Answer

a) \(P(X=2)\approx 0.2889\), or \(28.89\%\) b) \(P(X\ge 1)\approx 0.9010\), or \(90.10\%\) c) \(P(X>13)\approx 0.0450\), or \(4.50\%\)
52686012
A digital memory chip stores data in blocks of \(20\) bits, where each bit is either \(0\) or \(1\). a) How many different \(20\)-bit blocks contain exactly \(2\) ones? b) During transmission, each bit independently has a constant error probability of \(p=0.01\). Find the probability that a \(20\)-bit block has at most \(1\) error. c) State the conditions under which the number of errors in a block can be modeled with a binomial distribution.

Hints

- How many ways can you choose \(2\) positions from \(20\) positions? - Which error counts are included in “at most \(1\)”? - Use the binomial probability formula for each included error count. - Recall the conditions for a binomial setting.

Solution

1. For part a), choose the \(2\) positions containing ones: \(\binom{20}{2}=190\). 2. For part b), let \(X\) be the number of bit errors. Then \(X\) has a binomial distribution with \(n=20\) and \(p=0.01\). 3. \(P(X\le 1)=P(X=0)+P(X=1)\). 4. \(P(X=0)=(0.99)^{20}\approx 0.8179\), and \(P(X=1)=20(0.01)(0.99)^{19}\approx 0.1652\). 5. Therefore, \(P(X\le 1)\approx 0.9831\). 6. A binomial model requires a fixed number of bits, two outcomes for each bit, the same error probability for every bit, and independent error outcomes.

Answer

a) There are \(190\) such blocks. b) The probability is approximately \(0.9831\), or \(98.31\%\). c) The model requires a fixed number of trials, two outcomes per trial, a constant error probability, and independent trials.
52687712
A logistics company labels storage bins with codes. Each code contains either one or two uppercase letters from \(\mathrm{A}\) through \(\mathrm{Z}\), followed by a two-digit number from \(01\) through \(99\). Assume all valid codes are equally likely. A code is called special when its first letter is \(\mathrm{A}\). 1. Find the total number of possible codes. 2. Find the probability \(p\) that a randomly generated code is special. 3. Twenty codes are independently selected with replacement. Find the probability that exactly \(3\) are special.

Hints

- Count the one-letter and two-letter portions separately. - Count how many allowable letter strings begin with \(\mathrm{A}\). - With replacement, each selection has the same two possible classifications: special or not special. - Use the binomial probability formula for an exact number of successes.

Solution

1. There are \(26\) one-letter strings and \(26^2=676\) two-letter strings. Each can be paired with any of \(99\) numbers, so the total is \((26+676)\cdot99=702\cdot99=69{,}498\). 2. A special code has either the one-letter string \(\mathrm{A}\) or one of the \(26\) two-letter strings beginning with \(\mathrm{A}\). Thus, there are \((1+26)\cdot99=2673\) special codes, and \(p=\frac{2673}{69498}=\frac{1}{26}\approx0.0385\). 3. Let \(X\) count the special codes among the \(20\) selections. Then \(X\) is binomial with \(n=20\) and \(p=\frac{1}{26}\). 4. \(P(X=3)=\binom{20}{3}\left(\frac{1}{26}\right)^3\left(\frac{25}{26}\right)^{17}\approx0.0333\).

Answer

1. \(69{,}498\) codes 2. \(p=\frac{1}{26}\approx0.0385\) 3. \(P(X=3)\approx0.0333\), or about \(3.33\%\)
52687812
An archive assigns each document a code consisting of one letter from \(\text{A}\), \(\text{B}\), \(\text{C}\), \(\text{D}\), or \(\text{E}\), followed by a three-digit string from \(000\) through \(999\). A code is called a priority code when its number begins with \(9\). 1. Show that the probability \(p\) that a randomly generated code is a priority code is \(0.1\). 2. Ten codes are generated independently. Find the probability that more than \(2\) are priority codes.

Hints

- Count all possible three-digit strings and those that begin with \(9\). - Which values are included in “more than \(2\)”? - Would a complement make the second probability shorter to calculate? - Include every term needed for the complementary cumulative probability.

Solution

1. There are \(5\cdot 1000=5000\) possible codes. For each letter, the numbers \(900\) through \(999\) give \(100\) priority codes, so there are \(5\cdot 100=500\) priority codes. 2. Therefore, \(p=\frac{500}{5000}=0.1\). 3. Let \(Y\) be the number of priority codes among the \(10\) generated. Then \(Y\) has a binomial distribution with \(n=10\) and \(p=0.1\). 4. \(P(Y>2)=1-P(Y\le 2)\). 5. \(P(Y\le 2)=\sum_{k=0}^{2}\binom{10}{k}(0.1)^k(0.9)^{10-k}\approx 0.9298\). 6. Therefore, \(P(Y>2)\approx 1-0.9298=0.0702\).

Answer

1. \(p=\frac{5\cdot 100}{5\cdot 1000}=0.1\) 2. The probability is approximately \(7.02\%\).
52688312
Five friends—Ava, Ben, Clara, David, and Elias—sit in a random order in a row of five movie theater seats. a) Describe a model using draws from a bag for this random process. Find the probability \(p\) that Ava and Ben sit next to each other. b) The five friends go to the movies on \(10\) different nights and choose a new random seating order each time. Find the probability that Ava and Ben sit next to each other on exactly \(4\) nights.

Hints

- How many ways can five people be arranged in a row? - Treat the two people who must sit together as one block. - Account for the two possible orders within that block. - For the repeated nights, use a model with independent trials and two outcomes.

Solution

1. A suitable model is a bag containing five distinct tiles, one for each person. Draw all five tiles without replacement; the draw order represents the seating order. 2. There are \(5!=120\) possible seating orders. Treat Ava and Ben as one block. The block and the other three people can be arranged in \(4!\) ways, and Ava and Ben can switch positions within the block. Thus, there are \(2\cdot4!=48\) favorable arrangements. 3. Therefore, \(p=\frac{48}{120}=0.4\). 4. Across \(10\) independent nights, let \(X\) count the nights when Ava and Ben sit together. Then \(X\) is binomial with \(n=10\) and \(p=0.4\). 5. \(P(X=4)=\binom{10}{4}(0.4)^4(0.6)^6=0.250822656\approx0.2508\).

Answer

a) Draw five distinct labeled tiles without replacement, using the draw order as the seat order. The probability is \(p=0.4\). b) \(P(X=4)\approx0.2508\), or about \(25.08\%\)
52691312
An online retailer knows that \(15\%\) of shipped packages are returned by customers. A random sample of \(25\) packages is selected, and the return outcomes can be modeled as independent Bernoulli trials. a) Find the probability that among the \(25\) packages: (1) exactly \(3\) are returned. (2) none are returned. (3) at least \(2\) are returned. b) During a promotion, the return rate is expected to increase to \(20\%\). Find the expected number of returns in a sample of \(25\) packages and compare it with the expected number when the return rate is \(15\%\).

Hints

- Identify \(n\), \(p\), and the required value or range of the random variable. - For an “at least” event, consider using a complement. - What is the complement of at least \(2\) returns? - Use \(E(X)=np\) for the expected value of a binomial random variable.

Solution

1. For part a), let \(X\) be the number of returns. Then \(X\) has a binomial distribution with \(n=25\) and \(p=0.15\). 2. \(P(X=3)=\binom{25}{3}(0.15)^3(0.85)^{22}\approx 0.2174\). 3. \(P(X=0)=(0.85)^{25}\approx 0.0172\). 4. \(P(X\ge 2)=1-[P(X=0)+P(X=1)]\). 5. Since \(P(X=1)=25\cdot0.15(0.85)^{24}\approx 0.0759\), \(P(X\ge 2)\approx 0.9069\). 6. For part b), when \(p=0.15\), \(E(X)=np=25\cdot0.15=3.75\). 7. When \(p=0.20\), \(E(X)=25\cdot0.20=5\). The expected number increases by \(1.25\) returns.

Answer

a) (1) Approximately \(21.74\%\) (2) Approximately \(1.72\%\) (3) Approximately \(90.69\%\) b) The expected number is \(3.75\) returns at a \(15\%\) rate and \(5\) returns at a \(20\%\) rate, an increase of \(1.25\).
52698212
A shipping company finds that each package independently has a \(15\%\) chance of requiring special handling because of its dimensions. A sample of \(20\) packages is inspected. a) Find the number of different sequences of special handling \((S)\) and normal handling \((N)\) that contain exactly \(4\) packages requiring special handling. b) Find the probability that exactly \(4\) packages require special handling. c) Find the probability that at most \(2\) packages require special handling.

Hints

- How many ways can \(4\) special-handling outcomes be placed among \(20\) positions? - Use the binomial probability formula for a fixed number of successes. - Which values are included in “at most \(2\)”? - Add the probabilities for \(0\), \(1\), and \(2\) successes.

Solution

1. The number of sequences in part a) is \(\binom{20}{4}=4845\). 2. Let \(X\) be the number of packages requiring special handling. Then \(X\) has a binomial distribution with \(n=20\) and \(p=0.15\). 3. \(P(X=4)=\binom{20}{4}(0.15)^4(0.85)^{16}\approx 0.1821\). 4. \(P(X\le 2)=P(X=0)+P(X=1)+P(X=2)\). 5. A cumulative binomial calculation gives \(P(X\le 2)\approx 0.4049\).

Answer

a) There are \(4845\) different sequences. b) The probability is approximately \(18.21\%\). c) The probability is approximately \(40.49\%\).
52698912
A fair six-sided die is rolled \(6\) times. Treat rolling a \(6\) as a success and any other result as a failure. Consider these two exact sequences: Sequence A: \((6, 6, 6, 6, 6, 6)\) Sequence B: \((6, \text{not }6, 6, \text{not }6, \text{not }6, 6)\) a) Without calculating, decide which sequence is more likely. Explain your reasoning. b) Calculate \(P(A)\) and \(P(B)\). c) Find the probability that exactly three of the six rolls show a \(6\). Explain how this probability differs from \(P(B)\).

Hints

- Compare how many times each sequence contains the less likely result. - For a specified order, multiply the probabilities along that one sequence. - Distinguish one exact sequence from an event that can occur in several orders. - When comparing sequences with different numbers of successes, the success probability matters.

Solution

1. Sequence B is more likely because it contains only three occurrences of the less likely outcome, a \(6\), while Sequence A contains six. 2. \(P(A)=\left(\frac16\right)^6=\frac{1}{46656}\approx0.0000214\). 3. \(P(B)=\left(\frac16\right)^3\left(\frac56\right)^3=\frac{125}{46656}\approx0.00268\). 4. For exactly three \(6\)s in any order, \(P(X=3)=\binom63\left(\frac16\right)^3\left(\frac56\right)^3\approx0.0536\). 5. Sequence B is only one specific arrangement. The event \(X=3\) includes all \(\binom63=20\) arrangements with three \(6\)s.

Answer

a) Sequence B is more likely because it contains fewer occurrences of the less likely result, a \(6\). b) \(P(A)\approx0.0000214\); \(P(B)\approx0.00268\) c) \(P(X=3)\approx0.0536\). This includes all \(20\) arrangements with three \(6\)s, not only Sequence B.
52700612
A container holds \(40\) components, \(6\) of which are defective. Five components are selected one at a time. Determine whether the experiment is binomial when the components are selected a) with replacement; b) without replacement. For the case that is binomial, find the probability of selecting exactly one defective component.

Hints

- How does replacement affect the success probability from one trial to the next? - What condition makes repeated trials independent? - For the binomial case, identify \(n\), \(k\), \(p\), and \(1-p\).

Solution

1. With replacement, each selection has constant defect probability \(p=\frac{6}{40}=0.15\), and the selections are independent. This is a binomial experiment with \(n=5\). 2. Without replacement, the composition of the container changes, so the defect probability changes and the selections are dependent. This is not a binomial experiment. 3. For the with-replacement case, \(P(X=1)=\binom{5}{1}(0.15)^1(0.85)^4\approx 0.3915\).

Answer

a) Binomial with \(n=5\) and \(p=0.15\). The probability of exactly one defective component is \(\binom{5}{1}(0.15)(0.85)^4\approx 0.3915\), or about \(39.15\%\). b) Not binomial because the probability changes without replacement.
52701112
A machine produces bolts that independently meet a quality standard with probability \(0.85\). Six bolts are selected and inspected in order. Find the probability of each event. a) Only the third bolt fails to meet the standard. b) All six bolts meet the standard. c) Only the first four bolts meet the standard. d) Exactly five of the six bolts meet the standard.

Hints

- Decide whether each part specifies exact positions or only a total number of successes. - Count the required successes and failures in each event. - For a specified order, multiply the probabilities in that order. - When only the number of successes matters, include a binomial coefficient.

Solution

1. Let success mean that a bolt meets the standard. Then \(p=0.85\), \(1-p=0.15\), and \(n=6\). 2. For part a), the positions are fixed, so \(P=(0.85)^5(0.15)\approx0.0666\). 3. For part b), \(P=(0.85)^6\approx0.3771\). 4. For part c), the first four bolts meet the standard and the last two do not, so \(P=(0.85)^4(0.15)^2\approx0.0117\). 5. For part d), the five successes can occur in any positions, so \(P(X=5)=\binom65(0.85)^5(0.15)\approx0.3993\).

Answer

a) \(0.0666\), or about \(6.66\%\) b) \(0.3771\), or about \(37.71\%\) c) \(0.0117\), or about \(1.17\%\) d) \(0.3993\), or about \(39.93\%\)
52702612
In a manufacturing process, each component independently has a \(5\%\) chance of being defective. A sample of \(100\) components is selected. Let \(Y\) be the number of defective components in the sample. Describe each expression in context and find its value. a) \(\sum_{k=0}^{10}\binom{100}{k}(0.05)^k(0.95)^{100-k}\) b) \(1-\sum_{k=0}^{2}\binom{100}{k}(0.05)^k(0.95)^{100-k}\) c) \(\sum_{k=4}^{7}\binom{100}{k}(0.05)^k(0.95)^{100-k}\)

Hints

- Identify \(n\) and \(p\) from the context. - What does each term inside the sum represent? - For an expression of the form \(1-P(\text{event})\), identify the complementary event. - In part c), check which integer values are included in the summation limits.

Solution

1. The random variable \(Y\) has a binomial distribution with \(n=100\) and \(p=0.05\). 2. Part a) is \(P(Y\le 10)\), the probability that at most \(10\) components are defective. Its value is approximately \(0.9885\). 3. Part b) is \(1-P(Y\le 2)=P(Y\ge 3)\), the probability that at least \(3\) components are defective. Its value is approximately \(0.8817\). 4. Part c) is \(P(4\le Y\le 7)\), the probability that from \(4\) through \(7\) components are defective. Its value is \(P(Y\le 7)-P(Y\le 3)\approx 0.8720-0.2578=0.6142\).

Answer

a) At most \(10\) defective components; \(P(Y\le 10)\approx 0.9885\) b) At least \(3\) defective components; \(P(Y\ge 3)\approx 0.8817\) c) From \(4\) through \(7\) defective components; \(P(4\le Y\le 7)\approx 0.6142\)
52702712
A random variable \(X\) has a binomial distribution, and its cumulative distribution function is \(F(k)=P(X\le k)\). For integers \(a\) and \(b\) in \(\{0,1,\ldots,n\}\) with \(a<b\), explain why \(P(a<X\le b)=F(b)-F(a)\).

Hints

- Interpret a cumulative probability as a sum of point probabilities. - Split the event \(X\le b\) into two nonoverlapping parts. - Which values included in \(F(b)\) are not included in \(F(a)\)? - Subtract the common portion.

Solution

1. By definition, \(F(b)=P(X\le b)\) and \(F(a)=P(X\le a)\). 2. The event \(\{X\le b\}\) is the disjoint union of \(\{X\le a\}\) and \(\{a<X\le b\}\). 3. Therefore, additivity for disjoint events gives \(P(X\le b)=P(X\le a)+P(a<X\le b)\). 4. Subtracting \(P(X\le a)\) from both sides yields \(P(a<X\le b)=P(X\le b)-P(X\le a)=F(b)-F(a)\).

Answer

Because \(\{X\le b\}\) is the disjoint union of \(\{X\le a\}\) and \(\{a<X\le b\}\), \(F(b)=F(a)+P(a<X\le b)\). Therefore, \(P(a<X\le b)=F(b)-F(a)\).
52703012
Let \(X\sim\operatorname{Bin}(150,0.6)\). Write each probability using summation notation, and find its value. Round to four decimal places. a) \(P(X<85)\) b) \(P(X\ge 100)\) c) \(P(80\le X<110)\)

Hints

- Distinguish carefully between \(<\) and \(\le\). - “At least \(100\)” means values from \(100\) through \(150\). - For \(80\le X<110\), include \(80\) but stop at \(109\). - The lower and upper summation limits should match the smallest and largest included integer values. - The probabilities for all values from \(0\) through \(150\) sum to \(1\).

Solution

1. Because \(X\) is integer-valued, \(P(X<85)=P(X\le 84)\). 2. \(P(X<85)=\sum_{i=0}^{84}\binom{150}{i}(0.6)^i(0.4)^{150-i}\approx 0.1794\). 3. \(P(X\ge 100)=\sum_{i=100}^{150}\binom{150}{i}(0.6)^i(0.4)^{150-i}\approx 0.0555\). 4. For part c), the included values are \(80\) through \(109\). 5. \(P(80\le X<110)=\sum_{i=80}^{109}\binom{150}{i}(0.6)^i(0.4)^{150-i}\approx 0.9587\).

Answer

a) \(\sum_{i=0}^{84}\binom{150}{i}(0.6)^i(0.4)^{150-i}\approx 0.1794\) b) \(\sum_{i=100}^{150}\binom{150}{i}(0.6)^i(0.4)^{150-i}\approx 0.0555\) c) \(\sum_{i=80}^{109}\binom{150}{i}(0.6)^i(0.4)^{150-i}\approx 0.9587\)
52703312
Let \(X\sim\operatorname{Bin}(80,0.25)\). a) List all possible values of \(X\) that satisfy \(|X-20|\le 3\). b) Find each probability. (1) \(P(|X-20|\le 3)\) (2) \(P(|X-20|>3)\) (3) \(P(|X-18|<2)\)

Hints

- Rewrite each absolute-value inequality as a compound inequality. - Keep only integer values because \(X\) counts successes. - Express an interval probability as a difference of cumulative probabilities. - Use the complement rule for part b)(2).

Solution

1. The inequality \(|X-20|\le 3\) is equivalent to \(-3\le X-20\le 3\), so \(17\le X\le 23\). The possible values are \(17,18,19,20,21,22,23\). 2. Using the binomial cumulative distribution function, \(P(17\le X\le 23)=P(X\le 23)-P(X\le 16)\approx 0.8181-0.1841=0.6340\). 3. By the complement rule, \(P(|X-20|>3)=1-P(|X-20|\le 3)\approx 1-0.6340=0.3660\). 4. The inequality \(|X-18|<2\) is equivalent to \(16<X<20\), so \(X\) can be \(17,18,19\). Thus \(P(17\le X\le 19)=P(X\le 19)-P(X\le 16)\approx 0.4572-0.1841=0.2731\).

Answer

a) \(X\in\{17,18,19,20,21,22,23\}\) b) (1) \(P(|X-20|\le 3)\approx 0.6340\) (2) \(P(|X-20|>3)\approx 0.3660\) (3) \(P(|X-18|<2)\approx 0.2731\)
52703612
An archer hits the bullseye with probability \(0.70\) on each shot, independently of the other shots. The archer takes \(8\) shots. Find the probability that the archer: a) hits the bullseye exactly \(6\) times. b) hits the bullseye at least \(7\) times. c) hits the bullseye at most \(5\) times.

Hints

- Identify the probability of a bullseye on one shot. - How many shots are taken? - “At least \(7\)” includes the cases \(7\) and \(8\). - For “at most \(5\),” consider subtracting the excluded upper-tail cases from \(1\).

Solution

1. Let \(X\) be the number of bullseyes. Then \(X\) has a binomial distribution with \(n=8\) and \(p=0.70\). 2. \(P(X=6)=\binom{8}{6}(0.70)^6(0.30)^2\approx 0.2965\). 3. \(P(X\ge 7)=P(X=7)+P(X=8)\approx 0.1977+0.0576=0.2553\). 4. Use a complement for part c): \(P(X\le 5)=1-P(X\ge 6)\). 5. Since \(P(X\ge 6)\approx 0.2965+0.1977+0.0576=0.5518\), \(P(X\le 5)\approx 0.4482\).

Answer

a) The probability is approximately \(29.65\%\). b) The probability is approximately \(25.53\%\). c) The probability is approximately \(44.82\%\).
52704012
For each expression, describe the event it represents and find its probability. a) For \(X\sim\operatorname{Bin}(8,0.25)\), evaluate \(\sum_{i=3}^{5}P(X=i)\). b) For \(Y\sim\operatorname{Bin}(12,0.5)\), evaluate \(\sum_{i=4}^{6}P(Y=i)\).

Hints

- The lower and upper limits of a sum identify the included success counts. - Count how many point probabilities must be added. - An interval probability can also be found from two cumulative probabilities.

Solution

1. In part a), the sum represents the probability of at least \(3\) and at most \(5\) successes. Thus \(\sum_{i=3}^{5}\binom{8}{i}(0.25)^i(0.75)^{8-i}\approx 0.3172\). 2. In part b), the sum represents the probability of at least \(4\) and at most \(6\) successes. Thus \(\sum_{i=4}^{6}\binom{12}{i}(0.5)^i(0.5)^{12-i}=\frac{2211}{4096}\approx 0.5398\).

Answer

a) The probability of \(3\), \(4\), or \(5\) successes; approximately \(0.3172\) b) The probability of \(4\), \(5\), or \(6\) successes; \(\frac{2211}{4096}\approx 0.5398\)
52704712
A spinner has \(8\) equal sections labeled \(1\) through \(8\). The spinner is spun \(10\) times, and the spins are independent. Find the probability of each event. 1. A prime number appears exactly \(4\) times. 2. A prime number appears at most \(3\) times. 3. A number greater than \(6\) appears at least \(8\) times. 4. A prime number appears on only the first \(3\) spins, and no prime number appears on any later spin.

Hints

- Identify the prime numbers from \(1\) through \(8\). - Distinguish between events where order does not matter and events with one required sequence. - For “at most” or “at least,” list the included values before calculating. - For one fixed sequence, multiply the probabilities in the required order without a binomial coefficient.

Solution

1. The prime numbers from \(1\) through \(8\) are \(2\), \(3\), \(5\), and \(7\), so the probability of a prime is \(p=\frac{4}{8}=0.5\). 2. For part 1, \(P(X=4)=\binom{10}{4}(0.5)^{10}\approx 0.2051\). 3. For part 2, \(P(X\le 3)=\sum_{k=0}^{3}\binom{10}{k}(0.5)^{10}=0.171875\approx 0.1719\). 4. For part 3, the numbers greater than \(6\) are \(7\) and \(8\), so \(p=\frac{2}{8}=0.25\). 5. \(P(X\ge 8)=\sum_{k=8}^{10}\binom{10}{k}(0.25)^k(0.75)^{10-k}\approx 0.000416\). 6. In part 4, a specific sequence is required: \(3\) prime results followed by \(7\) nonprime results. Its probability is \((0.5)^3(0.5)^7=(0.5)^{10}\approx 0.000977\).

Answer

1. \(P\approx 0.2051\), or \(20.51\%\) 2. \(P\approx 0.1719\), or \(17.19\%\) 3. \(P\approx 0.0004\), or approximately \(0.04\%\) 4. \(P\approx 0.0010\), or approximately \(0.10\%\)
52704812
A marble is drawn repeatedly from a bag with replacement. The probability of drawing a red marble is \(p\). Interpret each expression by stating the number of draws \(n\), the red-marble probability \(p\), and one event \(E\) represented by the expression. 1. \(\binom{15}{4}(0.2)^4(0.8)^{11}\) 2. \(\sum_{k=0}^{2}\binom{12}{k}(0.4)^k(0.6)^{12-k}\) 3. \((0.3)^3(0.7)^7\) 4. \(1-\sum_{k=0}^{13}\binom{20}{k}(0.9)^k(0.1)^{20-k}\)

Hints

- Use the exponents and any binomial coefficient to identify the number of draws. - What does the absence of a binomial coefficient imply about order? - Interpret a summation as a cumulative binomial probability. - For an expression beginning with \(1-\), identify the complementary event.

Solution

1. The first expression has \(n=15\) and \(p=0.2\). It represents drawing exactly \(4\) red marbles. 2. The second expression has \(n=12\) and \(p=0.4\). It represents drawing at most \(2\) red marbles. 3. The exponents in the third expression total \(10\), so \(n=10\) and \(p=0.3\). Because there is no binomial coefficient, the expression represents one specified sequence with \(3\) red draws and \(7\) non-red draws, such as red on the first \(3\) draws and non-red on the last \(7\). 4. The fourth expression has \(n=20\) and \(p=0.9\). The sum is \(P(X\le 13)\), so the full expression is \(P(X\ge 14)\), representing at least \(14\) red draws.

Answer

1. \(n=15\), \(p=0.2\); \(E\): exactly \(4\) red marbles are drawn. 2. \(n=12\), \(p=0.4\); \(E\): at most \(2\) red marbles are drawn. 3. \(n=10\), \(p=0.3\); \(E\): one specified sequence of \(3\) red draws and \(7\) non-red draws. 4. \(n=20\), \(p=0.9\); \(E\): at least \(14\) red marbles are drawn.
52705012
A multiple-choice test has \(20\) questions. Each question has five answer choices, exactly one of which is correct. A student guesses independently and randomly on every question. Determine the most likely number \(k\) of correct answers and the least likely number \(k\) of correct answers. Justify the least likely value by comparing the probabilities at the endpoints.

Hints

- Find the probability of one correct answer. - Relate the mode of a binomial distribution to \((n+1)p\). - Check the two endpoint probabilities, \(P(X=0)\) and \(P(X=n)\). - Compare all incorrect answers with all correct answers.

Solution

1. The number of correct answers is binomial with \(n=20\) and \(p=\frac15=0.2\). 2. Since \((n+1)p=21(0.2)=4.2\) is not an integer, the unique mode is \(\lfloor4.2\rfloor=4\). 3. The smallest binomial probability must occur at an endpoint. Compare \(P(X=0)=(0.8)^{20}\approx0.0115\) and \(P(X=20)=(0.2)^{20}\approx1.05\times10^{-14}\). 4. Because \((0.2)^{20}<(0.8)^{20}\), \(k=20\) is least likely.

Answer

The most likely number is \(k=4\). The least likely number is \(k=20\).
52705312
A gardener plants \(20\) seeds of a certain flower variety. Each seed germinates independently with probability \(0.85\). Let \(X\) be the number of seeds that germinate. Find the probability of each event. a) Exactly \(18\) seeds germinate. b) At most \(16\) seeds germinate. c) At least \(15\) seeds germinate. d) More than \(16\) but fewer than \(20\) seeds germinate.

Hints

- Identify the probability distribution for independent trials with two outcomes. - Translate “at most,” “at least,” and the open interval in part d) into values of \(X\). - Use a complement for “at least \(15\).” - List the included integer values carefully for part d).

Solution

1. The random variable \(X\) has a binomial distribution with \(n=20\) and \(p=0.85\). 2. \(P(X=18)=\binom{20}{18}(0.85)^{18}(0.15)^2\approx 0.2293\). 3. A cumulative binomial calculation gives \(P(X\le 16)\approx 0.3523\). 4. \(P(X\ge 15)=1-P(X\le 14)\approx 0.9327\). 5. In part d), the included values are \(17\), \(18\), and \(19\). Thus, \(P(17\le X\le 19)\approx 0.6090\).

Answer

a) \(P(X=18)\approx 22.93\%\) b) \(P(X\le 16)\approx 35.23\%\) c) \(P(X\ge 15)\approx 93.27\%\) d) \(P(17\le X\le 19)\approx 60.90\%\)
52705912
Let \(X\) be a binomial random variable with \(n=30\) and \(p=0.6\). Use a calculator or spreadsheet to determine: a) every value \(k\in\{0, 1, \ldots, 30\}\) for which \(P(X=k)\ge0.1\); b) every value \(k\in\{0, 1, \ldots, 30\}\) for which \(P(X=k)<0.05\).

Hints

- Expect the largest probabilities near the mean. - Use a table feature to view many individual probabilities efficiently. - Distinguish carefully between \(\ge\) and \(<\). - Check the probabilities immediately inside and outside each boundary.

Solution

1. For each \(k\), calculate \(P(X=k)=\binom{30}{k}(0.6)^k(0.4)^{30-k}\). The probabilities are largest near the mean \(np=18\). 2. Around the first cutoff, \(P(X=15)\approx0.0783\), \(P(X=16)\approx0.1101\), \(P(X=20)\approx0.1152\), and \(P(X=21)\approx0.0823\). Therefore, \(P(X=k)\ge0.1\) for \(k\in\{16, 17, 18, 19, 20\}\). 3. Around the second cutoff, \(P(X=14)\approx0.0489\), \(P(X=15)\approx0.0783\), \(P(X=22)\approx0.0505\), and \(P(X=23)\approx0.0263\). 4. Therefore, \(P(X=k)<0.05\) for \(k\in\{0, 1, \ldots, 14\}\cup\{23, 24, \ldots, 30\}\).

Answer

a) \(k\in\{16, 17, 18, 19, 20\}\) b) \(k\in\{0, 1, \ldots, 14\}\cup\{23, 24, \ldots, 30\}\)
52706012
A binomial random variable \(X\) has parameters \(n=100\) and \(p=0.25\). Use appropriate technology to find: a) the smallest nonnegative integer \(k\) such that \(P(X\le k)\ge 0.95\); b) the smallest nonnegative integer \(k\) such that \(P(X>k)\le 0.01\).

Hints

- Use a binomial cumulative distribution function to evaluate probabilities of the form \(P(X\le k)\). - Rewrite the condition in part b) using a complement. - The requested value is the first integer at which the inequality becomes true. - Check both the proposed value and the preceding integer.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(100,0.25)\). 2. For part a), \(P(X\le 31)\approx 0.9307<0.95\), while \(P(X\le 32)\approx 0.9554\ge 0.95\). Therefore, the smallest value is \(k=32\). 3. For part b), rewrite the condition as \(P(X\le k)\ge 0.99\). 4. Since \(P(X\le 34)\approx 0.9836<0.99\) and \(P(X\le 35)\approx 0.9906\ge 0.99\), the smallest value is \(k=35\).

Answer

a) \(k=32\) b) \(k=35\)
52706612
Consider a binomial setting with \(n\) trials and success probability \(p=0.5\), such as repeated flips of a fair coin. Let \(X\) be the number of successes. a) Use \(P(X=k)=\binom nkp^k(1-p)^{n-k}\) to prove that \(P(X=k)=P(X=n-k)\) for every \(k\in\{0, 1, \ldots, n\}\). b) For \(n=6\), calculate \(P(X\ge4)\), and use the symmetry from part a) to show that it equals \(P(X\le2)\).

Hints

- Simplify the two probability factors when \(p=1-p=0.5\). - Recall the symmetry of binomial coefficients. - Pair the upper-tail probabilities with corresponding lower-tail probabilities.

Solution

1. When \(p=0.5\), \(P(X=k)=\binom nk(0.5)^k(0.5)^{n-k}=\binom nk(0.5)^n\). 2. Similarly, \(P(X=n-k)=\binom n{n-k}(0.5)^{n-k}(0.5)^k=\binom n{n-k}(0.5)^n\). 3. Since \(\binom nk=\binom n{n-k}\), it follows that \(P(X=k)=P(X=n-k)\). 4. For \(n=6\), \(P(X\ge4)=\left(\binom64+\binom65+\binom66\right)(0.5)^6=\frac{22}{64}=0.34375\). 5. The symmetry gives \(P(X=4)=P(X=2)\), \(P(X=5)=P(X=1)\), and \(P(X=6)=P(X=0)\). Therefore, \(P(X\ge4)=P(X\le2)\).

Answer

a) Because \(\binom nk=\binom n{n-k}\) and both probabilities contain the factor \((0.5)^n\), \(P(X=k)=P(X=n-k)\). b) \(P(X\ge4)=0.34375\), and symmetry shows that this equals \(P(X\le2)\).
52706912
A manufacturer produces LED light strips, each containing \(20\) individual LEDs. Each LED is defective with probability \(0.02\), independently of the others. Assume the light strips are also independent. a) Find the probability that a randomly selected light strip has no defective LEDs. b) A retailer receives \(15\) such light strips. Find the probability that exactly \(12\) strips are completely functional.

Hints

- First find the probability that one LED is functional. - Use the probability of a completely functional strip as the success probability in part b). - Identify the number of trials and success probability in each part.

Solution

1. For one strip, the probability that an LED is functional is \(0.98\). Therefore, the probability that all \(20\) LEDs are functional is \(p_S=(0.98)^{20}\approx0.667608\). 2. Let \(Y\) count the completely functional strips among the \(15\). Then \(Y\) is binomial with \(n=15\) and success probability \(p_S=(0.98)^{20}\). 3. \(P(Y=12)=\binom{15}{12}p_S^{12}(1-p_S)^3\approx0.1310\).

Answer

a) \((0.98)^{20}\approx0.6676\), or about \(66.76\%\) b) \(P(Y=12)\approx0.1310\), or about \(13.10\%\)
52707112
A glass-bottle manufacturer reports that each bottle independently has a \(4\%\) chance of having small inclusions and being classified as a second. A sample of \(200\) bottles is selected. Let \(X\) be the number of bottles with inclusions. Find the probability of each event. a) Exactly \(8\) bottles have inclusions. b) At most \(10\) bottles have inclusions. c) From \(5\) through \(12\) bottles, inclusive, have inclusions.

Hints

- Identify the binomial model and its parameters. - Distinguish between one exact value and a range of values. - Express an inclusive interval as the difference of two cumulative probabilities. - Translate words such as “at most” and “from ... through ...” carefully.

Solution

1. The random variable \(X\) has a binomial distribution with \(n=200\) and \(p=0.04\). 2. \(P(X=8)=\binom{200}{8}(0.04)^8(0.96)^{192}\approx 0.1425\). 3. A cumulative binomial calculation gives \(P(X\le 10)\approx 0.8200\). 4. \(P(5\le X\le 12)=P(X\le 12)-P(X\le 4)\approx 0.9401-0.0950=0.8451\).

Answer

a) \(P(X=8)\approx 14.25\%\) b) \(P(X\le 10)\approx 82.00\%\) c) \(P(5\le X\le 12)\approx 84.51\%\)
52707212
A multiple-choice test has \(30\) questions. Each question has \(4\) answer choices, exactly one of which is correct. A student answers every question by random guessing. Let \(X\) be the number of correct answers. Find each probability. a) More than \(10\) correct answers b) At least \(5\) but fewer than \(10\) correct answers

Hints

- What is the probability of a correct answer when guessing among four choices? - Translate “more than” into an inequality. - Use a complement when it reduces the number of terms. - Check whether each endpoint is included.

Solution

1. Since the probability of a correct guess is \(\frac{1}{4}=0.25\), \(X\sim\operatorname{Bin}(30,0.25)\). 2. For part a), use the complement: \(P(X>10)=1-P(X\le 10)\approx 1-0.8943=0.1057\). 3. For part b), the condition is \(5\le X\le 9\). Therefore, \(P(5\le X\le 9)=P(X\le 9)-P(X\le 4)\approx 0.8034-0.0979=0.7055\).

Answer

a) \(P(X>10)\approx 0.1057\), or about \(10.57\%\) b) \(P(5\le X<10)\approx 0.7055\), or about \(70.55\%\)
52707312
A seed company reports that a certain variety has a germination probability of \(0.85\). A gardener plants \(30\) seeds. For parts 1 and 2, assume the germination outcomes are independent and each seed has the same germination probability. 1. Find the probability that at most \(22\) seeds germinate. 2. Find the probability that at least \(28\) seeds germinate. 3. Explain the conditions under which a binomial model is appropriate in this situation.

Hints

- Which distribution applies to a fixed number of trials with two outcomes? - Relate “at most” and “at least” to cumulative probabilities. - What circumstances could cause different seeds to have different germination probabilities? - What does independence mean in this context?

Solution

1. Let \(X\) be the number of seeds that germinate. Then \(X\) has a binomial distribution with \(n=30\) and \(p=0.85\), provided the binomial conditions hold. 2. \(P(X\le 22)=\sum_{k=0}^{22}\binom{30}{k}(0.85)^k(0.15)^{30-k}\approx 0.0698\). 3. \(P(X\ge 28)=1-P(X\le 27)\approx 0.1514\). 4. The binomial model requires a fixed number of seeds, two outcomes for each seed, the same germination probability for each seed, and independent germination outcomes.

Answer

1. The probability is approximately \(6.98\%\). 2. The probability is approximately \(15.14\%\). 3. The model requires a fixed number of independent trials, two outcomes per trial, and a constant germination probability of \(0.85\).
52707512
A manufacturer reports that \(4\%\) of its LED bulbs have minor scratches on the housing. A random sample of \(250\) bulbs is inspected, and the inspections are treated as independent. Let \(X\) be the number of scratched bulbs in the sample. Find each probability. (1) At most \(5\) scratched bulbs (2) More than \(12\) scratched bulbs (3) At least \(8\) but at most \(15\) scratched bulbs

Hints

- Identify the binomial parameters from the context. - Translate “at most,” “more than,” and “at least” into inequalities. - Express an interval probability as a difference of cumulative probabilities. - Keep cumulative values unrounded until the final subtraction.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(250,0.04)\). 2. For part (1), \(P(X\le 5)\approx 0.0633\). 3. For part (2), use the complement: \(P(X>12)=1-P(X\le 12)\approx 1-0.7955=0.2045\). 4. For part (3), \(P(8\le X\le 15)=P(X\le 15)-P(X\le 7)\approx 0.7400\), using unrounded cumulative probabilities before subtracting.

Answer

(1) \(P(X\le 5)\approx 0.0633\) (2) \(P(X>12)\approx 0.2045\) (3) \(P(8\le X\le 15)\approx 0.7400\)
52707612
A study reports that each resident in a city independently has a \(15\%\) chance of regularly using a particular car-sharing service. A sample of \(80\) residents is selected. Find the probability that among those surveyed: (1) fewer than \(10\) regularly use the service. (2) at least \(15\) regularly use the service. (3) exactly \(12\) regularly use the service.

Hints

- Distinguish “fewer than \(10\)” from “at most \(10\).” - What is the complement of “at least \(15\)” for an integer-valued random variable? - Decide whether each part needs a cumulative probability or one exact binomial probability.

Solution

1. Let \(X\) be the number of regular users in the sample. Then \(X\) has a binomial distribution with \(n=80\) and \(p=0.15\). 2. “Fewer than \(10\)” means \(X\le 9\), so \(P(X<10)=P(X\le 9)\approx 0.2211\). 3. \(P(X\ge 15)=1-P(X\le 14)\approx 0.2126\). 4. \(P(X=12)=\binom{80}{12}(0.15)^{12}(0.85)^{68}\approx 0.1240\).

Answer

(1) \(P(X<10)\approx 0.2211\) (2) \(P(X\ge 15)\approx 0.2126\) (3) \(P(X=12)\approx 0.1240\)
52707712
An LED manufacturer reports that \(2.5\%\) of its bulbs are defective. A home-improvement store receives a shipment of \(80\) bulbs. Assume the number of defective bulbs is binomially distributed. Find the probability of each event. Event \(A\): Exactly \(2\) bulbs are defective. Event \(B\): More than \(5\%\) of the bulbs in the shipment are defective.

Hints

- Identify the number of trials and the defect probability. - Use the binomial formula for an exact number of defective bulbs. - Convert \(5\%\) of the shipment into a number of bulbs before interpreting Event \(B\). - For “more than,” consider using a complement.

Solution

1. Let \(X\) be the number of defective bulbs. Then \(X\) has a binomial distribution with \(n=80\) and \(p=0.025\). 2. \(P(A)=P(X=2)=\binom{80}{2}(0.025)^2(0.975)^{78}\approx 0.2741\). 3. Five percent of \(80\) is \(4\). Therefore, Event \(B\) is \(X>4\). 4. \(P(B)=P(X>4)=1-P(X\le 4)\approx 1-0.9496=0.0504\).

Answer

Event \(A\): \(P(A)\approx 0.2741\), or \(27.41\%\) Event \(B\): \(P(B)\approx 0.0504\), or \(5.04\%\)
52707812
In a quality-control process for glass bottles, \(15\%\) of the bottles have small air bubbles. A random sample of \(60\) bottles is selected. Let \(X\) be the number of bottles with air bubbles, and assume \(X\) is binomially distributed. Find the probability of each event. Event \(C\): At most \(5\) bottles have air bubbles. Event \(D\): From \(8\) through \(12\) bottles, inclusive, have air bubbles.

Hints

- Translate “at most” into the included values of \(X\). - Express an inclusive interval using cumulative probabilities. - When subtracting cumulative probabilities, use the value immediately below the lower endpoint.

Solution

1. The binomial parameters are \(n=60\) and \(p=0.15\). 2. \(P(C)=P(X\le 5)\approx 0.0968\). 3. \(P(D)=P(8\le X\le 12)\). 4. Using cumulative probabilities, \(P(D)=P(X\le 12)-P(X\le 7)\approx 0.8938-0.3047=0.5891\).

Answer

Event \(C\): \(P(C)\approx 0.0968\), or \(9.68\%\) Event \(D\): \(P(D)\approx 0.5891\), or \(58.91\%\)
52707912
A gardener buys a packet of \(15\) flower seeds. The manufacturer reports a germination probability of \(0.90\), and the seeds are assumed to germinate independently. a) Find the probability that at least \(14\) of the \(15\) seeds germinate. b) Another gardener needs exactly \(10\) plants and buys \(11\) seeds, saying, “Buying \(10\%\) extra guarantees that I will get \(10\) plants.” Evaluate this statement using a probability calculation.

Hints

- Which values are included in “at least \(14\)”? - Use the binomial probability formula for each included value. - For part b), calculate the probability of at least \(10\) successes in \(11\) trials. - A guarantee requires probability \(1\).

Solution

1. For part a), let \(X\) be the number of seeds that germinate. Then \(X\) has a binomial distribution with \(n=15\) and \(p=0.90\). 2. \(P(X\ge 14)=P(X=14)+P(X=15)\). 3. \(P(X\ge 14)=\binom{15}{14}(0.90)^{14}(0.10)+(0.90)^{15}\approx 0.5490\). 4. For part b), let \(Y\) be the number of germinating seeds among \(11\). Then \(Y\) has a binomial distribution with \(n=11\) and \(p=0.90\). 5. \(P(Y\ge 10)=P(Y=10)+P(Y=11)\approx 0.6974\). 6. A probability of about \(69.74\%\) is not a guarantee, so the statement is false.

Answer

a) The probability is approximately \(54.90\%\). b) The statement is false. The probability of getting at least \(10\) plants from \(11\) seeds is approximately \(69.74\%\), not \(100\%\).
52708012
An LED manufacturer finds that \(5\%\) of its bulbs are defective, independently of one another. a) An inspector selects a sample of \(50\) bulbs. Find the probability that at most \(3\) bulbs are defective. b) An employee claims, “In a sample of \(50\), exactly \(5\) defective bulbs is more likely than no defective bulbs.” Determine whether the claim is correct.

Hints

- Define a success as selecting a defective bulb. - Add the required exact probabilities for part a). - For part b), calculate and compare the two point probabilities directly. - Consider how a small success probability shapes the distribution.

Solution

1. Let \(X\) be the number of defective bulbs. Then \(X\) has a binomial distribution with \(n=50\) and \(p=0.05\). 2. \(P(X\le 3)=\sum_{k=0}^{3}\binom{50}{k}(0.05)^k(0.95)^{50-k}\approx 0.7604\). 3. For part b), \(P(X=5)=\binom{50}{5}(0.05)^5(0.95)^{45}\approx 0.0658\). 4. \(P(X=0)=(0.95)^{50}\approx 0.0769\). 5. Since \(0.0658<0.0769\), exactly \(5\) defective bulbs is less likely than no defective bulbs. The claim is false.

Answer

a) The probability is approximately \(76.04\%\). b) The claim is false because \(P(X=5)\approx 6.58\%\) is less than \(P(X=0)\approx 7.69\%\).
52708212
A biathlete shoots at a series of \(5\) targets in practice. Her average hit probability per shot is \(0.80\). a) State the theoretical conditions needed to model the number of hits with a binomial distribution having \(n=5\) and \(p=0.80\). b) Explain why independence of the shots might be questionable in an actual competition. c) Suppose a miss lowers her probability of hitting the immediately following target to \(0.60\). Which condition of a binomial model is violated?

Hints

- What conditions define a binomial experiment? - Could wind or weather affect several shots at once? - Could a previous result affect the athlete's mental state? - What does independence require about the probability of the next trial?

Solution

1. Each shot must have two relevant outcomes, hit or miss. The hit probability must remain \(0.80\) for all \(5\) shots, and the shots must be independent. 2. In competition, wind or other environmental conditions may affect several shots, and psychological pressure or fatigue may change performance over time. These effects can create dependence or a changing hit probability. 3. If the next-shot probability depends on whether the previous shot was a miss, the trials are not independent and the success probability is not constant. Therefore, the binomial model is not valid.

Answer

a) Two relevant outcomes per shot, a constant success probability \(p=0.80\), and independent shots. b) Weather, fatigue, or psychological effects may influence multiple shots or change the hit probability. c) Independence and constant \(p\) are violated because the next-shot probability depends on the previous result.
52708512
A company manufactures electronic components, and \(15\%\) receive a special quality seal. A random sample of \(12\) components is selected without replacement from a very large production lot. Find the probability that at least \(4\) selected components have the quality seal. Also explain the condition under which a binomial distribution is a reasonable approximation for this sample.

Hints

- Which values are included in “at least \(4\)”? - Would a complement reduce the number of terms? - What conditions are needed for a binomial model? - Why does sampling a small fraction of a very large population make the draws approximately independent?

Solution

1. Let \(X\) be the number of selected components with a quality seal. Because the sample is small relative to the very large lot, approximate \(X\) with a binomial distribution having \(n=12\) and \(p=0.15\). 2. \(P(X\ge 4)=1-P(X\le 3)\). 3. \(P(X\le 3)=\sum_{k=0}^{3}\binom{12}{k}(0.15)^k(0.85)^{12-k}\approx 0.9078\). 4. Therefore, \(P(X\ge 4)\approx 1-0.9078=0.0922\). 5. The approximation is reasonable when the sample is a small fraction of the population, so removing one component changes the success probability only negligibly.

Answer

The probability is approximately \(9.22\%\). A binomial approximation is reasonable when the sample is small relative to the population, making the success probability nearly constant across selections.
52708812
A spinner has blue, red, and green sections. The blue section has a central angle of \(108^\circ\). The spinner is spun \(20\) times, the spins are independent, and landing on blue counts as a success. a) State the parameters \(n\) and \(p\) of the binomial distribution and explain what they represent. b) Find the probability that the spinner never lands on blue. c) Find the probability of at most \(1\) blue result.

Hints

- Find the fraction of the full \(360^\circ\) circle represented by the blue section. - Which values are included in “at most \(1\)”? - Use the binomial probability formula for each required exact value.

Solution

1. The probability of blue is the fraction of the full circle occupied by the blue section: \(p=\frac{108}{360}=0.3\). 2. The number of spins is \(n=20\). Thus, \(n\) is the number of trials and \(p\) is the probability of blue on each spin. 3. Let \(X\) be the number of blue results. Then \(P(X=0)=(0.7)^{20}\approx 0.000798\). 4. \(P(X\le 1)=P(X=0)+P(X=1)\). 5. \(P(X=1)=20(0.3)(0.7)^{19}\approx 0.006839\). 6. Therefore, \(P(X\le 1)\approx 0.007637\).

Answer

a) \(n=20\), the number of spins, and \(p=0.3\), the probability of blue on each spin. b) \(P(X=0)\approx 0.0008\), or approximately \(0.08\%\) c) \(P(X\le 1)\approx 0.0076\), or approximately \(0.76\%\)
52709112
A basketball player makes a free throw with probability \(0.75\), independently of the other shots. The player takes \(4\) free throws. a) Match each event with the corresponding probability expression, and explain your reasoning. Events: \(A\): Exactly \(2\) shots are made. \(B\): All \(4\) shots are made. \(C\): The first made shot occurs on the fourth attempt. \(D\): The first \(2\) shots are missed, and the last \(2\) are made. \(E\): At least \(3\) shots are made. Expressions: (1) \((0.75)^4\) (2) \((0.25)^3(0.75)\) (3) \((0.25)^2(0.75)^2\) (4) \(\binom{4}{2}(0.75)^2(0.25)^2\) (5) \(\sum_{k=3}^{4}\binom{4}{k}(0.75)^k(0.25)^{4-k}\) b) Find the probability of Event \(E\).

Hints

- Decide whether each event specifies one sequence or only a total number of makes. - The binomial coefficient counts possible arrangements. - “At least” usually includes more than one exact value. - A fixed sequence is found by multiplying the probabilities in order.

Solution

1. Event \(A\) matches (4) because the binomial coefficient counts all arrangements of \(2\) makes and \(2\) misses. 2. Event \(B\) matches (1) because all \(4\) shots must be made. 3. Event \(C\) matches (2) because the required sequence is miss, miss, miss, make. 4. Event \(D\) matches (3) because one specific sequence of \(2\) misses followed by \(2\) makes is required. 5. Event \(E\) matches (5) because at least \(3\) makes means \(3\) or \(4\) makes. 6. \(P(E)=\binom{4}{3}(0.75)^3(0.25)+(0.75)^4=0.73828125\approx 0.7383\).

Answer

a) \(A\to(4)\), \(B\to(1)\), \(C\to(2)\), \(D\to(3)\), \(E\to(5)\) b) \(P(E)\approx 0.7383\), or \(73.83\%\)
52709212
A factory produces electronic components, and \(4\%\) are defective. Ten components are selected in order, and the selections can be modeled as independent Bernoulli trials. a) Find the probability that the first defective component occurs in the third position. b) Find the probability that at most \(1\) component in the sample is defective. c) Explain mathematically why the events “only the first component is defective” and “only the last component is defective” have the same probability.

Hints

- A first success in a specified position requires failures before that position and a success at that position. - Which values are included in “at most \(1\)”? - In part c), compare the factors in the two fixed-sequence probabilities.

Solution

1. For the first defect to occur in position \(3\), the first \(2\) components must be nondefective and the third must be defective. The remaining \(7\) outcomes are unrestricted. 2. The probability is \((0.96)^2(0.04)=0.036864\). 3. For part b), let \(X\) be the number of defective components. Then \(X\) has a binomial distribution with \(n=10\) and \(p=0.04\). 4. \(P(X\le 1)=(0.96)^{10}+10(0.04)(0.96)^9\approx 0.94185\). 5. Each event in part c) has exactly \(1\) defective outcome and \(9\) nondefective outcomes in a specified order. 6. Both probabilities are \((0.04)(0.96)^9\). Changing the position does not change the product because multiplication is commutative.

Answer

a) \(P=0.036864\), or approximately \(3.69\%\) b) \(P\approx 0.94185\), or approximately \(94.18\%\) c) Both events have probability \((0.04)(0.96)^9\), so they are equally likely.
52709312
A seed company states that a certain variety of tomato seed has a germination rate of \(85\%\). A gardener plants \(30\) seeds, and the germination outcomes are independent. a) Find the most likely number of seeds that germinate. b) Find the probability that at least \(24\) seeds germinate.

Hints

- Model germination as a binomial random variable. - Relate the most likely value to \((n+1)p\). - Translate “at least” into a range of values. - Consider using the complement for the cumulative probability.

Solution

1. Let \(X\) be the number of seeds that germinate. Then \(X\) is binomial with \(n=30\) and \(p=0.85\). 2. Since \((n+1)p=31\cdot0.85=26.35\) is not an integer, the unique mode is \(\lfloor26.35\rfloor=26\). 3. For part b), \(P(X\ge24)=\sum_{k=24}^{30}\binom{30}{k}(0.85)^k(0.15)^{30-k}\). 4. Using the complement, \(P(X\ge24)=1-P(X\le23)\approx1-0.1526=0.8474\).

Answer

a) \(26\) seeds b) \(P(X\ge24)\approx0.8474\), or about \(84.74\%\)
52709512
A battery manufacturer reports that \(94\%\) of its batteries last more than \(1000\) charge cycles. A random sample of \(50\) batteries is tested. Let \(X\) be the number that meet this standard. a) Assuming a binomial model, find the probability of each event. (1) Exactly \(48\) batteries meet the standard. (2) At most \(45\) batteries meet the standard. (3) At least \(46\), but fewer than \(50\), batteries meet the standard. b) State two conditions needed for a binomial model to be appropriate in this context.

Hints

- Translate “fewer than \(50\)” into included integer values. - Use cumulative probabilities to calculate an interval. - Recall the conditions for independent Bernoulli trials. - The number of trials must be fixed and the success probability constant.

Solution

1. The binomial parameters are \(n=50\) and \(p=0.94\). 2. \(P(X=48)=\binom{50}{48}(0.94)^{48}(0.06)^2\approx 0.2262\). 3. A cumulative binomial calculation gives \(P(X\le 45)\approx 0.1794\). 4. \(P(46\le X<50)=P(46\le X\le 49)=P(X\le 49)-P(X\le 45)\). 5. Since \(P(X\le 49)=1-(0.94)^{50}\approx 0.9547\), the interval probability is approximately \(0.9547-0.1794=0.7753\). 6. A binomial model requires independent battery outcomes and the same success probability for every battery.

Answer

a) (1) \(P(X=48)\approx 22.62\%\) (2) \(P(X\le 45)\approx 17.94\%\) (3) \(P(46\le X\le 49)\approx 77.53\%\) b) The battery outcomes must be independent, and each battery must have the same probability \(0.94\) of meeting the standard.
52709712
At a tour company's information booth, \(20\%\) of visitors are interested in an international trip. During one hour, \(80\) visitors stop at the booth. Assume their interests are independent, and let \(X\) be the number interested in an international trip. a) What is the minimum number of international-travel brochures the company should have so that the brochures are sufficient for all interested visitors with probability at least \(95\%\)? b) Suppose the company has \(18\) brochures. Find the probability that at least one brochure remains at the end of the hour.

Hints

- Identify the binomial parameters from the situation. - Translate “enough with probability at least \(95\%\)” into a cumulative-probability condition. - Determine exactly when at least one of \(18\) brochures remains. - Use a binomial cumulative distribution function to evaluate and compare consecutive integer values.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(80,0.2)\). 2. For part a), seek the smallest integer \(k\) such that \(P(X\le k)\ge 0.95\). 3. Since \(P(X\le 21)\approx 0.9340<0.95\) and \(P(X\le 22)\approx 0.9612\ge 0.95\), the minimum number of brochures is \(22\). 4. With \(18\) brochures, at least one remains exactly when fewer than \(18\) visitors want one. Thus \(P(X\le 17)\approx 0.6708\).

Answer

a) \(22\) brochures b) \(P(X\le 17)\approx 0.6708\), or about \(67.08\%\)
52709812
A home improvement store runs a promotion in which each customer rolls a fair six-sided die at checkout. A customer who rolls a \(6\) receives an instant-discount coupon. The die is rolled \(150\) times during the morning. a) What is the minimum number of coupons the store should prepare so that there are enough for all winners with probability at least \(90\%\)? b) The manager prepares \(30\) coupons. Find the probability that more customers win than there are coupons available.

Hints

- Determine the probability of rolling the winning number on one fair die roll. - For part a), translate the required confidence level into a cumulative-probability condition. - For part b), use a complement to find a probability of the form \(P(X>k)\).

Solution

1. Let \(X\) be the number of winners. Since a win occurs with probability \(p=\frac{1}{6}\), \(X\sim\operatorname{Bin}\left(150,\frac{1}{6}\right)\). 2. For part a), seek the smallest integer \(k\) such that \(P(X\le k)\ge 0.90\). 3. Since \(P(X\le 30)\approx 0.8842<0.90\) and \(P(X\le 31)\approx 0.9197\ge 0.90\), the store should prepare at least \(31\) coupons. 4. With \(30\) coupons, a shortage occurs when \(X>30\). Thus \(P(X>30)=1-P(X\le 30)\approx 1-0.8842=0.1158\).

Answer

a) \(31\) coupons b) \(P(X>30)\approx 0.1158\), or about \(11.58\%\)
52710112
A small movie theater has \(45\) seats for a documentary premiere. Based on past experience, each person with a reservation independently has a \(10\%\) chance of not attending. The theater accepts \(50\) reservations. Find the probability that every person who attends will have a seat.

Hints

- What attendance count allows everyone to have a seat? - What is the probability that one reservation holder attends? - Would counting no-shows make the event easier to express? - Use a cumulative binomial probability for the required range.

Solution

1. Let \(X\) be the number of people who attend. Assuming independent attendance decisions, \(X\) has a binomial distribution with \(n=50\) and \(p=0.90\). 2. Everyone has a seat when \(X\le 45\). 3. Equivalently, let \(Y\) be the number of no-shows. Then \(Y\) has a binomial distribution with \(n=50\) and \(p=0.10\), and everyone has a seat when \(Y\ge 5\). 4. \(P(Y\ge 5)=1-P(Y\le 4)\). 5. A cumulative binomial calculation gives \(P(Y\le 4)\approx 0.4312\). 6. Therefore, \(P(Y\ge 5)\approx 1-0.4312=0.5688\).

Answer

The probability that every attendee has a seat is approximately \(56.88\%\).
52710312
A multiple-choice test has \(5\) answer choices for each question, exactly one of which is correct. A test taker guesses independently and randomly on every question. Find the minimum number of questions needed so that the probability of answering at least \(1\) question correctly is: a) at least \(80\%\). b) at least \(95\%\).

Hints

- What is the probability of guessing one question correctly? - Express the probability of answering no questions correctly. - Use the complement for at least \(1\) correct answer. - Use logarithms to solve for an unknown exponent. - The number of questions must be a whole number.

Solution

1. The probability of guessing one question correctly is \(p=\frac{1}{5}=0.2\), so the probability of an incorrect answer is \(0.8\). 2. For \(n\) questions, \(P(X\ge 1)=1-(0.8)^n\). 3. For part a), require \(1-(0.8)^n\ge 0.80\), so \((0.8)^n\le 0.20\). 4. Thus, \(n\ge \frac{\ln(0.20)}{\ln(0.8)}\approx 7.213\). The minimum is \(8\) questions. 5. For part b), require \((0.8)^n\le 0.05\). 6. Thus, \(n\ge \frac{\ln(0.05)}{\ln(0.8)}\approx 13.425\). The minimum is \(14\) questions.

Answer

a) \(8\) questions b) \(14\) questions
52710512
An egg producer reports that at most \(2\%\) of the eggs in a shipment have cracked shells. A grocery store inspects a carton of \(20\) eggs and rejects the shipment if more than \(1\) egg in the sample is cracked. a) Assuming the true cracked-shell rate is exactly \(2\%\), find the probability that the shipment is rejected. Assume a binomial model is appropriate. b) Find the rejection probability if the true cracked-shell rate is \(1\%\).

Hints

- Translate “more than \(1\)” into values of the random variable. - Use the complement of the rejection event. - Substitute the appropriate value of \(p\) in each part. - A binomial model requires the success probability to remain approximately constant.

Solution

1. Let \(X\) be the number of cracked eggs in the sample. The shipment is rejected when \(X>1\). 2. For part a), \(X\) has a binomial distribution with \(n=20\) and \(p=0.02\). 3. \(P(X>1)=1-P(X\le 1)=1-[(0.98)^{20}+20(0.02)(0.98)^{19}]\approx 0.0599\). 4. For part b), use \(n=20\) and \(p=0.01\). 5. \(P(X>1)=1-[(0.99)^{20}+20(0.01)(0.99)^{19}]\approx 0.0169\).

Answer

a) The probability is approximately \(5.99\%\). b) The probability is approximately \(1.69\%\).
52710712
At a garden center, a certain type of seed germinates independently with probability \(0.90\). For each expression, describe a suitable random experiment and an event in context whose probability is represented by the expression. a) \(\binom{20}{18}(0.9)^{18}(0.1)^2\) b) \(1-[(0.1)^{10}+10(0.9)(0.1)^9]\) c) \(1-\sum_{k=0}^{16}\binom{20}{k}(0.9)^k(0.1)^{20-k}\)

Hints

- Interpret \(0.9\) as the success probability and use the exponents to identify counts. - Use the binomial coefficient or total exponents to identify \(n\). - An expression beginning with \(1-\) represents a complementary event. - Identify which exact probabilities are included in each sum.

Solution

1. In every expression, a success means that a seed germinates, with \(p=0.90\). 2. Part a) has \(n=20\) and \(k=18\). It represents planting \(20\) seeds and exactly \(18\) germinating. 3. In part b), the terms in brackets are \(P(X=0)+P(X=1)\) for \(n=10\). The complement is \(P(X\ge 2)\). It represents planting \(10\) seeds and at least \(2\) germinating. 4. In part c), the sum is \(P(X\le 16)\) for \(n=20\). The complement is \(P(X\ge 17)\). It represents planting \(20\) seeds and at least \(17\) germinating.

Answer

a) Plant \(20\) seeds; exactly \(18\) germinate. b) Plant \(10\) seeds; at least \(2\) germinate. c) Plant \(20\) seeds; at least \(17\) germinate.
52711112
An electronics manufacturer finds that each microchip independently has a \(5\%\) chance of being defective. The chips are shipped in packages of \(60\). a) Find the probability that more than \(10\%\) of the microchips in a randomly selected package are defective. b) During quality control, the chips in a package are tested one at a time. Describe an event in context whose probability is represented by \(\sum_{k=0}^{4}(0.95)^k(0.05)\). c) What minimum proportion of the manufacturer's microchips must be nondefective so that a sample of \(30\) chips contains no defective chips with probability at least \(80\%\)?

Hints

- In part a), determine the first whole-number count that is greater than \(10\%\) of \(60\). - In part b), interpret a term of the form \(q^kp\). - Express the probability that all \(30\) chips are nondefective in terms of the nondefective proportion. - Use a root or logarithms to solve the inequality in part c).

Solution

1. For part a), let \(X\sim\operatorname{Bin}(60,0.05)\). More than \(10\%\) of \(60\) means at least \(7\) defective chips. 2. Therefore, \(P(X\ge 7)=1-P(X\le 6)\approx 0.029694\). 3. In part b), the term \((0.95)^k(0.05)\) is the probability that the first defective chip occurs on test \(k+1\). Summing from \(k=0\) through \(4\) gives the probability that the first defective chip is found within the first \(5\) tests. 4. For part c), let \(q\) be the proportion of nondefective chips. The requirement is \(q^{30}\ge 0.80\). 5. Thus, \(q\ge (0.80)^{1/30}\approx 0.992589\).

Answer

a) The probability is approximately \(2.97\%\). b) The first defective chip is found during one of the first \(5\) tests. c) The proportion of nondefective chips must be at least approximately \(99.26\%\).
52711212
In an online game, each treasure chest independently has a \(12\%\) probability of containing a rare item. A player opens \(40\) chests. a) Find the probability that the player receives at most \(3\) rare items. b) Describe an event in context whose probability is \(1-(0.88)^{15}\). c) What is the minimum number of treasure chests a player must open so that the probability of receiving at least \(1\) rare item is at least \(99\%\)?

Hints

- In part a), decide whether you need one binomial probability or a cumulative probability. - What does \(0.88\) represent in this context? - In part c), use the complement event of receiving no rare items. - “At least one” is often easier to calculate using its complement.

Solution

1. For part a), let \(X\sim\operatorname{Bin}(40,0.12)\). 2. Then \(P(X\le 3)=\sum_{k=0}^{3}\binom{40}{k}(0.12)^k(0.88)^{40-k}\approx 0.276803\). 3. For part b), \((0.88)^{15}\) is the probability of receiving no rare items in \(15\) chests. Its complement is the probability of receiving at least \(1\) rare item. 4. For part c), require \(1-(0.88)^n\ge 0.99\), so \((0.88)^n\le 0.01\). 5. Solving gives \(n\ge \frac{\ln(0.01)}{\ln(0.88)}\approx 36.03\). Therefore, the minimum is \(37\) chests.

Answer

a) The probability is approximately \(27.68\%\). b) At least \(1\) rare item is found in \(15\) chests. c) The player must open at least \(37\) chests.
52711312
A manufacturer packages glass figurines in sets of \(5\). Historical data show that \(40\%\) of the figurines have minor flaws. Assume that flaws occur independently and that the sets are assembled independently. A retailer receives \(20\) sets. Find the probability that at most \(2\) of the sets contain only flawless figurines.

Hints

- First find the probability that one figurine is flawless. - What assumption allows you to multiply the probabilities for \(5\) flawless figurines? - Treat the \(20\) sets as a new sequence of Bernoulli trials. What counts as a success? - Which values are included in “at most \(2\)”?

Solution

1. The probability that one figurine is flawless is \(1-0.40=0.60\). 2. A set contains only flawless figurines with probability \(p_S=(0.60)^5=0.07776\). 3. Let \(X\) be the number of all-flawless sets among the \(20\) sets. Then \(X\sim\operatorname{Bin}(20,0.07776)\). 4. Therefore, \(P(X\le 2)=\sum_{k=0}^{2}\binom{20}{k}(0.07776)^k(0.92224)^{20-k}\approx 0.799737\).

Answer

The probability is approximately \(79.97\%\).
52711812
A gardener knows that \(80\%\) of a certain type of flower seed germinates, and seed outcomes are treated as independent. a) The gardener plants \(450\) seeds in a garden bed. How many seedlings should be expected? b) For a small pot, the gardener plants \(20\) seeds. Find the probability that exactly \(18\) germinate. c) Find the probability that at least \(19\) of the \(20\) seeds germinate.

Hints

- Identify the probability of germination. - Use \(np\) for the mean of a binomial random variable. - “At least \(19\)” includes which exact outcomes? - Use the binomial probability formula for each exact outcome.

Solution

1. The germination probability is \(p=0.8\). 2. For \(450\) seeds, the expected number is \(np=450\cdot 0.8=360\). 3. For \(20\) seeds, let \(X\sim\operatorname{Bin}(20,0.8)\). Then \(P(X=18)=\binom{20}{18}(0.8)^{18}(0.2)^2\approx 0.1369\). 4. At least \(19\) means \(19\) or \(20\): \(P(X\ge 19)=\binom{20}{19}(0.8)^{19}(0.2)+(0.8)^{20}\approx 0.0692\).

Answer

a) \(360\) seedlings b) \(P(X=18)\approx 0.1369\), or about \(13.69\%\) c) \(P(X\ge 19)\approx 0.0692\), or about \(6.92\%\)
52712912
A city survey estimates that \(35\%\) of eligible voters support a local ballot proposal. A random sample of \(150\) eligible voters is surveyed. The sample is small relative to the city’s voter population, so responses are modeled as approximately independent. Let \(X\) be the number who support the proposal. a) Find the probability that at least \(60\) people in the sample support the proposal. b) Find the probability that between \(45\) and \(60\) people, inclusive, support the proposal.

Hints

- Identify the binomial parameters from the survey. - Express “at least” using a complement and a cumulative probability. - Find an inclusive interval probability by subtracting two cumulative probabilities. - Check which endpoint belongs in each cumulative probability.

Solution

1. Model the count with \(X\sim\operatorname{Bin}(150,0.35)\). 2. For part a), use the complement: \(P(X\ge 60)=1-P(X\le 59)\approx 1-0.8840=0.1160\). 3. For part b), \(P(45\le X\le 60)=P(X\le 60)-P(X\le 44)\approx 0.9136-0.0841=0.8295\).

Answer

a) \(P(X\ge 60)\approx 0.1160\), or about \(11.60\%\) b) \(P(45\le X\le 60)\approx 0.8295\), or about \(82.95\%\)
52713612
At a greenhouse, seeds of a certain flower variety germinate independently with probability \(0.92\). A random sample of \(120\) seeds is planted under identical conditions. Let \(X\) be the number of seeds that germinate. a) Find \(P(X=110)\). b) Find the number of germinating seeds for which \(P(X=k)\) is greatest.

Hints

- Identify \(n\) and \(p\). - Use the binomial formula for one exact value. - The mode is close to the mean \(np\). - Use \((n+1)p\) to identify the most likely integer value.

Solution

1. The random variable \(X\) is binomial with \(n=120\) and \(p=0.92\). 2. \(P(X=110)=\binom{120}{110}(0.92)^{110}(0.08)^{10}\approx0.1295\). 3. Since \((n+1)p=121\cdot0.92=111.32\) is not an integer, the unique mode is \(\lfloor111.32\rfloor=111\).

Answer

a) \(P(X=110)\approx0.1295\), or about \(12.95\%\) b) \(111\) seeds
52713712
A diagnostic screening test produces a false-positive result for \(2\%\) of healthy people. In a study, \(500\) healthy people are tested independently. Let \(X\) be the number of false-positive results. a) Find the mean \(\mu\) and standard deviation \(\sigma\) of \(X\). Explain the meaning of the mean in this context. b) Find the probability that \(X\) is within one standard deviation of its mean.

Hints

- Interpret the binomial mean as a long-run average count. - Use the binomial standard-deviation formula. - Determine which integer counts lie within one standard deviation of the mean. - Use cumulative probabilities to evaluate the interval.

Solution

1. Since \(X\sim\operatorname{Bin}(500,0.02)\), the mean is \(\mu=np=500\cdot 0.02=10\). Over many groups of \(500\) healthy people, the average number of false-positive results would be about \(10\) per group. 2. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{500\cdot 0.02\cdot 0.98}=\sqrt{9.8}\approx 3.13\). 3. The interval \([\mu-\sigma,\mu+\sigma]\) is approximately \([6.87,13.13]\). The integer values in this interval are \(7\) through \(13\). 4. Therefore, \(P(7\le X\le 13)=P(X\le 13)-P(X\le 6)\approx 0.7391\).

Answer

a) \(\mu=10\) and \(\sigma\approx 3.13\). The mean indicates an average of about \(10\) false-positive results per group of \(500\) healthy people over many such studies. b) \(P(7\le X\le 13)\approx 0.7391\), or about \(73.91\%\)
52713812
A manufacturer knows that \(10\%\) of its glass marbles contain tiny air bubbles and are classified as seconds. A package contains \(300\) randomly selected marbles, and outcomes are treated as independent. Let \(Y\) be the number of marbles with air bubbles. a) Find the mean \(\mu\) and standard deviation \(\sigma\) of \(Y\). b) Find the probability that \(Y\) differs from its mean by at most two standard deviations.

Hints

- Identify the binomial parameters. - Translate “differs by at most two standard deviations” into an interval. - Keep only the integer values that lie in the interval. - Use a difference of cumulative probabilities.

Solution

1. Since \(Y\sim\operatorname{Bin}(300,0.1)\), the mean is \(\mu=np=300\cdot 0.1=30\). 2. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{300\cdot 0.1\cdot 0.9}=\sqrt{27}\approx 5.20\). 3. The interval \([\mu-2\sigma,\mu+2\sigma]\) is approximately \([19.61,40.39]\). The integer values in this interval are \(20\) through \(40\). 4. Therefore, \(P(20\le Y\le 40)=P(Y\le 40)-P(Y\le 19)\approx 0.9575\).

Answer

a) \(\mu=30\) and \(\sigma\approx 5.20\) b) \(P(20\le Y\le 40)\approx 0.9575\), or about \(95.75\%\)
52714012
The random variable \(Y\) has a binomial distribution with \(n=25\) and \(p=0.5\). a) Without calculating individual probabilities, justify why the probability distribution of \(Y\) has two adjacent values with the same maximum probability. b) State the two values and explain how the symmetry of the distribution supports your answer.

Hints

- What happens to the shape of a binomial distribution when \(p=0.5\)? - Where is the center of symmetry? - Find the expected value and examine the adjacent integers.

Solution

1. For a binomial distribution, if \((n+1)p\) is an integer \(m\), the two modes are \(m-1\) and \(m\). 2. Here, \((25+1)\cdot0.5=13\), so the two modes are \(12\) and \(13\). 3. Because \(p=0.5\), the distribution is symmetric about \(np=12.5\). The values \(12\) and \(13\) are equally far from \(12.5\), so \(P(Y=12)=P(Y=13)\).

Answer

a) Since \((n+1)p=13\) is an integer, the binomial distribution has equal maximum probabilities at the adjacent values \(13-1\) and \(13\). b) The values are \(12\) and \(13\). Because \(p=0.5\), the distribution is symmetric about \(12.5\), so these equally spaced values have equal probabilities.
52714112
The random variable \(X\) has a binomial distribution with \(n=100\) and success probability \(p\). Find all values of \(p\) with \(0<p<1\) for which \(P(X=45)<P(X=55)\).

Hints

- Write the binomial formula for each probability. - Compare the two binomial coefficients. - Identify the positive factors common to both sides of the inequality. - Remember that both bases of the powers are positive.

Solution

1. Write the two binomial probabilities and compare them: \(\binom{100}{45}p^{45}(1-p)^{55}<\binom{100}{55}p^{55}(1-p)^{45}\). 2. Since \(\binom{100}{45}=\binom{100}{55}\) and all factors are positive for \(0<p<1\), divide by the common positive factor \(\binom{100}{45}p^{45}(1-p)^{45}\). This gives \((1-p)^{10}<p^{10}\). 3. Both bases are positive, and raising positive numbers to the tenth power preserves their order. Thus, \(1-p<p\). 4. Solving gives \(1<2p\), so \(p>0.5\). Combining this with \(p<1\) gives \(0.5<p<1\).

Answer

\(0.5<p<1\), or \(p\in(0.5, 1)\).
52714212
A random variable \(Y\) has a binomial distribution with \(n=3\) and \(p\in(0, 1)\). Find all values of \(p\) for which the probability of no successes is at least as great as the probability of exactly one success.

Hints

- Translate “at least as great as” into an inequality. - Substitute \(n=3\) and the relevant values of \(k\) into the binomial probability formula. - Simplify by dividing by a positive power of \(1-p\). - Intersect your solution with the given domain for \(p\).

Solution

1. Translate the condition as \(P(Y=0)\ge P(Y=1)\). 2. Substitute the binomial probabilities: \(\binom{3}{0}p^0(1-p)^3\ge\binom{3}{1}p(1-p)^2\). 3. Since \(p<1\), the factor \((1-p)^2\) is positive. Dividing by it gives \(1-p\ge3p\). 4. Solving the inequality gives \(1\ge4p\), so \(p\le0.25\). 5. Combining this result with \(p\in(0, 1)\) gives \(0<p\le0.25\).

Answer

\(0<p\le0.25\), or \(p\in(0, 0.25]\).
52714912
A binomial random variable \(X\) has parameters \(n=80\) and \(p=0.3\). Find the narrowest interval of the form \([\mu-k,\mu+k]\), symmetric about the mean, that contains at least \(95\%\) of the probability. Assume \(k\) is a nonnegative integer.

Hints

- Find the mean, which is the center of the interval. - Express an interval probability as a difference of cumulative probabilities. - Test integer distances from the mean systematically. - Use unrounded cumulative values when subtracting.

Solution

1. The mean is \(\mu=np=80\cdot 0.3=24\). 2. Seek the smallest nonnegative integer \(k\) such that \(P(24-k\le X\le 24+k)\ge 0.95\). 3. For \(k=7\), \(P(17\le X\le 31)\approx 0.9337<0.95\). 4. For \(k=8\), \(P(16\le X\le 32)\approx 0.9628\ge 0.95\). 5. Therefore, \(k=8\), and the narrowest qualifying symmetric interval is \([16,32]\).

Answer

\([16,32]\)
52715012
A binomial random variable \(Y\) has parameters \(n=150\) and \(p=0.2\). Find the narrowest interval of the form \([\mu-c,\mu+c]\), symmetric about the mean, that contains at least \(90\%\) of the probability. Assume \(c\) is a nonnegative integer.

Hints

- Find the mean, which is the center of the interval. - A symmetric interval has endpoints the same distance from the mean. - Use a binomial cumulative distribution function to evaluate each interval probability. - When subtracting cumulative probabilities, make sure the lower endpoint remains included.

Solution

1. The mean is \(\mu=np=150\cdot 0.2=30\). 2. Seek the smallest nonnegative integer \(c\) such that \(P(30-c\le Y\le 30+c)\ge 0.90\). 3. For \(c=7\), \(P(23\le Y\le 37)\approx 0.8751<0.90\). 4. For \(c=8\), \(P(22\le Y\le 38)\approx 0.9182\ge 0.90\). 5. Therefore, \(c=8\), and the narrowest qualifying symmetric interval is \([22,38]\).

Answer

\([22,38]\)
52715112
At a customer service center, each call independently has a \(15\%\) chance of not being answered within the first \(20\) seconds. The center introduces a new staffing tool and evaluates it by selecting \(120\) calls and counting the calls that were not answered within \(20\) seconds. The tool is considered successful if the number of delayed calls is more than one standard deviation below the original expected number. Find the probability that the tool is classified as unsuccessful even though the actual delayed-call rate has decreased to \(8\%\).

Hints

- Model the number of delayed calls with a binomial distribution. - First calculate the mean and standard deviation using the original rate. - Determine the exact integer cutoff for a successful result. - Use the new rate when calculating the final probability, but keep the cutoff based on the original rate. - An unsuccessful classification is the complement of the success condition.

Solution

1. Under the original rate, \(n=120\) and \(p_0=0.15\). The mean is \(\mu_0=120\cdot0.15=18\), and the standard deviation is \(\sigma_0=\sqrt{120\cdot0.15\cdot0.85}\approx 3.9115\). 2. Success requires \(X<\mu_0-\sigma_0=18-3.9115\approx 14.0885\). Because \(X\) is an integer, the tool is classified as successful when \(X\le 14\). 3. When the actual delayed-call rate is \(0.08\), let \(X\sim\operatorname{Bin}(120,0.08)\). 4. The tool is classified as unsuccessful when \(X\ge 15\). Thus, \(P(X\ge 15)=1-P(X\le 14)\approx 0.056350\).

Answer

The probability is approximately \(5.64\%\).
52715312
An agricultural researcher studies the germination of a seed variety. In \(150\) trials, each using \(5\) seeds, the number \(k\) of seeds that germinated was recorded. <table> <tr><td>Number germinated \(k\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td></tr> <tr><td>Frequency \(n_k\)</td><td>\(4\)</td><td>\(21\)</td><td>\(50\)</td><td>\(48\)</td><td>\(22\)</td><td>\(5\)</td></tr> </table> 1. Find the mean number \(\bar{k}\) of germinated seeds per trial. 2. Assume the number germinated follows \(\operatorname{Bin}(5,p)\). Estimate \(p\) by setting the observed mean equal to the binomial mean. 3. Under the fitted model, find the probability that exactly \(2\) seeds germinate. Compare it with the observed relative frequency.

Hints

- Use a weighted mean for the frequency table. - Relate the binomial mean \(np\) to the observed mean. - Use the binomial formula for one exact count. - Divide the observed frequency by the total number of trials.

Solution

1. The weighted mean is \(\bar{k}=\frac{0\cdot4+1\cdot21+2\cdot50+3\cdot48+4\cdot22+5\cdot5}{150}=\frac{378}{150}=2.52\). 2. Since the binomial mean is \(5p\), set \(5p=2.52\). This gives \(p=0.504\). 3. The model probability is \(P(X=2)=\binom{5}{2}(0.504)^2(0.496)^3\approx 0.309960\). 4. The observed relative frequency is \(\frac{50}{150}=\frac{1}{3}\approx 0.333333\), about \(2.34\) percentage points higher than the model probability.

Answer

1. \(\bar{k}=2.52\) 2. \(p=0.504\) 3. Model: \(P(X=2)\approx 0.3100\) Observed: \(\frac{50}{150}\approx 0.3333\)
52715412
An electronics manufacturer tests components in groups of \(3\). For \(200\) groups, the number \(k\) of defective components per group was recorded. <table> <tr><td>Number defective \(k\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td></tr> <tr><td>Number of groups \(n_k\)</td><td>\(130\)</td><td>\(55\)</td><td>\(12\)</td><td>\(3\)</td></tr> </table> 1. Find the overall proportion of defective components in the sample. Use it as an estimate of the defect probability \(p\) for one component. 2. Assume the number of defects per group follows \(\operatorname{Bin}(3,p)\). Using the estimate from part 1, find the probability that a group contains at least \(1\) defective component. 3. Compare the model probability with the observed relative frequency of groups containing at least \(1\) defect.

Hints

- Distinguish the number of defects per component from the number per group. - Use the complement of no defects. - Combine all observed groups with \(k\ge 1\) to find the relative frequency.

Solution

1. The total number of defective components is \(0\cdot130+1\cdot55+2\cdot12+3\cdot3=88\). The total number tested is \(200\cdot3=600\), so \(p=\frac{88}{600}=\frac{11}{75}\approx 0.146667\). 2. For \(X\sim\operatorname{Bin}(3,\frac{11}{75})\), \(P(X\ge 1)=1-P(X=0)=1-\left(\frac{64}{75}\right)^3\approx 0.378622\). 3. The observed relative frequency is \(\frac{55+12+3}{200}=\frac{70}{200}=0.35\). The model probability is about \(2.86\) percentage points higher.

Answer

1. \(p=\frac{11}{75}\approx 0.1467\) 2. \(P(X\ge 1)\approx 0.3786\) 3. Model: approximately \(37.86\%\) Observed: \(35\%\)
52716112
A binomial random variable \(X\) has success probability \(p=0.5\) and expected value \(E(X)=6\). a) Determine the number of trials \(n\). b) Calculate \(P(X=4)\) and \(P(X=8)\), rounded to four decimal places. c) Use the binomial probability formula \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\) to explain why the histogram of any binomial distribution with \(p=0.5\) is symmetric about \(k=\frac{n}{2}\).

Hints

- Relate the expected value to the number of trials and the success probability. - Recall the symmetry identity for binomial coefficients. - Simplify the factors \(p^k\) and \((1-p)^{n-k}\) when \(p=0.5\). - Identify the midpoint between \(0\) and \(n\).

Solution

1. Since \(E(X)=np\), \(6=n\cdot0.5\), so \(n=12\). 2. \(P(X=4)=\binom{12}{4}(0.5)^4(0.5)^8=\binom{12}{4}(0.5)^{12}=\frac{495}{4096}\approx0.1208\). Because \(\binom{12}{8}=\binom{12}{4}\), \(P(X=8)\approx0.1208\) as well. 3. In general, when \(p=0.5\), \(P(X=k)=\binom{n}{k}(0.5)^n\). Since \(\binom{n}{k}=\binom{n}{n-k}\), it follows that \(P(X=k)=P(X=n-k)\). The values \(k\) and \(n-k\) are equally far from \(\frac{n}{2}\), establishing the symmetry.

Answer

a) \(n=12\) b) \(P(X=4)\approx0.1208\) and \(P(X=8)\approx0.1208\) c) For \(p=0.5\), \(P(X=k)=\binom{n}{k}(0.5)^n\). Because \(\binom{n}{k}=\binom{n}{n-k}\), \(P(X=k)=P(X=n-k)\), so the histogram is symmetric about \(k=\frac{n}{2}\).
52716212
Let \(X\sim\operatorname{Bin}(21,0.5)\). a) Without using a table or a calculator command for cumulative probabilities, explain why \(P(X\le 10)=0.5\). Use the symmetry of the distribution. b) Calculate \(P(X=10)\), and then use it to find \(P(X<10)\). Round each result to four decimal places.

Hints

- How many possible values can \(X\) take when \(n=21\)? - For a symmetric distribution with an even number of possible values, compare the sums over the lower and upper halves. - Pay attention to the difference between \(<\) and \(\le\). - Use the result from part a) to simplify part b).

Solution

1. Because \(p=0.5\), the distribution is symmetric about \(10.5\), so \(P(X=k)=P(X=21-k)\). 2. The values \(0\) through \(10\) pair with the values \(21\) through \(11\). Therefore, \(P(X\le 10)=P(X\ge 11)\). These two complementary events have equal probability, so each has probability \(0.5\). 3. \(P(X=10)=\binom{21}{10}(0.5)^{21}\approx 0.1682\). 4. Therefore, \(P(X<10)=P(X\le 10)-P(X=10)\approx 0.5-0.1682=0.3318\).

Answer

a) Symmetry pairs each outcome from \(0\) through \(10\) with an equally likely outcome from \(11\) through \(21\). The two halves are complementary, so \(P(X\le 10)=0.5\). b) \(P(X=10)\approx 0.1682\) \(P(X<10)\approx 0.3318\)
52716512
A manufacturing facility inspects a sample of \(500\) electronic components and finds \(15\) defective components. 1. Find the relative frequency of defective components. Identify the probability approach used when this relative frequency is treated as an estimate of the defect probability. 2. For planning purposes, a model assumes a constant defect probability of \(p=0.02\) and independent outcomes for \(20\) randomly selected components. Find the probability that exactly \(1\) component is defective. Identify the probability approach used in this model-based calculation.

Hints

- Divide the number in the category by the total sample size. - Observed results lead to an experimental probability estimate. - Use a binomial model for independent two-outcome trials. - Decide whether the probability comes from data or from an assumed mathematical model.

Solution

1. The relative frequency is \(\frac{15}{500}=0.03\), or \(3\%\). Using observed relative frequency as an estimate is an experimental or empirical probability approach. 2. Let \(X\sim\operatorname{Bin}(20,0.02)\). Then \(P(X=1)=\binom{20}{1}(0.02)(0.98)^{19}\approx 0.272493\). 3. This second calculation uses a theoretical, model-based probability because it begins with a specified probability model and parameter.

Answer

1. The relative frequency is \(0.03\), or \(3\%\). This is experimental or empirical probability. 2. The probability is approximately \(0.2725\), or \(27.25\%\). This is theoretical, model-based probability.
52719212
Decide whether each situation is a binomial experiment. If it is, give \(n\) and \(p\). a) A fair six-sided die is rolled \(20\) times. On each roll, record whether the result is prime. b) A spinner has \(10\) equal sectors numbered \(1\) through \(10\). It is spun \(15\) times, and the exact number shown on every spin is recorded.

Hints

- Can each trial be divided into exactly two relevant categories? - How many repetitions are made? - What is the success probability on one trial, and does it stay constant?

Solution

1. In a), each roll is classified as prime or not prime. The prime faces are \(2\), \(3\), and \(5\), so \(p=\frac{3}{6}=0.5\). The rolls are independent, so this is binomial with \(n=20\). 2. In b), the recorded outcome has \(10\) categories rather than two relevant categories. As stated, this is not a binomial experiment.

Answer

a) Binomial with \(n=20\) and \(p=0.5\). b) Not binomial because the recorded outcome has more than two categories.
52719312
A spinner has \(12\) equal sections labeled \(1\) through \(12\). The spinner is spun \(25\) times, and the spins are independent. a) Find the probability that (1) a multiple of \(3\) occurs at least \(5\) times, (2) the number \(12\) occurs more than \(2\) times, (3) a prime number occurs at most \(10\) times. b) The spinner has been spun \(25\) times without landing on \(12\). Find the probability that it lands on \(12\) on the \(26\)th spin.

Hints

- Determine a different success probability for each event. - When can a complement reduce the number of terms you must calculate? - How many prime numbers are there from \(1\) through \(12\)? - Does the result of one spin affect any later spin?

Solution

1. For part a)(1), the multiples of \(3\) are \(3,6,9,12\), so \(p=\frac{4}{12}=\frac{1}{3}\). If \(X\sim\operatorname{Bin}(25,\frac{1}{3})\), then \(P(X\ge 5)=1-P(X\le 4)\approx 0.953799\). 2. For part a)(2), let \(Y\sim\operatorname{Bin}(25,\frac{1}{12})\). Then \(P(Y>2)=1-P(Y\le 2)\approx 0.346704\). 3. For part a)(3), the prime labels are \(2,3,5,7,11\), so \(p=\frac{5}{12}\). If \(Z\sim\operatorname{Bin}(25,\frac{5}{12})\), then \(P(Z\le 10)\approx 0.518003\). 4. For part b), the spins are independent. Previous results do not change the probability of the next result, so the probability remains \(\frac{1}{12}\).

Answer

a) (1) \(P(X\ge 5)\approx 0.9538\) (2) \(P(Y>2)\approx 0.3467\) (3) \(P(Z\le 10)\approx 0.5180\) b) The probability is \(\frac{1}{12}\approx 0.0833\).
52719712
A manufacturer reports that each electronic component independently has a \(4\%\) chance of having a particular manufacturing defect. A technician selects a random sample of \(n\) components. What is the minimum sample size needed so that the probability of finding at least \(1\) component with this defect is at least \(95\%\)?

Hints

- What is the complement of “at least \(1\) component is defective”? - How can you express the probability of exactly \(0\) defects in \(n\) trials? - Use logarithms when the unknown is in an exponent. - The sample size must be a whole number.

Solution

1. The probability of finding no defective components in a sample of size \(n\) is \((0.96)^n\). 2. Require \(1-(0.96)^n\ge 0.95\), which is equivalent to \((0.96)^n\le 0.05\). 3. Taking logarithms gives \(n\ge \frac{\ln(0.05)}{\ln(0.96)}\approx 73.385\). 4. Because \(n\) must be a whole number, the minimum sample size is \(74\).

Answer

At least \(74\) components must be selected.
52719812
In a city, \(60\%\) of households have fiber-optic internet service. A random sample of \(100\) households is selected. The sample is small relative to the city’s household population, so households are modeled as approximately independent. Let \(X\) be the number in the sample that have fiber-optic service. 1. Find the mean \(\mu\) and standard deviation \(\sigma\) of \(X\). 2. Find the probability that \(X\) is within one standard deviation of its mean.

Hints

- Use the formulas for the mean and standard deviation of a binomial random variable. - Determine which integer values lie in the interval around the mean. - Express the interval probability as a difference of cumulative probabilities.

Solution

1. Since \(X\sim\operatorname{Bin}(100,0.6)\), the mean is \(\mu=np=100\cdot 0.6=60\), and the standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{100\cdot 0.6\cdot 0.4}=\sqrt{24}\approx 4.90\). 2. The interval \([\mu-\sigma,\mu+\sigma]\) is approximately \([55.10,64.90]\), so the included integer values are \(56\) through \(64\). 3. Therefore, \(P(56\le X\le 64)=P(X\le 64)-P(X\le 55)\approx 0.6416\).

Answer

1. \(\mu=60\) and \(\sigma\approx 4.90\) 2. \(P(56\le X\le 64)\approx 0.6416\), or about \(64.16\%\)
52721112
A computer-chip manufacturer finds that \(8\%\) of its chips are defective. Let \(X_n\) be the number of defective chips in a sample of size \(n\), so \(X_n\sim\operatorname{Bin}(n,0.08)\). a) A sample of \(20\) chips is inspected. Find the probability of each event. \(E_1\): None of the chips are defective. \(E_2\): At most \(2\) chips are defective. \(E_3\): Exactly \(17\) or \(18\) chips are nondefective. b) Find the minimum value of \(n\) needed so that the probability of finding at least \(1\) defective chip is at least \(99\%\).

Hints

- Translate a given number of nondefective chips into a number of defective chips. - Use a complement for an event involving at least \(1\) defect. - Use logarithms to solve an inequality with \(n\) in the exponent. - Express “at most \(2\)” as a sum of individual binomial probabilities.

Solution

1. For \(E_1\), \(P(X_{20}=0)=(0.92)^{20}\approx 0.188693\). 2. For \(E_2\), \(P(X_{20}\le 2)=\sum_{k=0}^{2}\binom{20}{k}(0.08)^k(0.92)^{20-k}\approx 0.787946\). 3. Having exactly \(17\) or \(18\) nondefective chips means having exactly \(3\) or \(2\) defective chips. Thus, \(P(E_3)=P(X_{20}=2)+P(X_{20}=3)\approx 0.412529\). 4. For part b), require \(1-(0.92)^n\ge 0.99\), so \((0.92)^n\le 0.01\). 5. Solving gives \(n\ge \frac{\ln(0.01)}{\ln(0.92)}\approx 55.230\). Therefore, the minimum is \(56\).

Answer

a) \(P(E_1)\approx 18.87\%\) \(P(E_2)\approx 78.79\%\) \(P(E_3)\approx 41.25\%\) b) The minimum sample size is \(56\).
52721312
A gardener plants \(12\) seeds of a rare flower. Each seed germinates with probability \(0.85\), independently of the others. Find the probability of each event. (1) Exactly \(9\) seeds germinate. (2) At least \(11\) seeds germinate. (3) Only the first three planted seeds fail to germinate, while all remaining seeds germinate.

Hints

- Decide whether each event specifies an order or only a total number of successes. - Break “at least” into the possible exact values. - Distinguish a binomial event from one specified sequence.

Solution

1. Let \(X\) be the number of seeds that germinate. Then \(X\) is binomial with \(n=12\), \(p=0.85\), and \(1-p=0.15\). 2. For event (1), \(P(X=9)=\binom{12}{9}(0.85)^9(0.15)^3\approx0.1720\). 3. For event (2), \(P(X\ge11)=P(X=11)+P(X=12)\approx0.4435\). 4. For event (3), the sequence is specified, so the probability is \((0.15)^3(0.85)^9\approx0.000782\).

Answer

(1) \(0.1720\), or about \(17.20\%\) (2) \(0.4435\), or about \(44.35\%\) (3) \(0.000782\), or about \(0.08\%\)
52721412
A quality-control sample contains \(15\) LED bulbs. Each bulb is independently defective with probability \(0.04\). Find the probability that: (1) exactly one bulb is defective; (2) more than two bulbs are defective; (3) only the last inspected bulb is defective.

Hints

- Identify \(n\) and \(p\). - Translate “more than two” into the excluded lower values. - Decide when a binomial coefficient is needed and when the order is fixed.

Solution

1. Let \(X\) be the number of defective bulbs. Then \(X\) is binomial with \(n=15\) and \(p=0.04\). 2. For event (1), \(P(X=1)=\binom{15}{1}(0.04)(0.96)^{14}\approx0.3388\). 3. For event (2), use the complement: \(P(X>2)=1-[P(X=0)+P(X=1)+P(X=2)]\approx0.0203\). 4. For event (3), the order is specified, so the probability is \((0.96)^{14}(0.04)\approx0.0226\).

Answer

(1) \(0.3388\), or about \(33.88\%\) (2) \(0.0203\), or about \(2.03\%\) (3) \(0.0226\), or about \(2.26\%\)
52721512
In a clinical study, each patient responds positively to a treatment with probability \(0.4\). Fifty patients are treated, and outcomes are treated as independent. Let \(X\) be the number of positive responses. Find the probability that \(X\) differs from its mean by at most \(4\).

Hints

- Find the expected number of positive responses. - Translate “differs by at most \(4\)” into an interval. - Express the interval probability as a difference of cumulative probabilities. - Check that both endpoints are included.

Solution

1. Since \(X\sim\operatorname{Bin}(50,0.4)\), the mean is \(\mu=np=50\cdot 0.4=20\). 2. A difference of at most \(4\) means \(16\le X\le 24\). 3. Using the binomial cumulative distribution function, \(P(16\le X\le 24)=P(X\le 24)-P(X\le 15)\approx 0.9022-0.0955=0.8067\).

Answer

\(P(16\le X\le 24)\approx 0.8067\), or about \(80.67\%\)
52721912
A fair coin is flipped independently \(8\) times. Let \(H\) represent heads and \(T\) represent tails. Consider these two sequences: Sequence 1: \((T, T, T, T, T, T, T, T)\) Sequence 2: \((H, T, T, H, T, H, H, T)\) a) Find the probability of each sequence. b) Many people think Sequence 2 is more likely because it appears more mixed. Explain why that intuition is mathematically incorrect. c) Let \(X\) be the number of tails in the \(8\) flips. Find \(P(X=4)\), and use it to explain why the event “exactly four tails” is much more likely than “eight tails.”

Hints

- Find the probability of one fully specified sequence. - Ask whether an apparently mixed pattern changes the probability of that exact sequence. - Count the sequences containing exactly four tails. - Distinguish one outcome sequence from an event containing many sequences.

Solution

1. Each exact sequence of eight fair coin flips has probability \((0.5)^8=\frac{1}{256}\approx0.0039\). Therefore, Sequences 1 and 2 are equally likely. 2. The flips are independent, and heads and tails have equal probability. The visual pattern of an exact sequence does not change its probability. 3. For exactly four tails, \(P(X=4)=\binom84(0.5)^4(0.5)^4=\frac{70}{256}\approx0.2734\). 4. The event “exactly four tails” contains \(\binom84=70\) equally likely sequences, while “eight tails” contains only one sequence.

Answer

a) Each sequence has probability \(\frac{1}{256}\approx0.0039\). b) Every exact eight-flip sequence is equally likely because the flips are independent and the coin is fair. c) \(P(X=4)=\frac{70}{256}\approx0.2734\). The event contains \(70\) sequences, compared with one sequence for eight tails.
52722012
A basketball player independently takes \(4\) free throws and makes each free throw with probability \(0.80\). Let \(M\) represent a make and \(F\) represent a miss. a) Find the probabilities of these sequences: \(S_1=(M, M, M, M)\) \(S_2=(M, F, M, M)\) b) Evaluate the statement: “In a binomial setting, every sequence with the same number of successes has the same probability, regardless of the positions of the successes.” Justify your answer using an example with three makes. c) Find the probability that the player makes exactly three free throws.

Hints

- Write the probabilities of a make and a miss. - For a specified sequence, multiply the probabilities in order. - Ask whether changing the order of the same factors changes their product. - Interpret \(\binom nk\) as the number of sequences with \(k\) successes.

Solution

1. The make probability is \(p=0.8\), and the miss probability is \(0.2\). 2. \(P(S_1)=(0.8)^4=0.4096\), and \(P(S_2)=(0.8)^3(0.2)=0.1024\). 3. The statement is true. Every sequence with exactly three makes and one miss has probability \((0.8)^3(0.2)\), regardless of the order of the factors. For example, \(P(F, M, M, M)=0.2\cdot(0.8)^3=0.1024=P(M, F, M, M)\). 4. There are \(\binom43=4\) sequences with exactly three makes, so \(P(X=3)=\binom43(0.8)^3(0.2)=0.4096\).

Answer

a) \(P(S_1)=0.4096\); \(P(S_2)=0.1024\) b) The statement is true. Each sequence with three makes and one miss has probability \((0.8)^3(0.2)=0.1024\). c) \(P(X=3)=0.4096\)
52788312
A gardener buys \(20\) seeds of a rare plant. The seeds germinate independently, but the germination probability \(p\) for each seed under the storage conditions is unknown. Find the minimum threshold for \(p\) that makes the probability that at least one seed germinates at least \(95\%\). Give an exact expression and a decimal approximation.

Hints

- Use the complement of “at least one success.” - Write the probability of exactly zero successes. - Rearrange the inequality so that the unknown probability is isolated. - Use a twentieth root to undo the twentieth power.

Solution

1. Let \(X\sim\operatorname{Bin}(20,p)\) be the number of seeds that germinate. 2. Use the complement: \(P(X\ge 1)=1-P(X=0)=1-(1-p)^{20}\). 3. The requirement \(1-(1-p)^{20}\ge 0.95\) is equivalent to \((1-p)^{20}\le 0.05\). 4. Taking the twentieth root gives \(1-p\le \sqrt[20]{0.05}\), so \(p\ge 1-\sqrt[20]{0.05}\). 5. The threshold is \(1-\sqrt[20]{0.05}\approx 0.139108\), or approximately \(13.9108\%\).

Answer

\(p\ge 1-\sqrt[20]{0.05}\). The threshold is approximately \(0.139108\), or \(13.9108\%\).
52788412
A manufacturer tests a random sample of \(50\) microchips from a new production line. Assume chip defects are independent, and let \(p\) be the unknown defect rate. Find the minimum threshold for \(p\) that makes the probability of finding at least one defective chip in the sample at least \(90\%\). Give an exact expression and a decimal approximation.

Hints

- First find the probability that no chips in the sample are defective. - Relate “no defective chips” to “at least one defective chip” using a complement. - Use a root or logarithm to solve for \(p\). - Preserve the direction of the “at least” inequality as you rearrange.

Solution

1. Let \(X\sim\operatorname{Bin}(50,p)\) be the number of defective chips in the sample. 2. Use the complement: \(P(X\ge 1)=1-P(X=0)=1-(1-p)^{50}\). 3. The requirement \(1-(1-p)^{50}\ge 0.90\) is equivalent to \((1-p)^{50}\le 0.10\). 4. Taking the fiftieth root gives \(1-p\le \sqrt[50]{0.10}\), so \(p\ge 1-\sqrt[50]{0.10}\). 5. The threshold is \(1-\sqrt[50]{0.10}\approx 0.0450074\), or approximately \(4.5007\%\).

Answer

\(p\ge 1-\sqrt[50]{0.10}\). The threshold is approximately \(0.0450074\), or \(4.5007\%\).
52790012
Decide whether a binomial model is appropriate for each situation. a) In a large city survey, \(100\) people are randomly selected and asked whether they own an electric vehicle. About \(4\%\) of the city's residents own one. b) An archer takes \(12\) practice shots. As the session continues, fatigue lowers the archer's concentration. c) A spinner has three equal sectors colored blue, yellow, and red. It is spun \(20\) times, and each spin is classified as yellow or not yellow.

Hints

- When can sampling without replacement be approximated by independent trials? - Look for language indicating that success chances change over time. - Can every trial be classified into success or failure?

Solution

1. In a), a binomial model is a reasonable approximation because the city population is very large compared with the sample. The ownership probability can be treated as approximately constant at \(p=0.04\), with \(n=100\). The random variable \(X\) counts electric-vehicle owners in the sample. 2. In b), fatigue causes the success probability to change over time. A binomial model is not appropriate. 3. In c), the spins are independent, and the probability of yellow is constant at \(p=\frac{1}{3}\). This is binomial with \(n=20\), and \(X\) counts yellow outcomes.

Answer

a) Yes, as an approximation: \(n=100\), \(p=0.04\), and \(X\) is the number of electric-vehicle owners in the sample. b) No; fatigue makes the success probability nonconstant. c) Yes: \(n=20\), \(p=\frac{1}{3}\), and \(X\) is the number of yellow outcomes.
52790312
A chip manufacturer reports that \(5\%\) of its computer chips are defective. A distributor tests \(400\) chips from a shipment, and outcomes are treated as independent. Let \(X\) be the number of defective chips. a) Find the mean \(\mu\) and standard deviation \(\sigma\) of \(X\). b) Find the probability that \(X\) lies in the interval \([\mu-\sigma,\mu+\sigma]\).

Hints

- Identify \(n\) and \(p\) from the context. - Use the binomial formulas for the mean and standard deviation. - Determine the integer counts inside the interval. - Use cumulative probabilities to evaluate the interval.

Solution

1. Since \(X\sim\operatorname{Bin}(400,0.05)\), the mean is \(\mu=np=400\cdot 0.05=20\). 2. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{400\cdot 0.05\cdot 0.95}=\sqrt{19}\approx 4.36\). 3. The interval is approximately \([15.64,24.36]\), so the included integer values are \(16\) through \(24\). 4. Therefore, \(P(16\le X\le 24)=P(X\le 24)-P(X\le 15)\approx 0.6990\).

Answer

a) \(\mu=20\) and \(\sigma\approx 4.36\) b) \(P(16\le X\le 24)\approx 0.6990\), or about \(69.90\%\)
52790412
In a small town, \(64\%\) of eligible voters support a local initiative. A random sample of \(150\) eligible voters is surveyed, and responses are treated as independent. Let \(X\) be the number who support the initiative. a) Find the mean \(\mu\) and standard deviation \(\sigma\) of \(X\). b) Find the probability that \(X\) differs from its mean by more than one standard deviation.

Hints

- Use the binomial mean and standard-deviation formulas. - Translate “more than one standard deviation” into values outside an interval. - It may be easier to find the probability inside the interval first. - Check which integer values remain inside the interval.

Solution

1. Since \(X\sim\operatorname{Bin}(150,0.64)\), the mean is \(\mu=np=150\cdot 0.64=96\). 2. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{150\cdot 0.64\cdot 0.36}\approx 5.88\). 3. Values within one standard deviation satisfy approximately \(90.12\le X\le 101.88\), so the included integer values are \(91\) through \(101\). 4. The probability outside this interval is \(1-P(91\le X\le 101)\approx 1-0.6505=0.3495\). Equivalently, the event is \(X\le 90\) or \(X\ge 102\).

Answer

a) \(\mu=96\) and \(\sigma\approx 5.88\) b) \(P(X\le 90\text{ or }X\ge 102)\approx 0.3495\), or about \(34.95\%\)
52872712
A multiple-choice test has \(8\) questions. Each question has \(4\) answer choices, exactly one of which is correct. A student guesses independently and randomly on every question. a) Find the probability that the student answers exactly \(2\) questions correctly. b) Find the probability that the student answers at most \(1\) question correctly. c) Which event is more likely: exactly \(2\) correct answers or more than \(2\) correct answers? Support your conclusion by comparing the probabilities.

Hints

- First determine the probability of a correct answer on one question. - Which probability model applies to repeated independent trials with two outcomes? - What values are included in “at most \(1\)”? - Use a complement to calculate “more than \(2\)” efficiently.

Solution

1. Let \(X\sim\operatorname{Bin}(8,0.25)\). 2. For part a), \(P(X=2)=\binom{8}{2}(0.25)^2(0.75)^6\approx 0.311462\). 3. For part b), \(P(X\le 1)=(0.75)^8+8(0.25)(0.75)^7\approx 0.367081\). 4. For part c), \(P(X>2)=1-P(X\le 2)=1-[P(X\le 1)+P(X=2)]\approx 0.321457\). 5. Since \(0.321457>0.311462\), more than \(2\) correct answers is slightly more likely.

Answer

a) The probability is approximately \(31.15\%\). b) The probability is approximately \(36.71\%\). c) More than \(2\) correct answers is more likely: approximately \(32.15\%\), compared with approximately \(31.15\%\) for exactly \(2\) correct answers.
52872812
At a glassworks, each figurine independently has a \(10\%\) chance of being defective. A random sample of \(12\) figurines is selected for quality control. a) Find the probability that none of the selected figurines are defective. b) Find the probability that at least \(2\) figurines are defective. c) Which event is more likely: exactly \(1\) figurine is defective or more than \(2\) figurines are defective?

Hints

- Identify the probability that one figurine is defective. - For “at least \(2\),” consider using the complement. - Calculate both probabilities in part c) before comparing them.

Solution

1. Let \(X\sim\operatorname{Bin}(12,0.10)\). 2. For part a), \(P(X=0)=(0.90)^{12}\approx 0.282430\). 3. For part b), \(P(X\ge 2)=1-[P(X=0)+P(X=1)]\approx 0.340998\). 4. For part c), \(P(X=1)=12(0.10)(0.90)^{11}\approx 0.376573\). 5. Also, \(P(X>2)=P(X\ge 2)-P(X=2)\approx 0.110870\). Therefore, exactly \(1\) defective figurine is more likely.

Answer

a) The probability is approximately \(28.24\%\). b) The probability is approximately \(34.10\%\). c) Exactly \(1\) defective figurine is more likely: approximately \(37.66\%\), compared with approximately \(11.09\%\) for more than \(2\).
52873512
A basketball player independently makes each free throw with probability \(0.70\). During practice, the player attempts \(6\) free throws. Find the probability of each event. a) The player makes exactly \(4\) free throws. b) The player makes no free throws. c) The player makes at most \(1\) free throw. d) The player makes at least \(5\) free throws.

Hints

- First identify the probability of a made free throw and a missed free throw. - Does the probability change from one attempt to the next? - Determine which individual counts are included in “at most” and “at least.” - Use the binomial probability formula for each required count.

Solution

1. Let \(X\sim\operatorname{Bin}(6,0.70)\). 2. For part a), \(P(X=4)=\binom{6}{4}(0.70)^4(0.30)^2=0.324135\). 3. For part b), \(P(X=0)=(0.30)^6=0.000729\). 4. For part c), \(P(X\le 1)=P(X=0)+P(X=1)=0.010935\). 5. For part d), \(P(X\ge 5)=P(X=5)+P(X=6)=0.420175\).

Answer

a) The probability is approximately \(32.41\%\). b) The probability is approximately \(0.07\%\). c) The probability is approximately \(1.09\%\). d) The probability is approximately \(42.02\%\).
52873712
A multiple-choice test has \(6\) questions. Each question has two answer choices, exactly one of which is correct. A person guesses randomly and independently on every question. Find the probability that (1) exactly \(3\) answers are correct, (2) exactly \(5\) answers are correct, (3) more answers are correct than incorrect.

Hints

- Determine the probability of a correct answer with two choices. - Model the questions as independent Bernoulli trials. - For \(6\) questions, which correct-answer counts mean more correct than incorrect? - Use the binomial formula for each exact count.

Solution

1. Let \(X\sim\operatorname{Bin}(6,0.5)\). 2. For part (1), \(P(X=3)=\binom{6}{3}(0.5)^6=0.3125\). 3. For part (2), \(P(X=5)=\binom{6}{5}(0.5)^6=0.09375\). 4. More answers are correct than incorrect when \(X=4,5,\) or \(6\). Thus, \(P(X\ge 4)=0.234375+0.09375+0.015625=0.34375\).

Answer

(1) The probability is \(0.3125\), or \(31.25\%\). (2) The probability is \(0.09375\), or \(9.375\%\). (3) The probability is \(0.34375\), or \(34.375\%\).
52873812
A basketball player makes each free throw independently with probability \(p=0.70\). The player attempts \(4\) free throws. a) Find the probability that the player (1) makes exactly \(2\) free throws and (2) makes at least \(3\) free throws. b) After training, the player's success probability increases to \(p=0.75\). Recalculate the probabilities from part a) and compare them with the original values.

Hints

- Substitute the appropriate success probability into the same binomial model. - Which exact counts are included in “at least \(3\)”? - Add the probabilities for the required counts. - Consider whether a higher success probability should make a low number of successes more or less likely.

Solution

1. For \(p=0.70\), \(P(X=2)=\binom{4}{2}(0.70)^2(0.30)^2=0.2646\). 2. Also, \(P(X\ge 3)=\binom{4}{3}(0.70)^3(0.30)+(0.70)^4=0.6517\). 3. For \(p=0.75\), \(P(X=2)=\binom{4}{2}(0.75)^2(0.25)^2=0.2109375\). 4. Also, \(P(X\ge 3)=\binom{4}{3}(0.75)^3(0.25)+(0.75)^4=0.73828125\). 5. The probability of exactly \(2\) makes decreases, while the probability of at least \(3\) makes increases.

Answer

a) (1) \(P(X=2)=0.2646\) (2) \(P(X\ge 3)=0.6517\) b) (1) \(P(X=2)\approx 0.2109\) (2) \(P(X\ge 3)\approx 0.7383\) The probability of exactly \(2\) makes decreases, and the probability of at least \(3\) makes increases.
52873912
A fair six-sided die is rolled independently \(10\) times. Rolling a \(6\) is considered a success. Find the probability that (1) exactly \(2\) rolls are \(6\)s, (2) at most \(1\) roll is a \(6\), (3) at least \(3\) rolls are \(6\)s.

Hints

- First determine the probability of rolling a \(6\) on one roll. - Use the binomial formula for an exact number of successes. - What counts are included in “at most \(1\)”? - Use a complement for “at least \(3\).”

Solution

1. Let \(X\sim\operatorname{Bin}(10,\frac{1}{6})\). 2. For part (1), \(P(X=2)=\binom{10}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^8\approx 0.290710\). 3. For part (2), \(P(X\le 1)=\left(\frac{5}{6}\right)^{10}+10\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^9\approx 0.484517\). 4. For part (3), \(P(X\ge 3)=1-P(X\le 2)\approx 0.224773\).

Answer

(1) The probability is approximately \(0.2907\), or \(29.07\%\). (2) The probability is approximately \(0.4845\), or \(48.45\%\). (3) The probability is approximately \(0.2248\), or \(22.48\%\).
52874012
A gardener plants \(12\) flower seeds of one variety. Each seed germinates with probability \(80\%\), independently of the others. Find the probability that (1) exactly \(10\) seeds germinate, (2) more than \(10\) seeds germinate, (3) at most \(2\) seeds do not germinate.

Hints

- Identify the number of trials and the germination probability. - Which exact counts are included in “more than \(10\)”? - Translate at most \(2\) failures into a condition on the number of germinated seeds. - Combine results from earlier parts when useful.

Solution

1. Let \(X\sim\operatorname{Bin}(12,0.80)\), where \(X\) is the number of seeds that germinate. 2. For part (1), \(P(X=10)=\binom{12}{10}(0.80)^{10}(0.20)^2\approx 0.283468\). 3. For part (2), \(P(X>10)=P(X=11)+P(X=12)\approx 0.274878\). 4. At most \(2\) seeds fail to germinate means at least \(10\) germinate. Thus, \(P(X\ge 10)\approx 0.558346\).

Answer

(1) The probability is approximately \(0.2835\), or \(28.35\%\). (2) The probability is approximately \(0.2749\), or \(27.49\%\). (3) The probability is approximately \(0.5583\), or \(55.83\%\).
52874112
A spinner has a win probability of \(0.25\) on each spin. It is spun independently \(10\) times. Find the probability of winning exactly \(3\) times or exactly \(4\) times.

Hints

- Check that the spins can be modeled as independent trials. - Identify the two exact values included in the event. - Use the binomial formula for each exact value. - Add the probabilities of mutually exclusive events.

Solution

1. Let \(X\) be the number of wins. Then \(X\) is binomial with \(n=10\) and \(p=0.25\). 2. \(P(X=3)=\binom{10}{3}(0.25)^3(0.75)^7\approx0.2503\). 3. \(P(X=4)=\binom{10}{4}(0.25)^4(0.75)^6\approx0.1460\). 4. The two events are mutually exclusive, so \(P(X=3\text{ or }X=4)=P(X=3)+P(X=4)\approx0.3963\).

Answer

\(P(X=3\text{ or }X=4)\approx0.3963\), or about \(39.63\%\)
52874412
Two fair six-sided dice are rolled together, and this experiment is repeated independently \(8\) times. Find the probability that a sum of exactly \(11\) occurs fewer than \(2\) times.

Hints

- Count the outcomes that produce a sum of \(11\). - Find the probability of that sum on one roll of the two dice. - Which values are included in “fewer than \(2\)”? - Use the binomial formula for those values.

Solution

1. A sum of \(11\) occurs with \((5, 6)\) or \((6, 5)\), so the success probability is \(p=\frac{2}{36}=\frac{1}{18}\). 2. Let \(X\sim\operatorname{Bin}(8,\frac{1}{18})\). 3. Fewer than \(2\) occurrences means \(X=0\) or \(X=1\). 4. Therefore, \(P(X<2)=\left(\frac{17}{18}\right)^8+8\left(\frac{1}{18}\right)\left(\frac{17}{18}\right)^7\approx 0.930899\).

Answer

The probability is approximately \(93.09\%\).
52874612
A manufacturer claims that only \(1\%\) of its USB drives are defective. A distributor tests \(100\) drives and observes \(3\) defective drives. Model the tested drives as independently defective with probability \(0.01\). Find the probability, under the claim, of observing at least \(3\) defective drives. The claim will be rejected if this probability is at most \(5\%\). Evaluate the result using this rule.

Hints

- Use the complement to calculate “at least \(3\).” - Which values belong to the complement event? - Compare the result with the \(5\%\) cutoff.

Solution

1. Under the claim, let \(X\sim\operatorname{Bin}(100,0.01)\). 2. The probability is \(P(X\ge 3)=1-P(X\le 2)\). 3. Therefore, \(P(X\ge 3)=1-\sum_{k=0}^{2}\binom{100}{k}(0.01)^k(0.99)^{100-k}\approx 0.079373\). 4. Since \(7.94\%>5\%\), the claim is not rejected under the stated rule.

Answer

The probability is approximately \(7.94\%\). Because this is greater than \(5\%\), the manufacturer''s claim is not rejected.
52875112
A survey reports that \(40\%\) of households in a city regularly use a streaming service. A market-research study randomly selects \(15\) households. Assume the household responses are independent. Find the probability that a) exactly \(6\) households use a streaming service, b) at most \(2\) households use a streaming service, c) at least \(13\) households use a streaming service.

Hints

- Check whether the conditions for a binomial model are reasonable. - Identify the number of trials and the success probability. - “At most” and “at least” include several possible counts. - Decide whether a direct sum or a complement is more efficient.

Solution

1. Let \(X\sim\operatorname{Bin}(15,0.40)\). 2. For part a), \(P(X=6)=\binom{15}{6}(0.40)^6(0.60)^9\approx 0.206598\). 3. For part b), \(P(X\le 2)=\sum_{k=0}^{2}\binom{15}{k}(0.40)^k(0.60)^{15-k}\approx 0.027114\). 4. For part c), \(P(X\ge 13)=\sum_{k=13}^{15}\binom{15}{k}(0.40)^k(0.60)^{15-k}\approx 0.000279\).

Answer

a) The probability is approximately \(20.66\%\). b) The probability is approximately \(2.71\%\). c) The probability is approximately \(0.0279\%\).
52875512
An archer hits the center ring of a target with probability \(0.4\) on each shot. She takes \(10\) independent shots. Let \(X\) be the number of center-ring hits. Find each probability. a) Exactly \(4\) hits b) At least \(2\) hits c) Between \(3\) and \(5\) hits, inclusive

Hints

- Identify the binomial parameters. - Use the binomial probability formula for an exact count. - For “at least,” consider the complement. - List the included values for an inclusive interval.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(10,0.4)\). 2. For part a), \(P(X=4)=\binom{10}{4}(0.4)^4(0.6)^6\approx 0.2508\). 3. For part b), use the complement: \(P(X\ge 2)=1-P(X\le 1)\approx 0.9536\). 4. For part c), \(P(3\le X\le 5)=P(X=3)+P(X=4)+P(X=5)\approx 0.6665\).

Answer

a) \(P(X=4)\approx 0.2508\), or about \(25.08\%\) b) \(P(X\ge 2)\approx 0.9536\), or about \(95.36\%\) c) \(P(3\le X\le 5)\approx 0.6665\), or about \(66.65\%\)
52875612
In a manufacturing process, \(5\%\) of electronic components are defective. A random sample of \(15\) components is selected, and defect outcomes are treated as independent. Let \(X\) be the number of defective components. Find each probability. a) Exactly \(1\) component is defective. b) More than \(2\) components are defective. c) At least \(1\) but at most \(3\) components are defective.

Hints

- Identify the binomial parameters. - Check whether each boundary value is included. - Use a complement for “more than \(2\).”

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(15,0.05)\). 2. \(P(X=1)=\binom{15}{1}\cdot 0.05\cdot (0.95)^{14}\approx 0.3658\). 3. Use the complement for part b): \(P(X>2)=1-P(X\le 2)\approx 0.0362\). 4. For part c), \(P(1\le X\le 3)=P(X=1)+P(X=2)+P(X=3)\approx 0.5312\).

Answer

a) \(P(X=1)\approx 0.3658\), or about \(36.58\%\) b) \(P(X>2)\approx 0.0362\), or about \(3.62\%\) c) \(P(1\le X\le 3)\approx 0.5312\), or about \(53.12\%\)
52875712
A player wins each spin of a game wheel with probability \(0.2\). The wheel is spun \(10\) times independently. Let \(X\) be the number of wins. Find the probability of winning at least once but no more than three times.

Hints

- Model each spin as a win-or-no-win trial. - List the win counts included in the event. - Use the binomial formula for each exact count. - Add the probabilities of the disjoint cases.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(10,0.2)\). 2. The required event is \(1\le X\le 3\). 3. Therefore, \(P(1\le X\le 3)=\sum_{k=1}^{3}\binom{10}{k}\cdot (0.2)^k\cdot (0.8)^{10-k}\approx 0.7718\).

Answer

\(P(1\le X\le 3)\approx 0.7718\), or about \(77.18\%\)
52875812
A basketball player makes a free throw with probability \(0.6\). He takes \(15\) independent free throws. Let \(X\) be the number made. Find the probability that he makes between \(8\) and \(10\) free throws, inclusive.

Hints

- Identify the exact values included in the interval. - Confirm that the trials have constant success probability and are independent. - Use the binomial formula for each included value. - Add the point probabilities.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(15,0.6)\). 2. The required probability is \(P(8\le X\le 10)=P(X=8)+P(X=9)+P(X=10)\). 3. The three probabilities are approximately \(0.1771\), \(0.2066\), and \(0.1859\). Their sum is \(0.5696\).

Answer

\(P(8\le X\le 10)\approx 0.5696\), or about \(56.96\%\)
52876712
In a large city, \(60\%\) of commuters drive to work. A random sample of \(20\) commuters is selected, and responses are treated as independent. Let \(X\) be the number who drive. a) What value of \(X\) is most likely? b) Find the probability that at most \(8\) or at least \(16\) commuters in the sample drive.

Hints

- Identify the binomial parameters. - Use the binomial mode rule involving \((n+1)p\). - The two ranges in part b) do not overlap. - Express “at least \(16\)” with a complement.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(20,0.6)\). 2. Since \((n+1)p=21\cdot 0.6=12.6\) is not an integer, the unique mode is \(\lfloor 12.6\rfloor=12\). 3. The events \(X\le 8\) and \(X\ge 16\) are disjoint, so their probabilities add: \(P(X\le 8\text{ or }X\ge 16)=P(X\le 8)+1-P(X\le 15)\). 4. Using cumulative probabilities, the result is approximately \(0.0565+0.0510=0.1075\).

Answer

a) \(X=12\) b) \(P(X\le 8\text{ or }X\ge 16)\approx 0.1075\), or about \(10.75\%\)
52876812
A gardener knows that \(20\%\) of a certain type of flower seed does not germinate. She plants \(20\) seeds, and outcomes are treated as independent. Let \(X\) be the number that do not germinate. a) What value of \(X\) is most likely? b) Show that the probability that at most \(1\) seed or at least \(8\) seeds do not germinate is about \(10\%\).

Hints

- Identify the binomial parameters. - Use the binomial mode rule involving \((n+1)p\). - Find the two disjoint tail probabilities separately. - Use a complement for “at least \(8\).”

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(20,0.2)\). 2. Since \((n+1)p=21\cdot 0.2=4.2\) is not an integer, the unique mode is \(\lfloor 4.2\rfloor=4\). 3. The events in part b) are disjoint. The first probability is \(P(X\le 1)=(0.8)^{20}+\binom{20}{1}\cdot 0.2\cdot (0.8)^{19}\approx 0.0692\). 4. The second probability is \(P(X\ge 8)=1-P(X\le 7)\approx 0.0321\). 5. Therefore, \(P(X\le 1\text{ or }X\ge 8)\approx 0.0692+0.0321=0.1013\), or about \(10.13\%\).

Answer

a) \(X=4\) b) \(P(X\le 1\text{ or }X\ge 8)\approx 0.1013\), or about \(10.13\%\)
52877812
A spinner has \(3\) equal sections labeled \(1\), \(2\), and \(3\). A player suspects that the spinner is biased so that \(3\) occurs less often than it should. In \(30\) independent spins, the spinner lands on \(3\) only \(4\) times. 1. Assuming the spinner is fair, find the probability that \(3\) occurs at most \(4\) times. 2. A result is considered significantly rare when the probability of that result or a more extreme result is less than \(5\%\). Under this criterion, does the observation support the player's suspicion? Explain.

Hints

- What is the probability of one section on a fair three-section spinner? - Which counts are included in “at most \(4\)”? - Use a cumulative binomial probability. - Compare the result with the stated significance cutoff.

Solution

1. Under a fair spinner, let \(X\sim\operatorname{Bin}(30,\frac{1}{3})\). 2. The probability is \(P(X\le 4)=\sum_{k=0}^{4}\binom{30}{k}\left(\frac{1}{3}\right)^k\left(\frac{2}{3}\right)^{30-k}\approx 0.012230\). 3. Since \(1.22\%<5\%\), the observation is significantly rare under the stated criterion and supports the suspicion that \(3\) occurs less often than expected.

Answer

1. \(P(X\le 4)\approx 1.22\%\) 2. Yes. The observation supports the suspicion because \(1.22\%<5\%\).
53081912
A spinner has four equal sections: one red winning section and three blue losing sections. Thus, \(P(\text{red})=0.25\). Assume successive spins are independent. Determine whether each interpretation is correct. (1) If the spinner is spun \(4\) times, it is certain to land on red exactly once. (2) If the spinner does not land on red in the first \(10\) spins, the probability of red on the \(11\)th spin increases to make up for the earlier results. (3) Over a very large number of spins, such as \(10{,}000\), the relative frequency of red will likely be close to \(0.25\). (4) In \(40\) spins, getting exactly \(10\) red results is more likely than getting exactly \(11\) red results.

Hints

- Distinguish a theoretical probability from the outcome of a small set of trials. - Decide whether one spin affects the next spin. - Think about what happens to relative frequency as the number of trials grows. - Compare the binomial probabilities for exactly \(10\) and exactly \(11\) successes.

Solution

1. Statement (1) is false. A probability of \(0.25\) does not guarantee an exact result in a small number of trials. The probability of exactly one red result in \(4\) spins is \(\binom{4}{1}(0.25)(0.75)^3\approx 0.4219\). 2. Statement (2) is false. The spins are independent, so the probability of red remains \(0.25\) on every spin. 3. Statement (3) is true. By the law of large numbers, the relative frequency tends to become more stable near the theoretical probability as the number of trials increases. 4. Statement (4) is true. For \(X\sim\operatorname{Bin}(40,0.25)\), \(\frac{P(X=11)}{P(X=10)}=\frac{40-10}{11}\cdot\frac{0.25}{0.75}=\frac{10}{11}<1\). Therefore, \(P(X=10)>P(X=11)\).

Answer

(1) False (2) False (3) True (4) True
53082012
An electronics manufacturer has found that \(5\%\) of its microchips are defective. Assume that chip defects are independent and that every chip has the same probability of being defective. Determine whether each statement is correct. (1) In a sample of \(100\) chips, the expected number of defective chips is \(5\). (2) The probability that none of the chips in a package of \(20\) is defective is \((0.95)^{20}\). (3) If \(3\) of \(20\) sampled chips are defective, this proves that the stated \(5\%\) defect rate must be wrong because the sample defect rate is \(15\%\). (4) A person stakes \(\$1\) on a randomly selected chip being defective. For a fair wager, the other person should stake \(\$19\) on the chip not being defective.

Hints

- Recall the mean of a binomial random variable. - Use independence to find the probability that all \(20\) chips are nondefective. - A sample proportion can differ from a population probability because of random variation. - A fair wager has an expected net gain of \(0\).

Solution

1. Statement (1) is true. For \(X\sim\operatorname{Bin}(100,0.05)\), \(E(X)=np=100\cdot0.05=5\). 2. Statement (2) is true. Each chip is nondefective with probability \(0.95\), so independence gives \(P(X=0)=(0.95)^{20}\). 3. Statement (3) is false. Random samples vary. In fact, \(P(X=3)=\binom{20}{3}(0.05)^3(0.95)^{17}\approx 0.0596\), so this result is possible under the stated model and does not by itself disprove the stated defect rate. 4. Statement (4) is true. If the bettor wins \(\$19\) when the chip is defective and loses \(\$1\) otherwise, the expected net gain is \(0.05\cdot19-0.95\cdot1=0\), so the wager is fair.

Answer

(1) True (2) True (3) False (4) True
53085612
A spinner has \(10\) equal sections labeled \(1\) through \(10\) and is spun twice. The random variable \(X\) is the number of spins that land on a multiple of \(3\). a) Create a probability distribution table for \(X\). b) Find the probability that the first or second result, or both, is divisible by \(3\). c) Compare the result from part b) with the probability that exactly one result is divisible by \(3\). Which event is more likely?

Hints

- Count the multiples of \(3\) from \(1\) through \(10\). - Model the two independent spins with a binomial distribution. - Translate “the first or second result, or both” into a condition on \(X\). - Use the table or a complement to find the probability of at least one success.

Solution

1. The multiples of \(3\) are \(3\), \(6\), and \(9\), so the success probability is \(p=\frac{3}{10}=0.3\). 2. The two spins are independent, so \(X\) is binomial with \(n=2\) and \(p=0.3\). Thus, \(P(X=0)=(0.7)^2=0.49\), \(P(X=1)=\binom{2}{1}(0.3)(0.7)=0.42\), and \(P(X=2)=(0.3)^2=0.09\). 3. The event in part b) is \(X\ge1\). Therefore, \(P(X\ge1)=1-P(X=0)=1-0.49=0.51\). 4. Exactly one divisible result has probability \(P(X=1)=0.42\). Since \(0.51>0.42\), at least one divisible result is more likely.

Answer

a) <table> <tr><td>\(k\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td></tr> <tr><td>\(P(X=k)\)</td><td>\(0.49\)</td><td>\(0.42\)</td><td>\(0.09\)</td></tr> </table> b) \(P(X\ge1)=0.51\) c) At least one result divisible by \(3\) is more likely because \(0.51>0.42\).
53087512
A basketball player independently takes \(5\) free throws and makes each free throw with probability \(p=0.7\). Find the probability that the player: 1. makes exactly \(3\) free throws; 2. makes at least \(4\) free throws.

Hints

- Model the number of makes with a binomial random variable. - Identify \(n\), \(p\), and the required number of successes. - Break “at least \(4\)” into exact values. - Use the binomial formula for each exact number of successes.

Solution

1. Let \(X\) be the number of made free throws. Then \(X\) is binomial with \(n=5\) and \(p=0.7\). 2. \(P(X=3)=\binom53(0.7)^3(0.3)^2=0.3087\). 3. \(P(X\ge4)=P(X=4)+P(X=5)\). 4. \(P(X=4)=\binom54(0.7)^4(0.3)=0.36015\), and \(P(X=5)=(0.7)^5=0.16807\). 5. Therefore, \(P(X\ge4)=0.36015+0.16807=0.52822\).

Answer

1. \(0.3087\) 2. \(0.52822\)
53088512
A spinner has \(5\) equal sections, and \(1\) section is blue. The spinner is spun \(6\) times independently. Let \(X\) be the number of times it lands on blue. Find each probability. (1) Exactly \(2\) times (2) At most \(1\) time (3) At least \(1\) time

Hints

- Identify the number of trials and the probability of blue on one spin. - How many paths contain exactly \(2\) blue results? - “At most \(1\)” includes which exact counts? - Use a complement for “at least \(1\).”

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(6,0.2)\). 2. \(P(X=2)=\binom{6}{2}\cdot (0.2)^2\cdot (0.8)^4=0.24576\). 3. \(P(X\le 1)=P(X=0)+P(X=1)=0.65536\). 4. Use the complement for part (3): \(P(X\ge 1)=1-P(X=0)=1-(0.8)^6=0.737856\).

Answer

(1) \(0.24576\), or \(24.576\%\) (2) \(0.65536\), or \(65.536\%\) (3) \(0.737856\), or \(73.7856\%\)
53088612
In a microchip manufacturing process, \(5\%\) of chips are defective. A random sample of \(10\) chips is tested, and outcomes are treated as independent. Let \(X\) be the number of defective chips. Find each probability. (1) Exactly \(1\) chip is defective. (2) No chips are defective. (3) More than \(2\) chips are defective.

Hints

- Identify the defect probability and sample size. - Use the binomial formula for an exact count. - For “more than \(2\),” consider the complement. - Do not round intermediate probabilities too early.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(10,0.05)\). 2. \(P(X=1)=\binom{10}{1}\cdot 0.05\cdot (0.95)^9\approx 0.3151\). 3. \(P(X=0)=(0.95)^{10}\approx 0.5987\). 4. Use the complement for part (3): \(P(X>2)=1-P(X\le 2)\approx 0.0115\), using unrounded probabilities in the subtraction.

Answer

(1) \(P(X=1)\approx 0.3151\), or about \(31.51\%\) (2) \(P(X=0)\approx 0.5987\), or about \(59.87\%\) (3) \(P(X>2)\approx 0.0115\), or about \(1.15\%\)
53088712
A multiple-choice test has \(8\) questions. Each question has \(4\) answer choices, exactly one of which is correct. A student guesses randomly on every question. Let \(X\) be the number of correct answers. a) Find the probability of exactly \(3\) correct answers. b) Find the probability of more than \(6\) correct answers.

Hints

- Does the probability of a correct guess change from question to question? - Use a binomial model for the number of correct answers. - “More than \(6\)” includes which two values?

Solution

1. Since a random guess is correct with probability \(0.25\), \(X\sim\operatorname{Bin}(8,0.25)\). 2. \(P(X=3)=\binom{8}{3}\cdot (0.25)^3\cdot (0.75)^5\approx 0.2076\). 3. More than \(6\) means \(7\) or \(8\) correct answers. Thus \(P(X>6)=P(X=7)+P(X=8)=\frac{25}{65536}\approx 0.000381\).

Answer

a) \(P(X=3)\approx 0.2076\), or about \(20.76\%\) b) \(P(X>6)=\frac{25}{65536}\approx 0.000381\), or about \(0.0381\%\)
53089512
A person claims to be able to predict whether a face-down playing card is red or black. In a test, \(12\) cards are shown one at a time. After each guess, the card is replaced and the deck is shuffled. Assume the person has no special ability and guesses randomly. Let \(X\) be the number of correct color guesses. a) Find the probability of exactly \(9\) correct guesses. b) Find the probability of at least \(10\) correct guesses.

Hints

- Decide whether the guesses are independent. - Identify the number of trials and the success probability. - Use the binomial formula for an exact count. - “At least \(10\)” includes three exact outcomes.

Solution

1. Each guess is correct with probability \(0.5\), and the trials are independent, so \(X\sim\operatorname{Bin}(12,0.5)\). 2. \(P(X=9)=\binom{12}{9}\cdot (0.5)^{12}=\frac{220}{4096}\approx 0.0537\). 3. For at least \(10\) correct guesses, add the probabilities for \(10\), \(11\), and \(12\): \(P(X\ge 10)=\frac{66+12+1}{4096}=\frac{79}{4096}\approx 0.0193\).

Answer

a) \(P(X=9)=\frac{220}{4096}\approx 0.0537\), or about \(5.37\%\) b) \(P(X\ge 10)=\frac{79}{4096}\approx 0.0193\), or about \(1.93\%\)
53089712
Ten equally skilled runners compete in a race with no ties. A spectator randomly predicts the runners who will finish first and second, in order. a) Find the probability \(p\) that one random prediction is completely correct. b) Sixty people make independent random predictions. Find the probability that exactly one prediction is correct.

Hints

- Count the ordered possibilities for two different runners in first and second place. - Model each prediction as correct or incorrect. - Identify \(n\), \(p\), and \(k\) for the binomial probability.

Solution

1. There are \(10\cdot9=90\) possible ordered pairs for first and second place. 2. Exactly one ordered pair is correct, so \(p=\frac{1}{90}\). 3. Let \(X\) be the number of correct predictions among the \(60\). Then \(X\) is binomial with \(n=60\) and \(p=\frac{1}{90}\). 4. \(P(X=1)=\binom{60}{1}\left(\frac{1}{90}\right)\left(\frac{89}{90}\right)^{59}\approx0.3448\).

Answer

a) \(p=\frac{1}{90}\approx0.0111\) b) \(P(X=1)\approx0.3448\), or about \(34.48\%\)
53089812
A factory produces electronic components with a defect rate of \(5\%\), and component statuses are independent. A quality-control sample of \(20\) components is selected. Let \(X\) be the number of defective components in the sample. a) Find the probability that exactly \(2\) components are defective. b) Find the probability that at most one component is defective. c) Calculate \(\binom{20}{2}\) and explain its meaning in this context.

Hints

- Use the binomial formula for one exact value. - “At most one” includes no defects and exactly one defect. - Interpret a binomial coefficient as a count of position choices.

Solution

1. The random variable \(X\) is binomial with \(n=20\) and \(p=0.05\). 2. For part a), \(P(X=2)=\binom{20}{2}(0.05)^2(0.95)^{18}\approx0.1887\). 3. For part b), \(P(X\le1)=P(X=0)+P(X=1)=(0.95)^{20}+\binom{20}{1}(0.05)(0.95)^{19}\approx0.7358\). 4. \(\binom{20}{2}=\frac{20\cdot19}{2}=190\). It counts the possible pairs of positions occupied by the two defective components among the \(20\) sampled components.

Answer

a) \(P(X=2)\approx0.1887\) b) \(P(X\le1)\approx0.7358\) c) \(\binom{20}{2}=190\). It counts the \(190\) possible position pairs for exactly two defective components.
53090012
A multiple-choice test has \(10\) questions. Each question has \(4\) answer choices, exactly one of which is correct. A student guesses randomly on every question. Let \(X\) be the number of correct answers. a) Find the probability of exactly \(3\) correct answers. b) How many correct/incorrect patterns contain exactly \(2\) correct answers? Use a binomial coefficient. c) Find the probability of at least \(1\) correct answer.

Hints

- Model each question as an independent correct-or-incorrect trial. - Use the binomial probability formula for part a). - A binomial coefficient counts the possible correct-answer positions in part b). - Use the complement of no correct answers for part c).

Solution

1. Since a random guess is correct with probability \(0.25\), \(X\sim\operatorname{Bin}(10,0.25)\). 2. \(P(X=3)=\binom{10}{3}\cdot (0.25)^3\cdot (0.75)^7\approx 0.2503\). 3. To obtain exactly \(2\) correct answers, choose the \(2\) correct-answer positions among the \(10\) questions. The number of correct/incorrect patterns is \(\binom{10}{2}=45\). 4. Use the complement for part c): \(P(X\ge 1)=1-P(X=0)=1-(0.75)^{10}\approx 0.9437\).

Answer

a) \(P(X=3)\approx 0.2503\), or about \(25.03\%\) b) \(\binom{10}{2}=45\) correct/incorrect patterns c) \(P(X\ge 1)\approx 0.9437\), or about \(94.37\%\)
53090412
A Bernoulli process has \(n=10\) independent trials with success probability \(p=0.5\). The random variable \(X\) is the number of successes. a) Determine \(P(X=2)\) and \(P(X=8)\). b) Use the binomial probability formula and a property of binomial coefficients to explain why the two probabilities are equal.

Hints

- Substitute the given values into the binomial probability formula. - Compare the success and failure probabilities when \(p=0.5\). - Recall the symmetry identity for \(\binom{n}{k}\).

Solution

1. \(P(X=2)=\binom{10}{2}(0.5)^2(0.5)^8=45(0.5)^{10}=\frac{45}{1024}\approx0.0439\). 2. \(P(X=8)=\binom{10}{8}(0.5)^8(0.5)^2=45(0.5)^{10}=\frac{45}{1024}\approx0.0439\). 3. The identity \(\binom{10}{2}=\binom{10}{8}\), together with \(p=1-p=0.5\), makes the corresponding binomial probability terms equal.

Answer

a) \(P(X=2)=P(X=8)=\frac{45}{1024}\approx0.0439\) b) Since \(\binom{10}{2}=\binom{10}{8}\) and both power products equal \((0.5)^{10}\), the probabilities are equal.
53090612
In a game show, a player chooses \(4\) numbers from \(18\). The player wins the grand prize only if all \(4\) chosen numbers match the \(4\) numbers drawn. Find the probability that a player wins the grand prize at least once in \(1000\) independent rounds, using one entry per round. Give the result as a percent rounded to two decimal places.

Hints

- First find the probability of winning one round. - For “at least once,” consider the complement of no wins. - Model the rounds as independent Bernoulli trials.

Solution

1. There are \(\binom{18}{4}=3060\) possible sets of \(4\) numbers, so the probability of winning one round is \(p=\frac{1}{3060}\). 2. The probability of no wins in \(1000\) independent rounds is \(\left(1-\frac{1}{3060}\right)^{1000}\). 3. Therefore, the probability of at least one win is \(1-\left(\frac{3059}{3060}\right)^{1000}\approx 0.278809\).

Answer

The probability is approximately \(27.88\%\).
53091112
A basketball player makes a free throw with probability \(0.75\). During practice, he takes \(6\) independent free throws. Let \(X\) be the number made. a) Find the probability that he makes exactly \(4\) free throws. b) Find the probability that he makes at least \(5\) free throws.

Hints

- Identify the success probability and number of trials. - Use the binomial formula for exactly \(4\) makes. - “At least \(5\)” includes two exact outcomes.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(6,0.75)\). 2. \(P(X=4)=\binom{6}{4}\cdot (0.75)^4\cdot (0.25)^2\approx 0.2966\). 3. At least \(5\) means \(5\) or \(6\). Thus \(P(X\ge 5)=\binom{6}{5}\cdot (0.75)^5\cdot 0.25+(0.75)^6\approx 0.5339\).

Answer

a) \(P(X=4)\approx 0.2966\), or about \(29.66\%\) b) \(P(X\ge 5)\approx 0.5339\), or about \(53.39\%\)
53091712
A spinner has \(20\) equal sections: \(10\) blue, \(6\) red, and \(4\) yellow. The spinner is spun \(15\) times independently. Find each probability. a) Exactly \(5\) blue results b) No yellow results c) More than \(2\) red results

Hints

- Treat the named color as success and all other colors as failure. - The success probability changes with the color being studied. - For “more than \(2\),” use the complement through \(2\). - The spin probabilities remain constant across trials.

Solution

1. For blue, the success probability is \(\frac{10}{20}=0.5\). Thus \(P(\text{exactly }5\text{ blue})=\binom{15}{5}\cdot (0.5)^{15}\approx 0.0916\). 2. The probability of not landing on yellow is \(0.8\), so \(P(\text{no yellow})=(0.8)^{15}\approx 0.0352\). 3. For red, the success probability is \(\frac{6}{20}=0.3\). Using the complement, \(P(\text{more than }2\text{ red})=1-P(X\le 2)\approx 0.8732\), where \(X\sim\operatorname{Bin}(15,0.3)\).

Answer

a) Approximately \(0.0916\), or \(9.16\%\) b) Approximately \(0.0352\), or \(3.52\%\) c) Approximately \(0.8732\), or \(87.32\%\)
53096912
A basketball player makes a free throw with probability \(p=0.75\). During practice, the player takes \(4\) free throws. The random variable \(X\) is the number of made free throws. a) Create a probability distribution table for \(X\). b) Find the probability that the player makes at least \(3\) free throws.

Hints

- Identify the possible values of \(X\). - Use a binomial model for repeated trials with two outcomes. - Apply the formula for exactly \(k\) successes. - Translate “at least \(3\)” into the relevant values of \(X\).

Solution

1. The number of made shots is binomial with \(n=4\) and \(p=0.75\). 2. Using \(P(X=k)=\binom{4}{k}(0.75)^k(0.25)^{4-k}\), the probabilities are \(P(X=0)=\frac{1}{256}\approx0.0039\), \(P(X=1)=\frac{12}{256}\approx0.0469\), \(P(X=2)=\frac{54}{256}\approx0.2109\), \(P(X=3)=\frac{108}{256}\approx0.4219\), and \(P(X=4)=\frac{81}{256}\approx0.3164\). 3. \(P(X\ge3)=P(X=3)+P(X=4)=\frac{108}{256}+\frac{81}{256}=\frac{189}{256}\approx0.7383\).

Answer

a) <table border="1"> <tr><td>\(k\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td></tr> <tr><td>\(P(X=k)\)</td><td>\(0.0039\)</td><td>\(0.0469\)</td><td>\(0.2109\)</td><td>\(0.4219\)</td><td>\(0.3164\)</td></tr> </table> b) \(P(X\ge3)=\frac{189}{256}\approx0.7383\)
53097012
Historically, \(10\%\) of the microchips produced by a factory are defective. A sample of \(5\) chips is selected from current production. The random variable \(Y\) is the number of defective chips in the sample. a) Determine the probability distribution of \(Y\). b) Calculate the expected value \(E(Y)\).

Hints

- Identify the distribution for a fixed number of trials with a constant defect probability. - List all possible numbers of defective chips in a sample of \(5\). - Use the direct expected-value formula for a binomial random variable.

Solution

1. Model \(Y\) as binomial with \(n=5\) and \(p=0.1\). 2. Using \(P(Y=k)=\binom{5}{k}(0.1)^k(0.9)^{5-k}\), the probabilities are \(P(Y=0)=0.59049\), \(P(Y=1)=0.32805\), \(P(Y=2)=0.0729\), \(P(Y=3)=0.0081\), \(P(Y=4)=0.00045\), and \(P(Y=5)=0.00001\). 3. For a binomial random variable, \(E(Y)=np=5\cdot0.1=0.5\).

Answer

a) <table border="1"> <tr><td>\(k\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td></tr> <tr><td>\(P(Y=k)\)</td><td>\(0.59049\)</td><td>\(0.32805\)</td><td>\(0.0729\)</td><td>\(0.0081\)</td><td>\(0.00045\)</td><td>\(0.00001\)</td></tr> </table> b) \(E(Y)=0.5\)
53098912
A spinner has four equal sections: one green section labeled “win” and three red sections labeled “loss.” The spinner is spun \(4\) times. a) Determine the probability distribution of the random variable \(X\), the number of wins. b) Determine the probability distribution of the random variable \(Y\), the number of losses. c) Compare the distributions. What relationship holds between \(P(X=k)\) and \(P(Y=4-k)\)?

Hints

- Relate the number of losses to the number of wins in \(4\) spins. - When the focus changes from wins to losses, the success and failure probabilities switch roles. - Recall the symmetry identity for binomial coefficients.

Solution

1. For \(X\), use a binomial distribution with \(n=4\) and \(p=0.25\). The probabilities for \(k=0,1,2,3,4\) are \(0.3164\), \(0.4219\), \(0.2109\), \(0.0469\), and \(0.0039\), respectively. 2. For \(Y\), use a binomial distribution with \(n=4\) and \(p=0.75\). The probabilities for \(k=0,1,2,3,4\) are \(0.0039\), \(0.0469\), \(0.2109\), \(0.4219\), and \(0.3164\), respectively. 3. Since every outcome with \(k\) wins has exactly \(4-k\) losses, \(P(X=k)=P(Y=4-k)\). The distributions contain the same probabilities in reverse order.

Answer

a) \(P(X=0)\approx0.3164\); \(P(X=1)\approx0.4219\); \(P(X=2)\approx0.2109\); \(P(X=3)\approx0.0469\); \(P(X=4)\approx0.0039\). b) \(P(Y=0)\approx0.0039\); \(P(Y=1)\approx0.0469\); \(P(Y=2)\approx0.2109\); \(P(Y=3)\approx0.4219\); \(P(Y=4)\approx0.3164\). c) The distributions are reflections of each other, and \(P(X=k)=P(Y=4-k)\).
53099012
A seed test finds that each seed of a certain type germinates independently with probability \(0.90\). A gardener plants \(10\) of these seeds. Let \(X\) be the number of seeds that germinate and \(Y\) be the number that do not germinate. a) Find the probability that at least \(9\) seeds germinate. b) Find the probability that at most \(1\) seed does not germinate. c) Explain without calculation why the answers to parts a) and b) must be equal.

Hints

- Relate each possible value of \(X\) to the corresponding value of \(Y\). - Germination and failure to germinate are complementary outcomes. - Restate “at least \(9\) of \(10\)” in terms of the number of failures.

Solution

1. For part a), \(X\sim\operatorname{Bin}(10,0.90)\), so \(P(X\ge 9)=P(X=9)+P(X=10)\approx 0.736099\). 2. For part b), \(Y\sim\operatorname{Bin}(10,0.10)\), so \(P(Y\le 1)=P(Y=0)+P(Y=1)\approx 0.736099\). 3. The events are identical: at least \(9\) of \(10\) seeds germinate exactly when at most \(1\) seed fails to germinate.

Answer

a) \(P(X\ge 9)\approx 0.7361\), or \(73.61\%\). b) \(P(Y\le 1)\approx 0.7361\), or \(73.61\%\). c) Both events describe \(0\) or \(1\) failure among \(10\) seeds, so they have the same probability.
53099212
At a garden center, \(10\%\) of tulip bulbs do not sprout. Among the bulbs that sprout, \(60\%\) bloom red, \(30\%\) bloom yellow, and \(10\%\) bloom white. One bulb is selected at random. a) Explain why recording the four outcomes—does not sprout, red, yellow, or white—is not a Bernoulli trial. b) Describe two different ways to define success so that the experiment becomes a Bernoulli trial. Find the success probability for each definition.

Hints

- How many relevant outcomes can a Bernoulli trial have? - Can several outcomes be grouped into one success category? - A bulb must sprout before its flower color can be observed.

Solution

1. A Bernoulli trial has exactly two relevant categories, success and failure. Recording four separate outcomes does not meet that condition. 2. One definition is “success means the bulb sprouts.” Then \(p=1-0.10=0.90\). 3. Another definition is “success means the bulb blooms red.” The bulb must first sprout and then bloom red, so \(p=0.90\cdot 0.60=0.54\).

Answer

a) The observation has four separate outcome categories rather than two. b) Example 1: success is “the bulb sprouts,” so \(p=0.90\). Example 2: success is “the bulb blooms red,” so \(p=0.54\).
53099912
A fair eight-sided die labeled \(1\) through \(8\) is rolled independently \(10\) times. Let \(X\) be the number of times an \(8\) is rolled. Find the probability of each event. (1) At least \(1\) eight is rolled. (2) More than \(2\) eights are rolled. (3) At most \(9\) eights are rolled. (4) Every roll is an eight.

Hints

- For each event, decide whether a direct calculation or a complement is simpler. - What does “at least \(1\)” mean in terms of the complement? - Use the binomial formula for exact counts. - How are “at most \(n-1\)” and “exactly \(n\)” related?

Solution

1. Let \(X\sim\operatorname{Bin}(10,\frac{1}{8})\). 2. For part (1), \(P(X\ge 1)=1-\left(\frac{7}{8}\right)^{10}\approx 0.736924\). 3. For part (2), \(P(X>2)=1-P(X\le 2)\approx 0.119502\). 4. For part (3), \(P(X\le 9)=1-P(X=10)=1-\left(\frac{1}{8}\right)^{10}\approx 0.9999999991\). 5. For part (4), \(P(X=10)=\left(\frac{1}{8}\right)^{10}\approx 9.313\times 10^{-10}\).

Answer

(1) \(P(X\ge 1)\approx 0.7369\) (2) \(P(X>2)\approx 0.1195\) (3) \(P(X\le 9)\approx 0.9999999991\) (4) \(P(X=10)\approx 9.313\times 10^{-10}\)
53100012
A spinner has \(5\) equal sections, exactly \(1\) of which is red. The spinner is spun independently \(12\) times. Let \(Y\) be the number of times it lands on red. Find the probability of each event. (1) At least \(1\) red result. (2) More than \(3\) red results. (3) At most \(10\) red results. (4) No red results.

Hints

- Identify \(n\) and \(p\). - For a “more than” event, a complement may be efficient. - Distinguish carefully among “less than,” “at most,” “more than,” and “at least.” - Use a binomial cumulative probability when appropriate.

Solution

1. Let \(Y\sim\operatorname{Bin}(12,0.20)\). 2. For part (1), \(P(Y\ge 1)=1-(0.80)^{12}\approx 0.931281\). 3. For part (2), \(P(Y>3)=1-P(Y\le 3)\approx 0.205431\). 4. For part (3), \(P(Y\le 10)=1-[P(Y=11)+P(Y=12)]\approx 0.9999997993\). 5. For part (4), \(P(Y=0)=(0.80)^{12}\approx 0.068719\).

Answer

(1) \(P(Y\ge 1)\approx 0.9313\) (2) \(P(Y>3)\approx 0.2054\) (3) \(P(Y\le 10)\approx 0.9999998\) (4) \(P(Y=0)\approx 0.0687\)
53100112
A machine produces parts, and \(20\%\) are defective. A sample of \(6\) parts is selected independently from ongoing production. The random variable \(X\) is the number of defective parts in the sample. a) Give the probability distribution of \(X\) in a table. b) Find each probability: (1) At most \(2\) parts are defective. (2) At least \(1\) part is defective.

Hints

- Identify \(n\) and \(p\). - Use the binomial probability formula for individual values. - Translate “at most” into a sum of appropriate probabilities. - Use a complement to simplify “at least one.”

Solution

1. Model \(X\) as binomial with \(n=6\) and \(p=0.2\). Use \(P(X=k)=\binom{6}{k}(0.2)^k(0.8)^{6-k}\). 2. Rounded to four decimal places, the distribution values are \(0.2621\), \(0.3932\), \(0.2458\), \(0.0819\), \(0.0154\), \(0.0015\), and \(0.0001\) for \(k=0\) through \(6\), respectively. 3. \(P(X\le2)=P(X=0)+P(X=1)+P(X=2)\approx0.2621+0.3932+0.2458=0.9011\). 4. Using a complement, \(P(X\ge1)=1-P(X=0)\approx1-0.2621=0.7379\).

Answer

a) <table border="1"> <tr><td>\(k\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td></tr> <tr><td>\(P(X=k)\)</td><td>\(0.2621\)</td><td>\(0.3932\)</td><td>\(0.2458\)</td><td>\(0.0819\)</td><td>\(0.0154\)</td><td>\(0.0015\)</td><td>\(0.0001\)</td></tr> </table> b) (1) \(P(X\le2)\approx0.9011\) (2) \(P(X\ge1)\approx0.7379\)
53100312
A multiple-choice test has \(10\) questions. Each question has \(2\) answer choices, exactly one of which is correct. A student guesses randomly on every question. Let \(X\) be the number of correct answers. a) Find the probability of exactly \(5\) correct answers. b) Find the probability of at least \(8\) correct answers. c) Is exactly \(5\) correct answers or more than \(5\) correct answers more likely? Justify your decision without calculating every point probability.

Hints

- Identify the binomial parameters. - “At least \(8\)” includes which exact values? - Use the symmetry of a binomial distribution with \(p=0.5\). - Compare the center probability with the combined probability on one side.

Solution

1. Since each random guess is correct with probability \(0.5\), \(X\sim\operatorname{Bin}(10,0.5)\). 2. \(P(X=5)=\binom{10}{5}\cdot (0.5)^{10}=\frac{252}{1024}\approx 0.2461\). 3. \(P(X\ge 8)=\left[\binom{10}{8}+\binom{10}{9}+\binom{10}{10}\right]\cdot (0.5)^{10}=\frac{56}{1024}=0.0547\). 4. Because \(p=0.5\), the distribution is symmetric about \(5\), so \(P(X<5)=P(X>5)\). Therefore, \(P(X>5)=\frac{1-P(X=5)}{2}=\frac{386}{1024}\approx 0.3770\), which is greater than \(P(X=5)\approx 0.2461\).

Answer

a) \(P(X=5)\approx 0.2461\), or about \(24.61\%\) b) \(P(X\ge 8)=0.0547\), or about \(5.47\%\) c) More than \(5\) correct answers is more likely: \(P(X>5)\approx 0.3770\), compared with \(P(X=5)\approx 0.2461\).
53100412
A faulty computer system generates a random sequence of \(8\) independent bits, where each bit is \(1\) with probability \(0.5\). Let \(X\) be the number of ones in the bit sequence. a) First consider a shorter sequence of \(3\) independent bits. Create a table showing the probability distribution of \(X\). b) For the original \(8\)-bit sequence, find the probability that the number of zeros is strictly greater than the number of ones.

Hints

- List all possible values for the number of ones in a \(3\)-bit sequence. - How many ones can an \(8\)-bit sequence have if it contains more zeros than ones? - When \(p=0.5\), each binomial term contains the same factor \((0.5)^n\).

Solution

1. For a \(3\)-bit sequence, \(X\sim\operatorname{Bin}(3,0.5)\). The probabilities are \(P(X=0)=0.125\), \(P(X=1)=0.375\), \(P(X=2)=0.375\), and \(P(X=3)=0.125\). 2. For \(8\) bits, more zeros than ones means at most \(3\) ones. 3. Therefore, \(P(X\le 3)=\frac{\binom{8}{0}+\binom{8}{1}+\binom{8}{2}+\binom{8}{3}}{2^8}=\frac{93}{256}\approx 0.363281\).

Answer

a) <table> <tr><td>\(k\)</td><td>0</td><td>1</td><td>2</td><td>3</td></tr> <tr><td>\(P(X=k)\)</td><td>\(0.125\)</td><td>\(0.375\)</td><td>\(0.375\)</td><td>\(0.125\)</td></tr> </table> b) The probability is \(\frac{93}{256}\approx 0.3633\), or \(36.33\%\).
53100512
Consider \(5\) independent Bernoulli trials with success probability \(p=\frac{1}{3}\). a) Determine the probability distribution of the random variable \(X\), the number of successes. Give the probabilities as fractions. b) Explain the relationship between the distribution of \(X\) and the binomial theorem using \(\left(\frac{1}{3}+\frac{2}{3}\right)^5\).

Hints

- Use the formula for exactly \(k\) successes. - Identify the failure probability. - Recall the general expansion of \((a+b)^n\). - Connect the sum of the expansion terms to the required sum of a probability distribution.

Solution

1. Here, \(n=5\), \(p=\frac{1}{3}\), and \(1-p=\frac{2}{3}\). 2. Using \(P(X=k)=\binom{5}{k}\left(\frac{1}{3}\right)^k\left(\frac{2}{3}\right)^{5-k}\), the probabilities are \(\frac{32}{243}\), \(\frac{80}{243}\), \(\frac{80}{243}\), \(\frac{40}{243}\), \(\frac{10}{243}\), and \(\frac{1}{243}\) for \(k=0\) through \(5\), respectively. 3. The binomial expansion \((p+q)^n=\sum_{k=0}^{n}\binom{n}{k}p^kq^{n-k}\) contains exactly the probability terms \(P(X=k)\). Since \(p+q=1\), \(\left(\frac{1}{3}+\frac{2}{3}\right)^5=1\), matching the fact that all probabilities in the distribution add to \(1\).

Answer

a) \(P(X=0)=\frac{32}{243}\); \(P(X=1)=\frac{80}{243}\); \(P(X=2)=\frac{80}{243}\); \(P(X=3)=\frac{40}{243}\); \(P(X=4)=\frac{10}{243}\); \(P(X=5)=\frac{1}{243}\). b) The terms in the expansion of \(\left(\frac{1}{3}+\frac{2}{3}\right)^5\) are the probabilities \(P(X=k)\). Their sum is \(1\).
53100612
A multiple-choice test has \(4\) questions, each with \(5\) answer choices and exactly one correct answer. A student guesses independently on every question. The random variable \(X\) is the number of correct answers. a) Create a probability distribution table for \(X\). b) Find the probability of at least two correct answers. c) Which binomial coefficient counts the answer patterns with exactly two correct answers? Evaluate it.

Hints

- Find the probability of guessing one question correctly. - Translate “at least two” into the possible values of \(X\). - Decide whether direct addition or a complement is more efficient. - Identify the factor in the binomial formula that counts arrangements.

Solution

1. The number of correct answers is binomial with \(n=4\), \(p=\frac{1}{5}=0.2\), and \(1-p=0.8\). 2. The distribution is \(P(X=0)=0.4096\), \(P(X=1)=0.4096\), \(P(X=2)=0.1536\), \(P(X=3)=0.0256\), and \(P(X=4)=0.0016\). 3. \(P(X\ge2)=P(X=2)+P(X=3)+P(X=4)=0.1536+0.0256+0.0016=0.1808\). 4. The coefficient is \(\binom{4}{2}=\frac{4\cdot3}{2\cdot1}=6\).

Answer

a) <table> <tr><td>\(k\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td></tr> <tr><td>\(P(X=k)\)</td><td>\(0.4096\)</td><td>\(0.4096\)</td><td>\(0.1536\)</td><td>\(0.0256\)</td><td>\(0.0016\)</td></tr> </table> b) \(P(X\ge2)=0.1808\) c) \(\binom{4}{2}=6\)
53100712
A manufacturer finds that each Bluetooth headphone independently develops a battery defect during the warranty period with probability \(0.15\). A sample of \(12\) headphones is selected. Find the probability that a) exactly \(3\) headphones have a battery defect, b) at most \(2\) headphones have a battery defect, c) more than \(4\) headphones have a battery defect.

Hints

- Identify the parameters of the binomial distribution. - What range of values is included in “at most \(2\)”? - For “more than \(4\),” consider the complement. - Decide whether each part requires one probability or a sum.

Solution

1. Let \(X\sim\operatorname{Bin}(12,0.15)\). 2. For part a), \(P(X=3)=\binom{12}{3}(0.15)^3(0.85)^9\approx 0.171976\). 3. For part b), \(P(X\le 2)\approx 0.735818\). 4. For part c), \(P(X>4)=1-P(X\le 4)\approx 0.023922\).

Answer

a) \(P(X=3)\approx 17.20\%\) b) \(P(X\le 2)\approx 73.58\%\) c) \(P(X>4)\approx 2.39\%\)
53100812
In a large city, \(40\%\) of commuters regularly ride a bicycle to work. A sample of \(15\) commuters is surveyed. Assume the commuter responses are independent. Find the probability that a) exactly \(6\) regularly ride a bicycle to work, b) fewer than \(5\) regularly ride a bicycle to work, c) at least \(8\) regularly ride a bicycle to work.

Hints

- Identify the success probability and number of trials. - Translate verbal conditions into inequalities involving \(X\). - Use cumulative binomial probabilities for ranges of outcomes. - For an “at least” event, consider its complement.

Solution

1. Let \(X\sim\operatorname{Bin}(15,0.40)\). 2. For part a), \(P(X=6)=\binom{15}{6}(0.40)^6(0.60)^9\approx 0.206598\). 3. For part b), \(P(X<5)=P(X\le 4)\approx 0.217278\). 4. For part c), \(P(X\ge 8)=1-P(X\le 7)\approx 0.213103\).

Answer

a) \(P(X=6)\approx 20.66\%\) b) \(P(X<5)\approx 21.73\%\) c) \(P(X\ge 8)\approx 21.31\%\)
53100912
A target shooter independently takes \(5\) shots in each series and hits the target with probability \(0.80\) on each shot. A five-shot series is called highly successful if the shooter gets at least \(4\) hits. The shooter completes \(12\) independent five-shot series. Find the probability that: 1. one series is highly successful; 2. exactly \(8\) of the \(12\) series are highly successful; 3. none of the \(12\) series is highly successful.

Hints

- First find the success probability for one five-shot series. - Use that result as the success probability for the \(12\) repeated series. - Identify the binomial model at each level. - Distinguish “exactly,” “at least,” and “none.”

Solution

1. Let \(X\) be the number of hits in one series. Then \(P(X\ge4)=P(X=4)+P(X=5)=\binom54(0.8)^4(0.2)+(0.8)^5=0.73728\). 2. Let \(Y\) be the number of highly successful series among the \(12\). Then \(Y\) is binomial with \(n=12\) and \(p=0.73728\). 3. \(P(Y=8)=\binom{12}{8}(0.73728)^8(1-0.73728)^4\approx0.2059\). 4. \(P(Y=0)=(1-0.73728)^{12}\approx1.08\times10^{-7}\).

Answer

1. \(0.73728\) 2. \(P(Y=8)\approx0.2059\), or about \(20.59\%\) 3. \(P(Y=0)\approx1.08\times10^{-7}\)
53101112
In a large city, each teenager independently owns a tablet with probability \(0.80\). A sample of \(50\) teenagers is selected. Let \(X\) be the number who own a tablet, and let \(Y=50-X\) be the number who do not own a tablet. Use \(Y\sim\operatorname{Bin}(50,0.20)\) to find the probability of each event. (1) Exactly \(42\) teenagers own a tablet. (2) At least \(38\) teenagers own a tablet. (3) At most \(44\) teenagers own a tablet.

Hints

- Use the complementary count to replace a success probability greater than \(0.5\) with one less than \(0.5\). - Translate each condition on \(X\) into a condition on \(Y\). - Track carefully which endpoint values are included. - The numbers who own and do not own a tablet always sum to \(50\).

Solution

1. Since \(Y=50-X\), \(P(X=42)=P(Y=8)\approx 0.116922\). 2. Also, \(P(X\ge 38)=P(Y\le 12)\approx 0.813943\). 3. Finally, \(P(X\le 44)=P(Y\ge 6)=1-P(Y\le 5)\approx 0.951973\).

Answer

(1) \(P(X=42)\approx 0.1169\) (2) \(P(X\ge 38)\approx 0.8139\) (3) \(P(X\le 44)\approx 0.9520\)
53101212
An online retailer finds that \(75\%\) of its customers place orders by smartphone. A random sample of \(100\) completed orders is analyzed, and order types are treated as independent. Let \(X\) be the number placed by smartphone. Find each probability. (1) Fewer than \(70\) smartphone orders (2) Between \(72\) and \(80\) smartphone orders, inclusive (3) More than \(80\) smartphone orders

Hints

- Translate strict inequalities carefully. - Use a difference of cumulative probabilities for the interval. - A binomial cumulative distribution function gives probabilities through an inclusive integer endpoint. - Use a complement for “more than \(80\).”

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(100,0.75)\). 2. \(P(X<70)=P(X\le 69)\approx 0.1038\). 3. \(P(72\le X\le 80)=P(X\le 80)-P(X\le 71)\approx 0.6929\). 4. \(P(X>80)=1-P(X\le 80)\approx 0.0995\).

Answer

(1) \(P(X<70)\approx 0.1038\) (2) \(P(72\le X\le 80)\approx 0.6929\) (3) \(P(X>80)\approx 0.0995\)
53101512
A manufacturer knows that \(5\%\) of its electronic components are defective. A random sample of \(200\) components is selected, and defect outcomes are treated as independent. Let \(X\) be the number of defective components. Find each probability. (1) At most \(10\) components are defective. (2) At least \(8\) and at most \(12\) components are defective. (3) More than \(15\) components are defective.

Hints

- Identify the binomial model. - Translate each verbal boundary into an inequality. - Use a difference of cumulative probabilities for the interval. - Use a complement for the upper tail.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(200,0.05)\). 2. \(P(X\le 10)\approx 0.5831\). 3. \(P(8\le X\le 12)=P(X\le 12)-P(X\le 7)\approx 0.5832\). 4. Use the complement for part (3): \(P(X>15)=1-P(X\le 15)\approx 0.0444\).

Answer

(1) \(P(X\le 10)\approx 0.5831\) (2) \(P(8\le X\le 12)\approx 0.5832\) (3) \(P(X>15)\approx 0.0444\)
53101612
A multiple-choice test has \(50\) questions. Each question has \(4\) answer choices, exactly one of which is correct. A student guesses randomly on every question. Let \(X\) be the number of correct answers. Find each probability. (1) At least \(15\) correct answers (2) Fewer than \(10\) correct answers (3) Between \(10\) and \(15\) correct answers, inclusive

Hints

- Find the success probability for a random guess. - Use a complement for “at least \(15\).” - “Fewer than \(10\)” means through \(9\). - Use a difference of cumulative probabilities for the interval.

Solution

1. Since a random guess is correct with probability \(0.25\), \(X\sim\operatorname{Bin}(50,0.25)\). 2. \(P(X\ge 15)=1-P(X\le 14)\approx 0.2519\). 3. \(P(X<10)=P(X\le 9)\approx 0.1637\). 4. \(P(10\le X\le 15)=P(X\le 15)-P(X\le 9)\approx 0.6732\).

Answer

(1) \(P(X\ge 15)\approx 0.2519\) (2) \(P(X<10)\approx 0.1637\) (3) \(P(10\le X\le 15)\approx 0.6732\)
53101712
A multiple-choice test has \(60\) questions. Each question has \(4\) answer choices, exactly one of which is correct. A student guesses randomly on every question. Let \(X\) be the number of correct answers. Find each probability. a) More than \(12\) correct answers b) At least \(10\) and at most \(20\) correct answers c) Fewer than \(40\) incorrect answers

Hints

- Find the probability of a correct guess. - Use complements for upper-tail events. - Use cumulative differences for an interval. - Convert the number incorrect into a condition on the number correct.

Solution

1. Since a random guess is correct with probability \(0.25\), \(X\sim\operatorname{Bin}(60,0.25)\). 2. \(P(X>12)=1-P(X\le 12)\approx 0.7684\). 3. \(P(10\le X\le 20)=P(X\le 20)-P(X\le 9)\approx 0.9007\). 4. Fewer than \(40\) incorrect answers means at least \(21\) correct answers. Therefore, \(P(X\ge 21)=1-P(X\le 20)\approx 0.0541\).

Answer

a) \(P(X>12)\approx 0.7684\) b) \(P(10\le X\le 20)\approx 0.9007\) c) \(P(X\ge 21)\approx 0.0541\)
53101812
In a region, \(12\%\) of households no longer have a home landline. A random sample of \(100\) households is selected, and household statuses are treated as independent. Let \(X\) be the number without a landline. a) Find each probability. (1) At most \(10\) households have no landline. (2) More than \(15\) households have no landline. (3) Between \(8\) and \(14\) households, inclusive, have no landline. b) Find the probability that at least \(90\) households still have a landline.

Hints

- Identify the binomial parameters. - Use a cumulative difference for the inclusive interval. - A household either has a landline or does not. - Relate the number with landlines to the number without landlines.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(100,0.12)\). 2. \(P(X\le 10)\approx 0.3337\). 3. \(P(X>15)=1-P(X\le 15)\approx 0.1414\). 4. \(P(8\le X\le 14)=P(X\le 14)-P(X\le 7)\approx 0.7079\). 5. At least \(90\) households with a landline is equivalent to at most \(10\) households without one. Therefore, the probability is also \(P(X\le 10)\approx 0.3337\).

Answer

a) (1) \(P(X\le 10)\approx 0.3337\) (2) \(P(X>15)\approx 0.1414\) (3) \(P(8\le X\le 14)\approx 0.7079\) b) \(P(\text{at least }90\text{ with a landline})\approx 0.3337\)
53102112
A gardener plants \(60\) seeds of a flower variety. Each seed germinates with probability \(0.75\), independently of the others. Let \(X\) be the number that germinate. Find each probability. 1. At most \(40\) seeds germinate. 2. More than \(50\) seeds germinate. 3. At least \(42\) and at most \(48\) seeds germinate.

Hints

- Identify the binomial parameters. - Use a complement for “more than \(50\).” - Use a cumulative difference for the interval. - Check that both interval endpoints are included.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(60,0.75)\). 2. \(P(X\le 40)\approx 0.0925\). 3. \(P(X>50)=1-P(X\le 50)\approx 0.0452\). 4. \(P(42\le X\le 48)=P(X\le 48)-P(X\le 41)\approx 0.7038\).

Answer

1. \(P(X\le 40)\approx 0.0925\) 2. \(P(X>50)\approx 0.0452\) 3. \(P(42\le X\le 48)\approx 0.7038\)
53102212
In a microchip manufacturing process, \(5\%\) of chips are defective. A random sample of \(100\) chips is tested, and defect outcomes are treated as independent. Let \(X\) be the number of defective chips. Find each probability. 1. Exactly \(5\) chips are defective. 2. At least \(3\) but fewer than \(8\) chips are defective. 3. More than \(10\) chips are defective.

Hints

- Distinguish an exact probability from a cumulative probability. - “Fewer than \(8\)” means the largest included value is \(7\). - Use a cumulative difference for the interval. - Use a complement for “more than \(10\).”

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(100,0.05)\). 2. \(P(X=5)=\binom{100}{5}\cdot (0.05)^5\cdot (0.95)^{95}\approx 0.1800\). 3. At least \(3\) but fewer than \(8\) means \(3\le X\le 7\). Thus \(P(3\le X\le 7)=P(X\le 7)-P(X\le 2)\approx 0.7538\). 4. Use the complement for part 3: \(P(X>10)=1-P(X\le 10)\approx 0.0115\).

Answer

1. \(P(X=5)\approx 0.1800\) 2. \(P(3\le X<8)\approx 0.7538\) 3. \(P(X>10)\approx 0.0115\)
53102312
An urn contains one red ball and two white balls. A ball is drawn repeatedly with replacement, and drawing red counts as a success. Consider these events: \(E_1\): Exactly \(1\) success in \(3\) draws. \(E_2\): Exactly \(2\) successes in \(3\) draws. \(E_3\): Exactly \(2\) successes in \(6\) draws. a) Calculate \(P(E_1)\) and \(P(E_2)\), and show that \(P(E_1)=2P(E_2)\). b) Determine whether \(P(E_1)\) is also exactly twice \(P(E_3)\).

Hints

- Find the success probability for one draw. - Use the binomial probability formula for an exact number of successes. - In part b), note that the number of trials changes. - Compare fractions using a common denominator.

Solution

1. The success probability is \(p=\frac{1}{3}\), and the failure probability is \(\frac{2}{3}\). 2. \(P(E_1)=\binom{3}{1}\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^2=\frac{4}{9}\). 3. \(P(E_2)=\binom{3}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)=\frac{2}{9}\). Therefore, \(P(E_1)=2P(E_2)\). 4. \(P(E_3)=\binom{6}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^4=\frac{80}{243}\). 5. Since \(P(E_1)=\frac{4}{9}=\frac{108}{243}\), it is not twice \(P(E_3)\). In fact, \(\frac{P(E_1)}{P(E_3)}=\frac{27}{20}=1.35\).

Answer

a) \(P(E_1)=\frac{4}{9}\) and \(P(E_2)=\frac{2}{9}\), so \(P(E_1)=2P(E_2)\). b) No. \(P(E_1)=\frac{108}{243}\) and \(P(E_3)=\frac{80}{243}\), so \(P(E_1)\ne2P(E_3)\).
53102512
Let \(X\sim\operatorname{Bin}(50,0.25)\). Find each probability. a) \(P(X\le 15)\) b) \(P(X\ge 10)\) c) \(P(8\le X\le 14)\)

Hints

- Express “at least” using a complement. - Use a difference of cumulative probabilities for the interval. - Check whether each boundary value is included before selecting cumulative endpoints. - Check the value immediately below the lower endpoint.

Solution

1. Using the binomial cumulative distribution function, \(P(X\le 15)\approx 0.8369\). 2. Use the complement for part b): \(P(X\ge 10)=1-P(X\le 9)\approx 0.8363\). 3. For part c), \(P(8\le X\le 14)=P(X\le 14)-P(X\le 7)\approx 0.7028\).

Answer

a) \(P(X\le 15)\approx 0.8369\) b) \(P(X\ge 10)\approx 0.8363\) c) \(P(8\le X\le 14)\approx 0.7028\)
53102612
Let \(X\sim\operatorname{Bin}(100,0.8)\). Find each probability. a) \(P(X\le 75)\) b) \(P(X>85)\) c) \(P(78<X<88)\)

Hints

- Use a complement for an upper-tail probability. - Identify the integer values inside the open interval. - Rewrite each event using cumulative probabilities of the form \(P(X\le k)\). - Express the interval as a difference of cumulative probabilities.

Solution

1. Using the binomial cumulative distribution function, \(P(X\le 75)\approx 0.1314\). 2. Use the complement for part b): \(P(X>85)=1-P(X\le 85)\approx 0.0804\). 3. Since \(X\) is integer-valued, \(78<X<88\) means \(79\le X\le 87\). Therefore, \(P(78<X<88)=P(X\le 87)-P(X\le 78)\approx 0.6287\).

Answer

a) \(P(X\le 75)\approx 0.1314\) b) \(P(X>85)\approx 0.0804\) c) \(P(78<X<88)\approx 0.6287\)
53102712
A basketball player independently makes each free throw with probability \(75\%\). During practice, the player attempts \(20\) free throws. Find the probability that (1) every free throw is made, (2) exactly \(15\) free throws are made, (3) more than \(15\) free throws are made.

Hints

- Use a binomial model for a fixed number of independent made-or-missed attempts. - Identify \(n\) and \(p\). - “More than \(15\)” includes the values \(16\) through \(20\); a complement may be useful.

Solution

1. Let \(X\sim\operatorname{Bin}(20,0.75)\). 2. For part (1), \(P(X=20)=(0.75)^{20}\approx 0.003171\). 3. For part (2), \(P(X=15)=\binom{20}{15}(0.75)^{15}(0.25)^5\approx 0.202331\). 4. For part (3), \(P(X>15)=1-P(X\le 15)\approx 0.414842\).

Answer

(1) The probability is approximately \(0.32\%\). (2) The probability is approximately \(20.23\%\). (3) The probability is approximately \(41.48\%\).
53102812
A factory produces electronic components. Each component is independently defective with probability \(0.05\). A sample of \(80\) components is selected for quality testing. Find the probability that the sample contains (1) no defective components, (2) exactly \(4\) defective components, (3) fewer than \(4\) defective components.

Hints

- Identify \(n\) and the probability that a component is defective. - Use the binomial formula for one exact count. - Which values are included in “fewer than \(4\)”?

Solution

1. Let \(X\sim\operatorname{Bin}(80,0.05)\). 2. For part (1), \(P(X=0)=(0.95)^{80}\approx 0.016515\). 3. For part (2), \(P(X=4)=\binom{80}{4}(0.05)^4(0.95)^{76}\approx 0.200431\). 4. For part (3), \(P(X<4)=P(X\le 3)\approx 0.428449\).

Answer

(1) The probability is approximately \(1.65\%\). (2) The probability is approximately \(20.04\%\). (3) The probability is approximately \(42.84\%\).
53102912
A basketball player attempts \(25\) independent shots and makes each shot with probability \(45\%\). Let \(X\) be the number of made shots. For each event, state its complement in words and find the probability of the complement. (1) \(E_1\): More than \(10\) made shots. (2) \(E_2\): At most \(14\) made shots.

Hints

- Identify the outcomes that do not belong to each event. - Pay attention to whether the boundary value is included. - Use cumulative binomial probabilities. - Express “at least \(k\)” using “at most \(k-1\)” when useful.

Solution

1. Let \(X\sim\operatorname{Bin}(25,0.45)\). 2. The complement of \(E_1\) is “at most \(10\) made shots.” Its probability is \(P(X\le 10)\approx 0.384260\). 3. The complement of \(E_2\) is “at least \(15\) made shots.” Its probability is \(P(X\ge 15)=1-P(X\le 14)\approx 0.095983\).

Answer

(1) \(\overline{E_1}\): At most \(10\) made shots; \(P(\overline{E_1})\approx 0.3843\). (2) \(\overline{E_2}\): At least \(15\) made shots; \(P(\overline{E_2})\approx 0.0960\).
53103012
In a survey of \(50\) people, each person independently supports a statement with probability \(0.2\). Let \(X\) be the number who support it. For each event, describe the complement and find the complement's probability. (1) \(E_1\): At least \(12\) people support the statement. (2) \(E_2\): More than \(5\) but at most \(12\) people support the statement.

Hints

- The probabilities of an event and its complement add to \(1\). - The complement of an interval consists of the values below it or above it. - Translate each verbal condition into inequalities before calculating.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(50,0.2)\). 2. The complement of \(E_1\) is \(X\le 11\), meaning at most \(11\) people support the statement. Its probability is \(P(X\le 11)\approx 0.7107\). 3. Event \(E_2\) is \(6\le X\le 12\). Its complement is \(X\le 5\) or \(X\ge 13\). 4. First find \(P(E_2)=P(X\le 12)-P(X\le 5)\approx 0.7659\). Therefore, \(P(E_2^c)=1-P(E_2)\approx 0.2341\).

Answer

(1) \(E_1^c\): At most \(11\) people support the statement; \(P(E_1^c)\approx 0.7107\) (2) \(E_2^c\): At most \(5\) or at least \(13\) people support the statement; \(P(E_2^c)\approx 0.2341\)
53103212
A new neighborhood has \(150\) households that use the same internet provider. At a randomly selected time during the evening, each household independently uses the service at full bandwidth with probability \(0.20\). Let \(X\) be the number using full bandwidth at that time. What is the minimum number of households the network must be able to serve at full bandwidth simultaneously so that demand is met with probability at least \(95\%\)?

Hints

- Define the random variable and identify its possible values. - Translate the capacity requirement into a cumulative-probability inequality. - Use the mean and spread to identify reasonable values to test. - Check consecutive capacity values to verify the minimum.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(150,0.20)\). 2. Seek the smallest integer \(k\) such that \(P(X\le k)\ge 0.95\). 3. The mean and standard deviation are \(\mu=np=30\) and \(\sigma=\sqrt{np(1-p)}=\sqrt{24}\approx 4.899\), which help locate the threshold. 4. Using cumulative binomial probabilities, \(P(X\le 37)\approx 0.9341<0.95\), while \(P(X\le 38)\approx 0.9554\ge 0.95\). 5. Therefore, the network must support at least \(38\) households simultaneously at full bandwidth.

Answer

The network must support at least \(38\) households simultaneously.
53103312
In a quality-control process, each electronic component is assumed to be independently defective with probability \(0.06\). A sample of \(150\) components is selected. Let \(X\) be the number of defective components. Find the probability that a) exactly \(9\) components are defective, b) at most \(5\) components are defective, c) at least \(12\) components are defective.

Hints

- Use a binomial model for defective versus nondefective components. - Determine which values are included in “at most” and “at least.” - For “at least,” a complement may reduce the calculation. - Use binomial probability and cumulative probability functions as appropriate.

Solution

1. Let \(X\sim\operatorname{Bin}(150,0.06)\). 2. For part a), \(P(X=9)=\binom{150}{9}(0.06)^9(0.94)^{141}\approx 0.135891\). 3. For part b), \(P(X\le 5)\approx 0.108296\). 4. For part c), \(P(X\ge 12)=1-P(X\le 11)\approx 0.190893\).

Answer

a) \(P(X=9)\approx 13.59\%\) b) \(P(X\le 5)\approx 10.83\%\) c) \(P(X\ge 12)\approx 19.09\%\)
53103412
In a region, \(25\%\) of households use solar energy to generate electricity. A random sample of \(80\) households is selected, and household statuses are treated as independent. Let \(X\) be the number that use solar energy. Find each probability. a) More than \(20\) households use solar energy. b) Fewer than \(15\) households use solar energy. c) At least \(18\) but at most \(22\) households use solar energy.

Hints

- Distinguish “more than” from “at least.” - “Fewer than \(15\)” ends at \(14\). - Use a cumulative difference for the interval. - Use a complement for the upper tail.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(80,0.25)\). 2. \(P(X>20)=1-P(X\le 20)\approx 0.4403\). 3. \(P(X<15)=P(X\le 14)\approx 0.0740\). 4. \(P(18\le X\le 22)=P(X\le 22)-P(X\le 17)\approx 0.4811\).

Answer

a) \(P(X>20)\approx 0.4403\) b) \(P(X<15)\approx 0.0740\) c) \(P(18\le X\le 22)\approx 0.4811\)
53103512
A hotel has \(80\) rooms. Historical data show that \(10\%\) of guests with reservations do not arrive. For one night, the hotel accepts \(85\) reservations. Let \(X\) be the number of guests who arrive, and treat guest arrivals as independent. Find each probability. 1. At most \(80\) guests arrive, so everyone receives a room. 2. More than \(80\) guests arrive. 3. Exactly \(80\) guests arrive. 4. At least \(70\) but at most \(75\) guests arrive.

Hints

- Use the probability that a guest arrives, not the no-show probability, for \(p\). - Parts 1 and 2 are complementary events. - Use the binomial formula for exactly \(80\). - Use a cumulative difference for the interval.

Solution

1. Since each guest arrives with probability \(0.9\), \(X\sim\operatorname{Bin}(85,0.9)\). 2. \(P(X\le 80)\approx 0.9357\). 3. Use the complement for part 2: \(P(X>80)=1-P(X\le 80)\approx 0.0643\). 4. \(P(X=80)=\binom{85}{80}\cdot (0.9)^{80}\cdot (0.1)^5\approx 0.0717\). 5. \(P(70\le X\le 75)=P(X\le 75)-P(X\le 69)\approx 0.3338\).

Answer

1. \(P(X\le 80)\approx 0.9357\) 2. \(P(X>80)\approx 0.0643\) 3. \(P(X=80)\approx 0.0717\) 4. \(P(70\le X\le 75)\approx 0.3338\)
53103612
A computer-chip manufacturer reports that \(4\%\) of its chips are defective. A random sample of \(200\) chips is tested, and defect outcomes are treated as independent. Let \(X\) be the number of defective chips. Find each probability. 1. At most \(5\) chips are defective. 2. More than \(10\) chips are defective. 3. At least \(6\) but at most \(12\) chips are defective. 4. The number of defective chips equals the mean of \(X\).

Hints

- Identify the binomial model. - Find the mean before completing part 4. - For the interval, subtract the cumulative probability through \(5\). - Use a complement for “more than \(10\).”

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(200,0.04)\). 2. \(P(X\le 5)\approx 0.1856\). 3. \(P(X>10)=1-P(X\le 10)\approx 0.1800\). 4. \(P(6\le X\le 12)=P(X\le 12)-P(X\le 5)\approx 0.7545\). 5. The mean is \(\mu=np=200\cdot 0.04=8\). Therefore, \(P(X=8)=\binom{200}{8}\cdot (0.04)^8\cdot (0.96)^{192}\approx 0.1425\).

Answer

1. \(P(X\le 5)\approx 0.1856\) 2. \(P(X>10)\approx 0.1800\) 3. \(P(6\le X\le 12)\approx 0.7545\) 4. \(P(X=8)\approx 0.1425\)
53103712
A gardener plants \(60\) seeds of a flower variety. Each seed germinates with probability \(0.8\), independently of the others. Let \(X\) be the number that germinate. Find each probability. (1) Exactly \(50\) seeds germinate. (2) Fewer than \(45\) seeds germinate. (3) At least \(52\) seeds germinate. (4) Between \(46\) and \(54\) seeds germinate, inclusive.

Hints

- Identify the binomial parameters. - Translate “fewer than” carefully. - Use a complement for “at least.” - Use a cumulative difference for the interval.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(60,0.8)\). 2. \(P(X=50)=\binom{60}{50}\cdot (0.8)^{50}\cdot (0.2)^{10}\approx 0.1102\). 3. \(P(X<45)=P(X\le 44)\approx 0.1306\). 4. \(P(X\ge 52)=1-P(X\le 51)\approx 0.1268\). 5. \(P(46\le X\le 54)=P(X\le 54)-P(X\le 45)\approx 0.7814\).

Answer

(1) \(P(X=50)\approx 0.1102\) (2) \(P(X<45)\approx 0.1306\) (3) \(P(X\ge 52)\approx 0.1268\) (4) \(P(46\le X\le 54)\approx 0.7814\)
53103812
In a large population, \(12\%\) of people are left-handed. A random sample of \(150\) people is selected, and handedness outcomes are treated as independent. Let \(X\) be the number who are left-handed. Find each probability. (1) Exactly \(18\) people are left-handed. (2) At most \(15\) people are left-handed. (3) More than \(20\) people are left-handed. (4) At least \(10\) and at most \(25\) people are left-handed.

Hints

- Identify the binomial parameters and each target range. - Use a complement for “more than \(20\).” - Use a cumulative difference for the interval. - Check the value immediately below the lower endpoint.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(150,0.12)\). 2. \(P(X=18)\approx 0.0998\). 3. \(P(X\le 15)\approx 0.2715\). 4. \(P(X>20)=1-P(X\le 20)\approx 0.2587\). 5. \(P(10\le X\le 25)=P(X\le 25)-P(X\le 9)\approx 0.9543\).

Answer

(1) \(P(X=18)\approx 0.0998\) (2) \(P(X\le 15)\approx 0.2715\) (3) \(P(X>20)\approx 0.2587\) (4) \(P(10\le X\le 25)\approx 0.9543\)
53103912
A gardener plants \(80\) flower bulbs of one variety. Each bulb independently fails to sprout with probability \(0.15\). The gardener needs at least \(65\) flowers for a garden design. Find the probability that the \(80\) planted bulbs produce at least \(65\) flowers.

Hints

- State the condition needed for the gardener to have enough flowers. - Define a random variable for either sprouting bulbs or failed bulbs. - Translate at least \(65\) successes into a condition on failures. - Use a cumulative binomial probability.

Solution

1. Let \(Y\) be the number of bulbs that fail to sprout. Then \(Y\sim\operatorname{Bin}(80,0.15)\). 2. At least \(65\) bulbs sprout exactly when at most \(15\) fail to sprout. 3. Therefore, \(P(Y\le 15)\approx 0.862466\).

Answer

The probability is approximately \(86.25\%\).
53104112
A collectible card company reports that one out of every seven sealed packs contains a limited-edition card. A random sample of \(20\) packs is selected from a very large production run. a) Explain why this process can be modeled as a binomial setting. b) Identify \(n\) and \(p\) when \(X\) is the number of packs containing a limited-edition card. c) Find the probability that exactly \(3\) of the \(20\) packs contain a limited-edition card. Give the expression and round the result to four decimal places.

Hints

- Recall the conditions for a binomial setting. - Define success in this context. - Use the binomial formula for an exact number of successes. - Calculate the binomial coefficient for the possible positions of the three special packs.

Solution

1. Each pack has two outcomes for this question: it contains a limited-edition card or it does not. Because the production run is very large, the probability is approximately constant, and the selections can be treated as independent. 2. The parameters are \(n=20\) and \(p=\frac17\). 3. \(P(X=3)=\binom{20}{3}\left(\frac17\right)^3\left(\frac67\right)^{17}\). 4. Evaluating gives \(P(X=3)\approx0.2418\).

Answer

a) There are two outcomes per pack, the success probability is approximately constant, and the selections are approximately independent. b) \(n=20\); \(p=\frac17\) c) \(P(X=3)=\binom{20}{3}\left(\frac17\right)^3\left(\frac67\right)^{17}\approx0.2418\)
53105312
A spinner has four equal sections, exactly one of which is red. The spinner is spun \(40\) times. The random variable \(X\) is the number of times the spinner lands on red, so \(p=0.25\). a) Calculate the expected value \(\mu\). b) A binomial distribution has its greatest probability at \(X=\mu\) when \(\mu\) is an integer. Calculate \(P(X=10)\). c) Calculate \(P(X=9)\) and \(P(X=11)\), and compare them with \(P(X=10)\).

Hints

- Use the expected-value formula for a binomial random variable. - Apply the binomial probability formula to an exact number of successes. - A maximum probability must be at least as large as the probabilities at neighboring values.

Solution

1. The expected value is \(\mu=np=40\cdot0.25=10\). 2. Using the binomial probability formula, \(P(X=10)=\binom{40}{10}(0.25)^{10}(0.75)^{30}\approx0.1444\). 3. Similarly, \(P(X=9)=\binom{40}{9}(0.25)^9(0.75)^{31}\approx0.1397\), and \(P(X=11)=\binom{40}{11}(0.25)^{11}(0.75)^{29}\approx0.1312\). 4. Since \(0.1444>0.1397\) and \(0.1444>0.1312\), \(P(X=10)\) is greater than both neighboring probabilities.

Answer

a) \(\mu=10\) b) \(P(X=10)\approx0.1444\) c) \(P(X=9)\approx0.1397\) and \(P(X=11)\approx0.1312\); both are less than \(P(X=10)\).
53105412
A treatment is effective for a patient with probability \(p=0.7\). In a study, the treatment is given to \(15\) patients, and the outcomes are assumed to be independent. a) Find the expected value \(\mu\) and the failure probability \(q=1-p\). b) When \(\mu\) is not an integer, a mode \(k\) lies in the interval \([\mu-q, \mu+p]\). Find this interval and identify the integer \(k\) in it. c) Calculate \(P(X=k)\), \(P(X=k-1)\), and \(P(X=k+1)\) to verify that \(k\) is the most likely number of patients for whom the treatment is effective.

Hints

- First determine whether the expected value is an integer. - Calculate the two interval endpoints carefully. - Compare the candidate value with its immediate neighbors.

Solution

1. The expected value is \(\mu=np=15\cdot0.7=10.5\), and \(q=1-0.7=0.3\). 2. The interval is \([\mu-q, \mu+p]=[10.5-0.3, 10.5+0.7]=[10.2, 11.2]\). 3. The only integer in the interval is \(k=11\). 4. \(P(X=10)=\binom{15}{10}(0.7)^{10}(0.3)^5\approx0.2061\). 5. \(P(X=11)=\binom{15}{11}(0.7)^{11}(0.3)^4\approx0.2186\). 6. \(P(X=12)=\binom{15}{12}(0.7)^{12}(0.3)^3\approx0.1700\). 7. Since \(P(X=11)\) is greater than both neighboring probabilities, \(11\) is the mode.

Answer

a) \(\mu=10.5\); \(q=0.3\) b) \([10.2, 11.2]\); \(k=11\) c) \(P(X=10)\approx0.2061\), \(P(X=11)\approx0.2186\), and \(P(X=12)\approx0.1700\). Therefore, \(11\) is the most likely value.
53107112
At a large university, \(2.5\%\) of students have season tickets for the local basketball team. A lecture includes \(80\) randomly selected students, and ticket-holder statuses are treated as independent. Let \(X\) be the number who have season tickets. a) Find the probability that exactly \(2\) students have season tickets. b) Find the probability that at most \(3\) students have season tickets. c) Find the probability that at least \(1\) student has a season ticket.

Hints

- Identify the binomial parameters. - “At most \(3\)” includes four exact counts. - Use the complement of no season-ticket holders for part c).

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(80,0.025)\). 2. \(P(X=2)=\binom{80}{2}\cdot (0.025)^2\cdot (0.975)^{78}\approx 0.2741\). 3. \(P(X\le 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)\approx 0.8594\). 4. Use the complement for part c): \(P(X\ge 1)=1-P(X=0)=1-(0.975)^{80}\approx 0.8681\).

Answer

a) \(P(X=2)\approx 0.2741\), or about \(27.41\%\) b) \(P(X\le 3)\approx 0.8594\), or about \(85.94\%\) c) \(P(X\ge 1)\approx 0.8681\), or about \(86.81\%\)
53107212
A manufacturer reports that \(0.8\%\) of its microchips are defective. A random sample of \(300\) chips is tested, and defect outcomes are treated as independent. Let \(X\) be the number of defective chips. a) Find \(P(X=0)\), \(P(X=1)\), and \(P(X=2)\). b) Find the probability that more than \(3\) chips are defective. c) Suppose a sample of \(300\) chips is tested every day for \(365\) days. On how many days should the inspector expect to find no defective chips?

Hints

- Identify the defect probability. - “More than \(3\)” is not the same as “at least \(3\).” - For the expected number of days, multiply the daily probability by \(365\).

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(300,0.008)\). 2. The point probabilities are \(P(X=0)=(0.992)^{300}\approx 0.0898\), \(P(X=1)=300\cdot 0.008\cdot (0.992)^{299}\approx 0.2174\), and \(P(X=2)=\binom{300}{2}\cdot (0.008)^2\cdot (0.992)^{298}\approx 0.2621\). 3. Use the complement for part b): \(P(X>3)=1-P(X\le 3)\approx 0.2208\). 4. The expected number of days with no defective chips is \(365P(X=0)=365\cdot (0.992)^{300}\approx 32.79\).

Answer

a) \(P(X=0)\approx 0.0898\), \(P(X=1)\approx 0.2174\), and \(P(X=2)\approx 0.2621\) b) \(P(X>3)\approx 0.2208\) c) About \(33\) days
53109112
At a carnival prize booth, each ticket wins the grand prize with probability \(\frac{1}{200}\). A visitor buys \(200\) tickets. Assume the ticket outcomes are independent and have the same win probability. a) Find the probability that the visitor wins no grand prizes. Compare it with \(\frac1e\approx0.3679\). b) Find the probability that the visitor wins exactly one grand prize. c) Find the probability that the visitor wins at least two grand prizes.

Hints

- Identify \(n\) and \(p\). - Use the binomial formula for \(X=0\) and \(X=1\). - For at least two successes, subtract the cases with zero and one success from \(1\).

Solution

1. Let \(X\) be the number of grand prizes. Then \(X\) is binomial with \(n=200\) and \(p=\frac{1}{200}=0.005\). 2. \(P(X=0)=(0.995)^{200}\approx0.3670\), which differs from \(\frac1e\approx0.3679\) by about \(0.0009\). 3. \(P(X=1)=\binom{200}{1}(0.005)(0.995)^{199}=(0.995)^{199}\approx0.3688\). 4. Using the complement, \(P(X\ge2)=1-P(X=0)-P(X=1)\approx0.2642\).

Answer

a) \(P(X=0)\approx0.3670\); it is about \(0.0009\) less than \(\frac1e\) b) \(P(X=1)\approx0.3688\) c) \(P(X\ge2)\approx0.2642\)
53109912
At a coworking space, \(15\) members are present. Each member uses a soundproof phone booth for an average of \(15\) minutes during an hour. Assume the members' booth-use times are independent and uniformly distributed throughout the hour. At a randomly selected time, let \(X\) be the number of booths in use. a) Find the probability that at most \(4\) booths are needed. b) Find the probability that more than \(6\) booths are needed. c) What is the minimum number of booths needed so that all members can be accommodated with probability at least \(95\%\)?

Hints

- Convert one member's average booth-use time into a probability at a randomly selected time. - Model the number of simultaneous users with a binomial random variable. - Use a complement for “more than \(6\).” - For the minimum number of booths, test consecutive cumulative probabilities until the required confidence level is reached.

Solution

1. At a randomly selected time, one member is using a booth with probability \(p=\frac{15}{60}=0.25\). Thus \(X\sim\operatorname{Bin}(15,0.25)\). 2. For part a), \(P(X\le 4)\approx 0.6865\). 3. For part b), use the complement: \(P(X>6)=1-P(X\le 6)\approx 1-0.9434=0.0566\). 4. For part c), seek the smallest integer \(k\) such that \(P(X\le k)\ge 0.95\). Since \(P(X\le 6)\approx 0.9434<0.95\) and \(P(X\le 7)\approx 0.9827\ge 0.95\), the minimum is \(7\) booths.

Answer

a) \(P(X\le 4)\approx 0.6865\), or about \(68.65\%\) b) \(P(X>6)\approx 0.0566\), or about \(5.66\%\) c) \(7\) booths
53110012
In a warehouse, \(20\) employees work at packing stations. Each employee uses a powered pallet jack for an average of \(6\) minutes per hour. Assume the employees' use times are independent and uniformly distributed throughout the hour. At a randomly selected time, let \(X\) be the number of pallet jacks needed. a) Find the probability that exactly \(2\) pallet jacks are needed. b) Find the probability that more than \(4\) pallet jacks are needed. c) What is the minimum number of pallet jacks that must be available so that demand is met with probability at least \(99\%\)?

Hints

- Convert one employee's average use time into a probability at a randomly selected time. - Distinguish between an exact-count probability and a cumulative probability. - Use a complement for “more than \(4\).” - Test consecutive equipment counts with cumulative binomial probabilities to verify the minimum.

Solution

1. At a randomly selected time, one employee needs a pallet jack with probability \(p=\frac{6}{60}=0.1\). Thus \(X\sim\operatorname{Bin}(20,0.1)\). 2. For part a), \(P(X=2)=\binom{20}{2}\cdot(0.1)^2\cdot(0.9)^{18}\approx 0.2852\). 3. For part b), use the complement: \(P(X>4)=1-P(X\le 4)\approx 1-0.9568=0.0432\). 4. For part c), seek the smallest integer \(k\) such that \(P(X\le k)\ge 0.99\). Since \(P(X\le 5)\approx 0.9887<0.99\) and \(P(X\le 6)\approx 0.9976\ge 0.99\), the minimum is \(6\) pallet jacks.

Answer

a) \(P(X=2)\approx 0.2852\), or about \(28.52\%\) b) \(P(X>4)\approx 0.0432\), or about \(4.32\%\) c) \(6\) pallet jacks
53110212
A highway rest area has \(6\) fast-charging stations for electric vehicles. During a two-hour period, \(24\) vehicles arrive to charge. Assume that, at a randomly selected time during the period, each vehicle independently has probability \(\frac{20}{120}=\frac{1}{6}\) of needing a charging station. Find the probability that more than half of the charging stations are needed at that time.

Hints

- Determine the probability for one vehicle at the selected time. - Translate “more than half” into a whole-number count. - Model the number of vehicles using a binomial distribution. - Use a complement to calculate the upper-tail probability.

Solution

1. Let \(X\) be the number of vehicles needing a charger at the selected time. Then \(X\sim\operatorname{Bin}(24,\frac{1}{6})\). 2. More than half of \(6\) stations means at least \(4\) stations are needed. 3. Therefore, \(P(X\ge 4)=1-P(X\le 3)\approx 0.584487\).

Answer

The probability is approximately \(58.45\%\).
53110312
An office has \(50\) employees who share licenses for specialized software. Each employee uses the software for an average of \(15\) minutes per hour. Assume access times are independent and uniformly distributed throughout the hour. At a randomly selected time, let \(X\) be the number of employees using the software. What is the minimum number of licenses needed so that the probability that at least one employee cannot access the software is at most \(2\%\)?

Hints

- Convert one employee's average use time into a probability at a randomly selected time. - Define the random variable for the number of simultaneous users. - Express a license shortage as an inequality involving the number of users and licenses. - Use cumulative binomial probabilities to compare consecutive license counts.

Solution

1. At a randomly selected time, one employee is using the software with probability \(p=\frac{15}{60}=0.25\). Thus \(X\sim\operatorname{Bin}(50,0.25)\). 2. If \(k\) licenses are available, an access shortage occurs when \(X>k\). The requirement is \(P(X>k)\le 0.02\), or equivalently \(P(X\le k)\ge 0.98\). 3. Using cumulative binomial probabilities, \(P(X\le 18)\approx 0.9713<0.98\), while \(P(X\le 19)\approx 0.9861\ge 0.98\). 4. Therefore, the minimum number of licenses is \(19\).

Answer

The office needs at least \(19\) licenses.
53110412
A fitness studio offers a circuit class for \(40\) participants. One station has \(10\) identical rowing machines. Each participant uses a rowing machine for a total of \(10\) minutes during the \(60\)-minute class. Assume the participants' use times are independent and uniformly distributed throughout the class. At a randomly selected time, let \(X\) be the number who want to use a rowing machine. a) Find the probability that more than \(10\) participants want to use a rowing machine at the same time. b) What is the minimum total number of rowing machines needed so that the probability that someone must wait is at most \(5\%\)?

Hints

- Convert one participant's total use time into a probability at a randomly selected time. - Use the complement of “more than \(10\).” - For part b), test consecutive equipment counts until the cumulative probability reaches the required level.

Solution

1. At a randomly selected time, one participant wants a rowing machine with probability \(p=\frac{10}{60}=\frac{1}{6}\). Thus \(X\sim\operatorname{Bin}\left(40,\frac{1}{6}\right)\). 2. For part a), use the complement: \(P(X>10)=1-P(X\le 10)\approx 1-0.9416=0.0584\). 3. For part b), seek the smallest integer \(k\) such that \(P(X>k)\le 0.05\), or equivalently \(P(X\le k)\ge 0.95\). 4. Since \(P(X\le 10)\approx 0.9416<0.95\) and \(P(X\le 11)\approx 0.9739\ge 0.95\), the minimum is \(11\) rowing machines.

Answer

a) \(P(X>10)\approx 0.0584\), or about \(5.84\%\) b) \(11\) rowing machines
53110912
A medical office has a waiting room with \(10\) seats. The office is open for \(300\) minutes each morning and schedules \(80\) patients. Assume that each patient spends \(15\) minutes in the waiting room and that the waiting periods are independently and uniformly distributed over the morning. Thus, the probability that a given patient is in the waiting room at a randomly selected time is \(\frac{15}{300}=0.05\). Find the probability that more patients are in the waiting room than there are seats. Then determine whether \(10\) seats are sufficient under a requirement that overcrowding occur no more than \(1\%\) of the time.

Hints

- Find the probability that one patient is present at the selected time. - Model the number of patients present with a binomial distribution. - Express “more than \(10\)” using a cumulative probability. - Compare the result with the \(1\%\) limit.

Solution

1. Let \(X\) be the number of patients in the waiting room at the selected time. Then \(X\sim\operatorname{Bin}(80,0.05)\). 2. Overcrowding occurs when \(X>10\). 3. Therefore, \(P(X>10)=1-P(X\le 10)\approx 0.002099\), or approximately \(0.21\%\). 4. Since \(0.21\%<1\%\), the number of seats meets the stated requirement.

Answer

The probability of overcrowding is approximately \(0.21\%\). Because this is below \(1\%\), \(10\) seats are sufficient under the stated criterion.
53113012
A component passes a rigorous quality inspection with probability \(p=0.3\). In a batch of \(n=600\) independently tested components, let \(X\) be the number that pass. The mean is \(\mu=180\). Find the probability that \(X\) is outside the interval \([165,195]\).

Hints

- Identify the integer values that lie outside the interval. - You can add the probabilities in the two tails or subtract the interval probability from \(1\). - Express an inclusive interval probability using cumulative probabilities.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(600,0.3)\). 2. Being outside \([165,195]\) means \(X\le 164\) or \(X\ge 196\). 3. Using the binomial cumulative distribution function, \(P(X\le 164)\approx 0.0828\) and \(P(X\ge 196)=1-P(X\le 195)\approx 1-0.9156=0.0844\). 4. The two events are disjoint, so \(P(X\le 164)+P(X\ge 196)\approx 0.0828+0.0844=0.1672\).

Answer

\(P(X<165\text{ or }X>195)\approx 0.1672\), or about \(16.72\%\)
53113312
A binomial random variable \(X\) has parameters \(n=150\) and \(p=0.4\). 1. Find the mean \(\mu\) and standard deviation \(\sigma\). 2. Find the probability that \(X\) is within \(5\) of its mean.

Hints

- Use the formulas for the mean and standard deviation of a binomial random variable. - Translate “within \(5\)” into an inclusive interval centered at the mean. - Express the interval probability as a difference of cumulative probabilities. - Check which value must be used in the lower cumulative probability.

Solution

1. The mean is \(\mu=np=150\cdot 0.4=60\). The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{150\cdot 0.4\cdot 0.6}=\sqrt{36}=6\). 2. Being within \(5\) of the mean means \(55\le X\le 65\). 3. Using cumulative probabilities, \(P(55\le X\le 65)=P(X\le 65)-P(X\le 54)\approx 0.8206-0.1799=0.6407\).

Answer

1. \(\mu=60\) and \(\sigma=6\) 2. \(P(55\le X\le 65)\approx 0.6407\), or about \(64.07\%\)
53113512
A binomial random variable \(X\) has parameters \(n=200\) and \(p=0.4\). First find the mean \(\mu\). Then state how far each endpoint of the interval \([72,90]\) is from the mean and find \(P(72\le X\le 90)\).

Hints

- Use the formula for the mean of a binomial random variable. - Compare each endpoint with the mean separately. - To include the lower endpoint in a cumulative-probability subtraction, subtract through the integer immediately below it.

Solution

1. The mean is \(\mu=np=200\cdot 0.4=80\). 2. The lower endpoint, \(72\), is \(8\) below the mean. The upper endpoint, \(90\), is \(10\) above the mean, so the interval is not symmetric about the mean. 3. Write the interval probability as \(P(72\le X\le 90)=P(X\le 90)-P(X\le 71)\). 4. Using the binomial cumulative distribution function, \(P(X\le 90)\approx 0.9345\) and \(P(X\le 71)\approx 0.1094\). 5. Therefore, \(P(72\le X\le 90)\approx 0.9345-0.1094=0.8251\).

Answer

The mean is \(\mu=80\). The lower endpoint is \(8\) below the mean, and the upper endpoint is \(10\) above the mean. Also, \(P(72\le X\le 90)\approx 0.8251\), or about \(82.51\%\).
53113712
At a distribution center, automated scanners cannot read the barcode on \(2\%\) of packages. A random sample of \(400\) packages is inspected, and barcode outcomes are treated as independent. Let \(X\) be the number with unreadable barcodes. Find each probability. a) At most \(5\) packages have unreadable barcodes. b) More than \(10\) packages have unreadable barcodes. c) Between \(6\) and \(12\) packages, inclusive, have unreadable barcodes.

Hints

- Identify the binomial parameters in the situation. - Pay close attention to which endpoint values are included by “more than” and “between.” - Use a complement for an upper-tail probability and a difference of cumulative probabilities for an interval. - A cumulative binomial probability includes every integer count from \(0\) through its stated endpoint.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(400,0.02)\). 2. For part a), \(P(X\le 5)\approx 0.1885\). 3. For part b), use the complement: \(P(X>10)=1-P(X\le 10)\approx 1-0.8179=0.1821\). 4. For part c), \(P(6\le X\le 12)=P(X\le 12)-P(X\le 5)\approx 0.7497\), using unrounded cumulative probabilities.

Answer

a) \(P(X\le 5)\approx 0.1885\), or about \(18.85\%\) b) \(P(X>10)\approx 0.1821\), or about \(18.21\%\) c) \(P(6\le X\le 12)\approx 0.7497\), or about \(74.97\%\)
53113812
A pharmaceutical company reports that a certain side effect occurs in \(40\%\) of patients who take a medication. A study includes \(100\) patients, and side-effect outcomes are treated as independent. Let \(X\) be the number who experience the side effect. Find the probability that the number of patients with the side effect is: a) less than \(35\); b) at least \(45\); c) within \(5\) of the mean.

Hints

- Use the formula for the mean of a binomial random variable. - Translate “within \(5\)” into an inclusive interval centered at the mean. - Remember that “less than \(35\)” does not include \(35\). - Use a difference of cumulative probabilities for the interval.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(100,0.4)\). 2. For part a), \(P(X<35)=P(X\le 34)\approx 0.1303\). 3. For part b), use the complement: \(P(X\ge 45)=1-P(X\le 44)\approx 1-0.8211=0.1789\). 4. The mean is \(\mu=np=100\cdot 0.4=40\). Being within \(5\) of the mean means \(35\le X\le 45\). 5. Therefore, \(P(35\le X\le 45)=P(X\le 45)-P(X\le 34)\approx 0.8689-0.1303=0.7386\).

Answer

a) \(P(X<35)\approx 0.1303\), or about \(13.03\%\) b) \(P(X\ge 45)\approx 0.1789\), or about \(17.89\%\) c) \(P(35\le X\le 45)\approx 0.7386\), or about \(73.86\%\)
53114112
A factory produces components with a \(4\%\) defect rate. A random sample of \(500\) components is selected, and defect outcomes are treated as independent. Let \(X\) be the number of defective components in the sample. a) Find the probability that at most \(15\) components are defective. b) Find the probability that at least \(25\) components are defective. c) Find the probability that between \(18\) and \(22\) components, inclusive, are defective.

Hints

- Identify the binomial parameters in the situation. - Pay attention to whether words such as “at most” and “at least” include the stated endpoint. - Use a complement for the “at least” probability. - For an inclusive interval, subtract the cumulative probability through the integer immediately below the lower endpoint.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(500,0.04)\). 2. For part a), \(P(X\le 15)\approx 0.1513\). 3. For part b), use the complement: \(P(X\ge 25)=1-P(X\le 24)\approx 1-0.8478=0.1522\). 4. For part c), \(P(18\le X\le 22)=P(X\le 22)-P(X\le 17)\approx 0.7238-0.2924=0.4314\).

Answer

a) \(P(X\le 15)\approx 0.1513\), or about \(15.13\%\) b) \(P(X\ge 25)\approx 0.1522\), or about \(15.22\%\) c) \(P(18\le X\le 22)\approx 0.4314\), or about \(43.14\%\)
53114912
At a light-bulb factory, \(4\%\) of bulbs have a minor cosmetic defect. Assume defects occur independently. a) Find the probability that at least one bulb in a random box of \(25\) bulbs has a cosmetic defect. b) What is the minimum number of bulbs that must be inspected so that the probability of finding at least one bulb with a cosmetic defect is greater than \(90\%\)?

Hints

- Begin with the complement event that no inspected bulb has the defect. - Relate “at least one” to “none” using a complement. - Use logarithms to solve an inequality in which the unknown appears in an exponent. - Remember that dividing an inequality by a negative number reverses its direction.

Solution

1. The probability that one bulb does not have the defect is \(0.96\). 2. For part a), use the complement: \(P(X\ge 1)=1-P(X=0)=1-(0.96)^{25}\approx 0.6396\). 3. For part b), require \(1-(0.96)^n>0.90\), which is equivalent to \((0.96)^n<0.10\). 4. Taking logarithms gives \(n\ln(0.96)<\ln(0.10)\). Because \(\ln(0.96)<0\), dividing reverses the inequality: \(n>\frac{\ln(0.10)}{\ln(0.96)}\approx 56.4055\). 5. Therefore, the minimum integer sample size is \(57\). Indeed, the probability is about \(0.8983\) for \(n=56\) and \(0.9024\) for \(n=57\).

Answer

a) \(1-(0.96)^{25}\approx 0.6396\), or about \(63.96\%\) b) \(57\) bulbs
53115412
A computer company receives USB drives in packages of \(50\). Each drive is independently defective with probability \(0.10\). A customer needs \(42\) working drives for a training session and opens a new package. Find the probability that the package contains enough working drives.

Hints

- Decide whether to count working or defective drives. - How many drives may be defective while still leaving \(42\) working drives? - Use the corresponding cumulative binomial probability.

Solution

1. Let \(X\) be the number of defective drives. Then \(X\sim\operatorname{Bin}(50,0.10)\). 2. At least \(42\) drives work exactly when at most \(8\) are defective. 3. Therefore, \(P(X\le 8)\approx 0.942133\).

Answer

The probability is approximately \(94.21\%\).
53115712
At a regional blood donation center, \(15\%\) of donors have type O-negative blood. Assume the blood types of \(200\) expected donors are independent. Let \(X\) be the number with type O-negative blood. a) Find the probability that at most \(25\) donors have type O-negative blood. b) Find the probability that more than \(35\) donors have type O-negative blood. c) Find the probability that more than \(25\) but fewer than \(40\) donors have type O-negative blood.

Hints

- Identify the binomial parameters in the situation. - Pay attention to whether “more than” and “fewer than” include their stated endpoints. - Use a complement for the upper-tail probability and a difference of cumulative probabilities for the interval. - List the integer counts represented by the strict inequalities before choosing cumulative endpoints.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(200,0.15)\). 2. For part a), \(P(X\le 25)\approx 0.1876\). 3. For part b), use the complement: \(P(X>35)=1-P(X\le 35)\approx 1-0.8613=0.1387\). 4. For part c), the integer counts are \(26\) through \(39\). Thus \(P(25<X<40)=P(X\le 39)-P(X\le 25)\approx 0.9665-0.1876=0.7788\).

Answer

a) \(P(X\le 25)\approx 0.1876\), or about \(18.76\%\) b) \(P(X>35)\approx 0.1387\), or about \(13.87\%\) c) \(P(25<X<40)\approx 0.7788\), or about \(77.88\%\)
53115812
An online retailer emails a promotional offer to \(120\) customers. Each customer independently places an order with probability \(0.12\). Let \(X\) be the number of orders placed. a) Find the probability that fewer than \(10\) orders are placed. b) Find the probability that at least \(15\) customers place an order. c) Find the probability that between \(12\) and \(20\) orders, inclusive, are placed.

Hints

- Identify the binomial parameters from the situation. - Rewrite “fewer than \(10\)” using an inclusive integer endpoint. - Use a complement for the “at least” probability. - For an inclusive interval, subtract the cumulative probability below the lower endpoint from the cumulative probability at the upper endpoint.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(120,0.12)\). 2. For part a), \(P(X<10)=P(X\le 9)\approx 0.0782\). 3. For part b), use the complement: \(P(X\ge 15)=1-P(X\le 14)\approx 1-0.5255=0.4745\). 4. For part c), \(P(12\le X\le 20)=P(X\le 20)-P(X\le 11)\approx 0.9513-0.2110=0.7403\).

Answer

a) \(P(X<10)\approx 0.0782\), or about \(7.82\%\) b) \(P(X\ge 15)\approx 0.4745\), or about \(47.45\%\) c) \(P(12\le X\le 20)\approx 0.7403\), or about \(74.03\%\)
53116212
Historically, \(4\%\) of the microchips produced by a factory are defective. Ten chips are selected from a very large shipment. Because the population is very large, defect outcomes are modeled as approximately independent. a) Explain why a binomial model is reasonable even though the chips are selected without replacement. b) Find the probability that exactly one selected chip is defective. c) Find the probability that at most one selected chip is defective.

Hints

- How much does a small sample change a very large population? - “At most one” includes which values of \(X\)? - Compute the required point probabilities separately before adding them.

Solution

1. Removing \(10\) chips from a very large population changes the defect proportion by a negligible amount, so the defect probability can be treated as constant at \(p=0.04\), and the selections can be modeled as approximately independent. 2. Let \(X\sim\operatorname{Bin}(10,0.04)\). Then \(P(X=1)=\binom{10}{1}(0.04)(0.96)^9\approx 0.2770\). 3. Also, \(P(X=0)=(0.96)^{10}\approx 0.6648\). 4. Therefore, \(P(X\le 1)=P(X=0)+P(X=1)\approx 0.6648+0.2770=0.9418\).

Answer

a) The sample is negligible compared with the population, so the defect probability is approximately constant and the selections are approximately independent. b) \(P(X=1)\approx 0.2770\) c) \(P(X\le 1)\approx 0.9418\)
53116312
Let \(X\) be a binomial random variable with parameters \(n\) and \(p\). a) Use sigma notation to write the sum of \(P(X=k)\) for \(k=0,1,\ldots,n\). b) Use the binomial theorem \((a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^kb^{n-k}\) to explain why the probabilities in any binomial distribution add to \(1\) for every \(p\in[0, 1]\).

Hints

- Write the formula for a binomial point probability. - Compare its factors with the terms in the binomial theorem. - Substitute the success and failure probabilities for \(a\) and \(b\). - Recall the required total probability of a complete distribution.

Solution

1. The sum of all point probabilities is \(\sum_{k=0}^{n}P(X=k)=\sum_{k=0}^{n}\binom{n}{k}p^k(1-p)^{n-k}\). 2. In the binomial theorem, set \(a=p\) and \(b=1-p\). 3. The sum becomes \((p+(1-p))^n=1^n=1\). Therefore, every binomial distribution is normalized.

Answer

a) \(\sum_{k=0}^{n}\binom{n}{k}p^k(1-p)^{n-k}\) b) Substituting \(a=p\) and \(b=1-p\) into the binomial theorem gives \((p+(1-p))^n=1\).
53116412
Consider the expression \(T=\sum_{k=0}^{12}\binom{12}{k}(0.25)^k(0.75)^{12-k}\). a) Interpret this expression in the context of Bernoulli trials. Identify \(n\) and \(p\), and explain what \(T\) represents. b) Find the exact value of \(T\) without evaluating the terms individually. Use the binomial theorem. c) How would the value change if the sum ran from \(k=1\) to \(k=12\)? Give an expression for the new value.

Hints

- Compare the exponents and coefficient with the binomial probability formula. - Use the expansion of \((a+b)^n\) and note that the two bases add to \(1\). - If one term is omitted from a complete probability sum, subtract that term from \(1\). - The probabilities of all possible outcomes must add to \(1\).

Solution

1. Each summand has the binomial form \(\binom{n}{k}p^k(1-p)^{n-k}\), so \(n=12\) and \(p=0.25\). The sum includes every possible number of successes, so \(T\) is the total probability. 2. By the binomial theorem, \(T=(0.25+0.75)^{12}=1^{12}=1\). 3. Starting at \(k=1\) removes the \(k=0\) term, which is \((0.75)^{12}\). The new value is \(1-(0.75)^{12}\approx0.9683\).

Answer

a) \(n=12\), \(p=0.25\), and \(T\) is the sum of all probabilities in a binomial distribution. b) \(T=1\) c) \(1-(0.75)^{12}\approx0.9683\)
53116712
At a nursery, each seed of a certain flower variety independently germinates with probability \(p=0.80\). A total of \(20\) seeds are planted, and \(X\) is the number that germinate. Find \(P(X\le 14)\) by introducing a complementary binomial random variable with success probability at most \(0.5\).

Hints

- Count failures instead of successes. - How are the probabilities \(p\) and \(1-p\) related? - If at most \(14\) seeds germinate, how many must fail to germinate? - Convert an “at least” probability into a complement of an “at most” probability.

Solution

1. Let \(Y\) be the number of seeds that do not germinate. Then \(Y\sim\operatorname{Bin}(20,0.20)\). 2. The event \(X\le 14\) is equivalent to \(Y\ge 6\). 3. Therefore, \(P(X\le 14)=P(Y\ge 6)=1-P(Y\le 5)\approx 0.195792\).

Answer

\(P(X\le 14)\approx 0.1958\), or \(19.58\%\).
53118512
An online clothing retailer finds that each ordered item is independently returned because of fit with probability \(0.30\). A sample of \(50\) orders is examined. Find the probability that 1. exactly \(15\) items are returned, 2. fewer than \(10\) items are returned, 3. more than \(20\) items are returned.

Hints

- Identify the distribution and its parameters. - “Fewer than \(10\)” has an upper endpoint of \(9\). - Use a complement to calculate “more than \(20\).” - Check whether each boundary value is included.

Solution

1. Let \(X\sim\operatorname{Bin}(50,0.30)\). 2. For part 1, \(P(X=15)=\binom{50}{15}(0.30)^{15}(0.70)^{35}\approx 0.122347\). 3. For part 2, \(P(X<10)=P(X\le 9)\approx 0.040232\). 4. For part 3, \(P(X>20)=1-P(X\le 20)\approx 0.047764\).

Answer

1. \(P(X=15)\approx 12.23\%\) 2. \(P(X<10)\approx 4.02\%\) 3. \(P(X>20)\approx 4.78\%\)
53118612
A manufacturer finds that \(5\%\) of its microchips are defective. A random sample of \(200\) chips is inspected, and defect outcomes are treated as independent. Let \(X\) be the number of defective chips. Find each probability. 1. At most \(12\) chips are defective. 2. Between \(5\) and \(15\) chips, inclusive, are defective. 3. The number of defective chips is within \(2\) of the mean.

Hints

- Use the formula for the mean of a binomial random variable. - Express an interval probability as a difference of two cumulative probabilities. - Translate a maximum deviation from the mean into an inclusive interval. - To include the lower endpoint, subtract the cumulative probability through the integer immediately below it.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(200,0.05)\), and \(P(X\le 12)\approx 0.7965\). 2. For the inclusive interval, \(P(5\le X\le 15)=P(X\le 15)-P(X\le 4)\approx 0.9556-0.0264=0.9292\). 3. The mean is \(\mu=np=200\cdot 0.05=10\). Being within \(2\) of the mean means \(8\le X\le 12\). 4. Thus \(P(8\le X\le 12)=P(X\le 12)-P(X\le 7)\approx 0.7965-0.2133=0.5832\).

Answer

1. \(P(X\le 12)\approx 0.7965\), or about \(79.65\%\) 2. \(P(5\le X\le 15)\approx 0.9292\), or about \(92.92\%\) 3. \(P(8\le X\le 12)\approx 0.5832\), or about \(58.32\%\)
53119512
A bag contains \(25\) distinct numbered balls. A ball is drawn at random and replaced, and this process is repeated for a total of \(25\) draws. Focus on one specified ball, such as the ball numbered \(7\). a) Find the probability that this ball is drawn zero times, exactly once, and exactly twice. b) Find the probability that the ball is drawn more than twice. c) Find the expected number of times the ball is drawn.

Hints

- Determine whether replacement makes the trials independent with a constant probability. - Identify \(n\) and \(p\). - Use the binomial formula for the three exact values. - Use the complement of “at most two” for part b). - Recall the expected-value formula for a binomial random variable.

Solution

1. Let \(X\) be the number of times the specified ball is drawn. Because each draw is made with replacement, \(X\) is binomial with \(n=25\) and \(p=\frac{1}{25}=0.04\). 2. \(P(X=0)=(0.96)^{25}\approx0.3604\). 3. \(P(X=1)=\binom{25}{1}(0.04)(0.96)^{24}\approx0.3754\). 4. \(P(X=2)=\binom{25}{2}(0.04)^2(0.96)^{23}\approx0.1877\). 5. Using the complement, \(P(X>2)=1-P(X=0)-P(X=1)-P(X=2)\approx0.0765\). 6. The expected value is \(E(X)=np=25\cdot0.04=1\).

Answer

a) \(P(X=0)\approx0.3604\); \(P(X=1)\approx0.3754\); \(P(X=2)\approx0.1877\) b) \(P(X>2)\approx0.0765\) c) \(E(X)=1\)
53212412
The spinner shown has \(5\) equal sections: \(2\) blue and \(3\) yellow. The spinner is spun independently \(5\) times. Let \(X\) be the number of times it lands on blue. a) Find \(P(X\le 1)\). b) Find \(P(X\ge 3)\). Give each result as an exact decimal and as a percent.
Figure for problem 532124

Hints

- Explain why the spins form Bernoulli trials. - Find the probability of blue on one spin. - Which values are included in “at most \(1\)”? - Which values are included in “at least \(3\)”? - Calculate and add the relevant binomial probabilities.

Solution

1. The probability of blue is \(p=\frac{2}{5}=0.40\), so \(X\sim\operatorname{Bin}(5,0.40)\). 2. For part a), \(P(X\le 1)=P(X=0)+P(X=1)=(0.60)^5+5(0.40)(0.60)^4=0.33696\). 3. For part b), \(P(X\ge 3)=P(X=3)+P(X=4)+P(X=5)=0.31744\).

Answer

a) \(0.33696\), or \(33.696\%\) b) \(0.31744\), or \(31.744\%\)
53212612
The spinner shown has \(5\) equal sections. It is spun independently \(6\) times, and \(X\) is the number of times it lands on green. a) Explain why \(X\) has a binomial distribution, and state \(n\) and \(p\). b) Find the probability that the spinner lands on green exactly \(2\) times. c) Find the probability that the spinner lands on green at most \(1\) time.
Figure for problem 532126

Hints

- State the conditions for Bernoulli trials. - Count the green sections and the total sections. - Use the binomial formula for an exact number of successes. - “At most \(1\)” includes \(0\) and \(1\).

Solution

1. Each spin has two relevant outcomes, green or not green, with constant probability \(p=\frac{1}{5}=0.20\), and the spins are independent. Thus, \(X\sim\operatorname{Bin}(6,0.20)\). 2. For part b), \(P(X=2)=\binom{6}{2}(0.20)^2(0.80)^4=0.24576\). 3. For part c), \(P(X\le 1)=(0.80)^6+6(0.20)(0.80)^5=0.65536\).

Answer

a) \(X\sim\operatorname{Bin}(6,0.20)\) because the spins are independent and the probability of green is constant. b) \(P(X=2)=0.24576\), or \(24.576\%\). c) \(P(X\le 1)=0.65536\), or \(65.536\%\).
53215412
The spinner shown has \(6\) equal sections: \(3\) green, \(2\) yellow, and \(1\) red. The spinner is spun independently \(8\) times. a) Find the probability that it lands on green at most \(3\) times. b) Find the probability that it lands on red at least \(2\) times. Give each answer as a simplified fraction and as a percent rounded to two decimal places.
Figure for problem 532154

Hints

- Determine the probability of each specified color on one spin. - Model repeated independent spins with a binomial distribution. - Which counts are included in “at most \(3\)”? - For “at least \(2\),” consider the complement of \(0\) or \(1\).

Solution

1. For green, \(p_G=\frac{3}{6}=\frac{1}{2}\). If \(X\sim\operatorname{Bin}(8,\frac{1}{2})\), then \(P(X\le 3)=\frac{\binom{8}{0}+\binom{8}{1}+\binom{8}{2}+\binom{8}{3}}{2^8}=\frac{93}{256}\approx 0.363281\). 2. For red, \(p_R=\frac{1}{6}\). If \(Y\sim\operatorname{Bin}(8,\frac{1}{6})\), then \(P(Y\ge 2)=1-P(Y\le 1)=1-\left[\left(\frac{5}{6}\right)^8+8\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^7\right]=\frac{663991}{1679616}\approx 0.395323\).

Answer

a) \(\frac{93}{256}\approx 36.33\%\) b) \(\frac{663991}{1679616}\approx 39.53\%\)
53215912
The spinner shown has \(10\) equal sections. It is spun \(30\) times. Find each probability. Round each answer to the nearest hundredth of a percent. a) The spinner lands on red at most \(8\) times. b) The spinner lands on blue at least \(12\) times and at most \(18\) times.
Figure for problem 532159

Hints

- First determine the probability of landing on each requested color from the spinner. - Translate “at most” and “at least ... and at most” into inequalities for the count. - Use the binomial cumulative distribution function. - Express an inclusive interval probability as \(P(a\le X\le b)=P(X\le b)-P(X\le a-1)\).

Solution

1. The spinner has \(3\) red sections and \(5\) blue sections, so \(p_{\text{red}}=\frac{3}{10}=0.3\) and \(p_{\text{blue}}=\frac{5}{10}=0.5\). 2. Let \(R\) be the number of red results. Then \(R\sim\operatorname{Bin}(30,0.3)\), so \(P(R\le 8)\approx 0.4315=43.15\%\). 3. Let \(B\) be the number of blue results. Then \(B\sim\operatorname{Bin}(30,0.5)\). 4. Therefore, \(P(12\le B\le 18)=P(B\le 18)-P(B\le 11)\approx 0.7995=79.95\%\).

Answer

a) About \(43.15\%\) b) About \(79.95\%\)
53602312
A spinner has \(10\) equal sections: \(5\) blue, \(3\) yellow, and \(2\) red. The spinner is spun independently \(12\) times. Find the probability that a) it lands on blue at most \(4\) times, b) it lands on red at least \(2\) times.
Figure for problem 536023

Hints

- Determine the probability of each color on one spin. - Use a binomial model because the spins are independent and probabilities stay constant. - For “at most,” add probabilities from \(0\) through the endpoint. - For “at least,” a complement may be more efficient.

Solution

1. For blue, \(p_B=\frac{5}{10}=0.50\). If \(X\sim\operatorname{Bin}(12,0.50)\), then \(P(X\le 4)\approx 0.193848\). 2. For red, \(p_R=\frac{2}{10}=0.20\). If \(Y\sim\operatorname{Bin}(12,0.20)\), then \(P(Y\ge 2)=1-P(Y\le 1)\approx 0.725122\).

Answer

a) The probability is approximately \(19.38\%\). b) The probability is approximately \(72.51\%\).
53604812
The spinner shown has \(8\) equal sections and is spun independently \(10\) times. Let \(X\) be the number of times it lands on red. a) Explain why \(X\) has a binomial distribution, and state \(n\) and \(p\). b) Find \(P(X\le 2)\). c) The spinner is now spun independently \(4\) times. Find the probability that blue occurs exactly \(2\) times and that the two blue results are consecutive.
Figure for problem 536048

Hints

- State the conditions for a binomial model. - Add the probabilities for the included red counts. - For part c), list every valid sequence and use the multiplication and addition rules.

Solution

1. There are \(2\) red sections out of \(8\), so \(p=\frac{2}{8}=0.25\). The spins are independent with constant probability, so \(X\sim\operatorname{Bin}(10,0.25)\). 2. For part b), \(P(X\le 2)=\sum_{k=0}^{2}\binom{10}{k}(0.25)^k(0.75)^{10-k}\approx 0.525593\). 3. For part c), the valid patterns are BBNN, NBBN, and NNBB, where B means blue and N means not blue. 4. Each pattern has probability \((0.25)^2(0.75)^2\), so the total probability is \(3(0.25)^2(0.75)^2=0.10546875\).

Answer

a) \(X\sim\operatorname{Bin}(10,0.25)\). b) \(P(X\le 2)\approx 0.5256\), or \(52.56\%\). c) The probability is \(0.10546875\), or approximately \(10.55\%\).
53607312
The spinner shown has \(8\) equal sections and is spun independently \(3\) times. Find which expressions correctly calculate the probability that the spinner lands on yellow exactly twice. Explain your choices. (1) \(\left(\frac38\right)^2\frac58\) (2) \(3\cdot\frac38\left(\frac58\right)^2\) (3) \(\frac38\cdot\frac38\cdot\frac58+\frac38\cdot\frac58\cdot\frac38+\frac58\cdot\frac38\cdot\frac38\) (4) \(3\cdot\left(\frac38\right)^2\frac58\)
Figure for problem 536073

Hints

- Find the probabilities of yellow and not yellow from the sections. - List the possible orders containing exactly two yellow outcomes. - Find the probability of one specified order. - Add the probabilities of the mutually exclusive orders.

Solution

1. Three of the eight sections are yellow, so \(P(\text{yellow})=\frac38\), and \(P(\text{not yellow})=\frac58\). 2. Exactly two yellow outcomes can occur in three orders: yellow-yellow-not yellow, yellow-not yellow-yellow, and not yellow-yellow-yellow. 3. Each order has probability \(\left(\frac38\right)^2\frac58\). 4. Expression (3) adds the probabilities of the three orders separately. Expression (4) multiplies the probability of one order by \(3\). Both are correct. 5. Expression (1) includes only one order. Expression (2) represents exactly one yellow outcome and two not-yellow outcomes.

Answer

Expressions (3) and (4) are correct.
53608612
A spinner has \(6\) equal sections: \(2\) yellow and \(4\) blue, as shown. A researcher wants to test whether the spinner has been biased in favor of blue. The spinner is spun independently \(50\) times, and the number of yellow results is recorded. Assuming the spinner is fair, find the probability that yellow occurs at most \(10\) times. Round the result to three decimal places.
Figure for problem 536086

Hints

- Determine the probability of yellow from the section counts. - Use a binomial distribution for the number of yellow results. - “At most \(10\)” requires a cumulative probability. - Use a cumulative binomial function and round only the final result.

Solution

1. The probability of yellow is \(p=\frac{2}{6}=\frac{1}{3}\). Let \(X\sim\operatorname{Bin}(50,\frac{1}{3})\). 2. The required probability is \(P(X\le 10)=\sum_{k=0}^{10}\binom{50}{k}\left(\frac{1}{3}\right)^k\left(\frac{2}{3}\right)^{50-k}\approx 0.028440\). 3. Rounded to three decimal places, the probability is \(0.028\).

Answer

The probability is approximately \(0.028\).
53608912
A spinner has \(10\) equal sections: \(3\) purple sections labeled P and \(7\) orange sections labeled O. The spinner is spun independently \(8\) times. Let \(X\) be the number of purple results. For each expression, describe an event in context whose probability it represents. 1) \(\binom{8}{2}(0.30)^2(0.70)^6\) 2) \((0.70)^8\) 3) \(1-(0.70)^8\) 4) \(\sum_{k=0}^{2}\binom{8}{k}(0.30)^k(0.70)^{8-k}\)
Figure for problem 536089

Hints

- Match each color with its probability. - Compare the expressions with \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\). - Subtracting a probability from \(1\) gives a complement. - A summation combines several possible values of the random variable.

Solution

1. The probability of purple is \(0.30\), and the probability of orange is \(0.70\). 2. Expression 1 is the binomial probability \(P(X=2)\). 3. Expression 2 is the probability of orange on all \(8\) spins, so \(X=0\). 4. Expression 3 is the complement of no purple results, so \(X\ge 1\). 5. Expression 4 sums the probabilities for \(X=0,1,\) and \(2\), so \(X\le 2\).

Answer

1) Purple occurs exactly \(2\) times. 2) Purple never occurs; every result is orange. 3) Purple occurs at least once. 4) Purple occurs at most \(2\) times.
53610912
The spinner shown has \(8\) equal sections. a) The spinner is spun independently \(50\) times. Find the probability of each event. \(A\): Green occurs exactly \(20\) times. \(B\): Red occurs at most \(10\) times. b) Describe an event in context whose probability is \(\sum_{k=0}^{8}\binom{60}{k}(0.125)^k(0.875)^{60-k}\).
Figure for problem 536109

Hints

- Determine each color probability from the section counts. - Identify \(n\), \(p\), and the required value or range for each event. - “At most \(10\)” includes \(0\) through \(10\). - Match the expression in part b) to a cumulative binomial probability.

Solution

1. The spinner has \(3\) green, \(2\) red, \(2\) blue, and \(1\) yellow section. Thus, \(P(\text{green})=\frac{3}{8}=0.375\), \(P(\text{red})=0.25\), and \(P(\text{yellow})=0.125\). 2. For event \(A\), \(P(A)=\binom{50}{20}(0.375)^{20}(0.625)^{30}\approx 0.107230\). 3. For event \(B\), \(P(B)=\sum_{k=0}^{10}\binom{50}{k}(0.25)^k(0.75)^{50-k}\approx 0.262202\). 4. In part b), the success probability \(0.125\) corresponds to yellow, and the sum through \(8\) represents at most \(8\) successes in \(60\) spins.

Answer

a) \(P(A)\approx 10.72\%\) \(P(B)\approx 26.22\%\) b) In \(60\) spins, the spinner lands on yellow at most \(8\) times.
53745212
A seed supplier states that its wheat seeds have a \(95\%\) germination rate. A farmer suspects that the actual germination probability is only \(80\%\). To compare the two models, the farmer tests \(20\) seeds and observes that exactly \(17\) germinate. Assume seed outcomes are independent under each model. a) For Model A, \(p=0.95\), and Model B, \(p=0.80\), find the probability that exactly \(17\) of \(20\) seeds germinate. b) Which model makes the observed result more plausible? Explain. c) The diagram shows the first two stages of the sample under both models. Identify the missing probabilities \(p\) and \(q\) for Model B.
Figure for problem 537452

Hints

- Use the binomial probability formula for exactly \(17\) successes. - Compare the probability of the observed count under the two models. - The two branch probabilities at each stage must add to \(1\).

Solution

1. Use \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\). 2. For Model A, \(P(X=17)=\binom{20}{17}(0.95)^{17}(0.05)^3\approx 0.05958\). 3. For Model B, \(P(X=17)=\binom{20}{17}(0.80)^{17}(0.20)^3\approx 0.20536\). 4. The observed count is more probable under Model B, so Model B makes this particular result more plausible. This comparison alone does not prove that Model B is correct. 5. For Model B, \(p=0.80\) and \(q=1-p=0.20\).

Answer

a) Model A: \(0.05958\); Model B: \(0.20536\) b) Model B makes the observed result more plausible because exactly \(17\) germinations has the larger probability under that model. c) \(p=0.80\) and \(q=0.20\)
53745412
A multiple-choice test has \(3\) questions. Each question has \(4\) answer choices, and exactly one choice is correct. A student guesses randomly on every question. The random variable \(X\) is the number of questions answered correctly. a) Find the probability distribution of \(X\) and display it in a table. b) Explain how the probabilities are related to the terms in the binomial expansion of \((0.25+0.75)^3\).
Figure for problem 537454

Hints

- First find the probability of a correct answer when there are four choices. - A tree diagram can help you organize the paths for \(0\), \(1\), \(2\), or \(3\) correct answers. - Multiply probabilities along each path. - Compare the binomial coefficients with the number of paths that have the same number of correct answers.

Solution

1. Each question is an independent trial with \(n=3\), success probability \(p=0.25\), and failure probability \(q=0.75\). 2. Use \(P(X=k)=\binom{3}{k}p^kq^{3-k}\): \(P(X=0)=0.75^3=0.421875\). \(P(X=1)=3\cdot0.25\cdot0.75^2=0.421875\). \(P(X=2)=3\cdot0.25^2\cdot0.75=0.140625\). \(P(X=3)=0.25^3=0.015625\). 3. The probability distribution is: <table> <tr><td>\(k\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td></tr> <tr><td>\(P(X=k)\)</td><td>\(0.421875\)</td><td>\(0.421875\)</td><td>\(0.140625\)</td><td>\(0.015625\)</td></tr> </table> 4. In \((p+q)^3=p^3+3p^2q+3pq^2+q^3\), the terms give the probabilities of \(3\), \(2\), \(1\), and \(0\) correct answers, respectively. The binomial coefficients count the possible orders of correct and incorrect answers.

Answer

a) <table> <tr><td>\(k\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td></tr> <tr><td>\(P(X=k)\)</td><td>\(0.421875\)</td><td>\(0.421875\)</td><td>\(0.140625\)</td><td>\(0.015625\)</td></tr> </table> b) The terms \(0.25^3\), \(3\cdot0.25^2\cdot0.75\), \(3\cdot0.25\cdot0.75^2\), and \(0.75^3\) are \(P(X=3)\), \(P(X=2)\), \(P(X=1)\), and \(P(X=0)\), respectively.
52344412
The probability of an event \(B\) is given by \(P(B)=\left(\frac{2}{3}\right)^4+4\cdot\frac{1}{3}\cdot\left(\frac{2}{3}\right)^3+6\cdot\left(\frac{1}{3}\right)^2\cdot\left(\frac{2}{3}\right)^2\). Describe a random experiment that matches this expression, and state event \(B\) in words.

Hints

- Identify how many separate binomial probabilities are being added. - Use the exponents to determine the number of trials. - Use the bases of the powers to identify the success and failure probabilities. - A sum of several consecutive cases often corresponds to wording such as “at least” or “at most.”

Solution

1. The three terms are binomial probabilities for \(n=4\) trials with success probability \(p=\frac{2}{3}\). 2. The first term is \(P(X=4)\), the second is \(P(X=3)\), and the third is \(P(X=2)\). 3. Therefore, the sum represents \(P(X\ge 2)\). 4. One matching experiment is to place \(2\) red marbles and \(1\) blue marble in a bag, draw a marble \(4\) times with replacement, and count a red marble as a success. Event \(B\) is drawing a red marble at least \(2\) times.

Answer

One possible experiment is to draw a marble \(4\) times with replacement from a bag containing \(2\) red marbles and \(1\) blue marble. Event \(B\): A red marble is drawn at least \(2\) times.
52349812
A quality-control inspector checks electronic components, each of which independently has a \(12\%\) chance of being defective. Let \(X\) be the number of defective components in a sample of size \(n\). a) Find the minimum number of components the inspector must check so that the probability of finding at least one defective component is at least \(90\%\). b) Use systematic trial or a table to find the minimum value of \(n\) for which the probability of finding at most one defective component is less than \(10\%\).

Hints

- Write each condition as an inequality involving \(P(X)\). - For part a), use the complement of finding at least one defective component. - For part b), add the probabilities for \(X=0\) and \(X=1\), then test integer values of \(n\). - When solving the logarithmic inequality, remember that \(\ln(0.88)\) is negative.

Solution

1. The probability that a component is defective is \(p=0.12\), and the probability that it is not defective is \(0.88\). 2. For part a), require \(P(X\ge 1)\ge 0.90\). Using the complement gives \(1-(0.88)^n\ge 0.90\), so \((0.88)^n\le 0.10\). 3. Taking logarithms gives \(n\ge \frac{\ln(0.10)}{\ln(0.88)}\approx 18.012\). Because \(n\) is a whole number, the minimum is \(n=19\). 4. For part b), \(P(X\le 1)=(0.88)^n+n(0.12)(0.88)^{n-1}\). 5. Testing consecutive values gives \(P(X\le 1)\approx 0.1100\) when \(n=30\) and \(P(X\le 1)\approx 0.0994\) when \(n=31\). 6. Therefore, the minimum value is \(n=31\).

Answer

a) \(n=19\) b) \(n=31\)
52350512
A binomial random variable \(X\) has \(n=12\). Find the success probability \(p\) such that \(P(X\le 1)=0.15\). Round your answer to the nearest hundredth.

Hints

- Which exact values of \(X\) are included in \(X\le 1\)? - Use the binomial probability formula for each of those values. - The resulting equation is not convenient to solve algebraically, so use a numerical solver or a table. - Test values of \(p\) and observe how the cumulative probability changes.

Solution

1. The condition is \(P(X\le 1)=P(X=0)+P(X=1)=0.15\). 2. Substitute the binomial probabilities: \((1-p)^{12}+12p(1-p)^{11}=0.15\). 3. Solve this equation numerically for \(0\le p\le 1\). 4. The solution is \(p\approx 0.2546086\), which rounds to \(p\approx 0.25\).

Answer

\(p\approx 0.25\)
52350612
A binomial random variable \(X\) has \(n=20\). Find the success probability \(p\) if \(P(X\ge 15)=0.05\). Round your answer to the nearest hundredth.

Hints

- Rewrite the “at least” probability using a complement. - What value must \(P(X\le 14)\) have? - Use a binomial cumulative-probability function and vary \(p\) to locate the solution. - As \(p\) increases, the probability of a large number of successes also increases.

Solution

1. Use the complement: \(P(X\ge 15)=1-P(X\le 14)=0.05\). 2. Therefore, \(P(X\le 14)=0.95\). 3. Solve the binomial cumulative-probability equation numerically for \(p\). 4. The solution is \(p\approx 0.5444176\). This value gives \(P(X\ge 15)=0.05\). 5. Rounded to the nearest hundredth, \(p\approx 0.54\).

Answer

\(p\approx 0.54\)
52351012
A basketball player has a free-throw success probability of \(p\). Assume the outcomes of the free throws are independent. Find \(p\), rounded to the nearest hundredth, if the probability that the player makes at most \(1\) of \(4\) free throws is \(65\%\).

Hints

- Which values of \(X\) are included in “at most \(1\)”? - Write an equation by adding the probabilities of \(0\) and \(1\) success. - Use a numerical solver or a table because \(p\) appears in several powers.

Solution

1. Let \(X\) be the number of made free throws. Then \(X\) has a binomial distribution with \(n=4\) and success probability \(p\). 2. The condition is \(P(X\le 1)=0.65\). 3. Substitute the binomial probabilities: \((1-p)^4+4p(1-p)^3=0.65\). 4. Solve the equation numerically on \(0\le p\le 1\). 5. The solution is \(p\approx 0.3009635\), so \(p\approx 0.30\) to the nearest hundredth.

Answer

The success probability must be approximately \(0.30\), or \(30\%\).
52351512
An athlete wants the probability of making at least \(13\) successes in \(15\) independent attempts to be at least \(75\%\). What is the minimum success probability \(p\) for each attempt? Round to the nearest thousandth.

Hints

- Identify the probability distribution for the number of successes. - Which values of \(X\) are included in “at least \(13\)”? - How does \(P(X\ge 13)\) change as \(p\) increases? - Use a graphing calculator, table, or numerical solver to find the boundary value of \(p\).

Solution

1. Let \(X\) be the number of successes. Then \(X\) has a binomial distribution with \(n=15\) and success probability \(p\). 2. The condition is \(P(X\ge 13)\ge 0.75\). 3. At the boundary, solve \(P(X\ge 13)=\sum_{k=13}^{15}\binom{15}{k}p^k(1-p)^{15-k}=0.75\). 4. A numerical solution gives \(p\approx 0.8836624\). 5. Rounded to the nearest thousandth, the minimum required success probability is \(p\approx 0.884\).

Answer

The success probability must be at least approximately \(0.884\).
52351612
A quality-control inspector checks samples of \(20\) LED bulbs. Each bulb independently has probability \(p\) of being defective. A sample is accepted only when at most \(1\) bulb is defective. Find the boundary value of \(p\) for which a sample is accepted with probability \(90\%\). Round the boundary value to the nearest thousandth.

Hints

- Translate “at most \(1\) defective bulb” into possible values of \(X\). - Write an equation for the cumulative binomial probability. - Decide whether decreasing \(p\) raises or lowers the acceptance probability. - Use a binomial probability function and a numerical solver to locate the boundary value.

Solution

1. Let \(X\) be the number of defective bulbs in a sample. Then \(X\) has a binomial distribution with \(n=20\) and defect probability \(p\). 2. At the boundary, \(P(X\le 1)=0.90\). 3. This gives \((1-p)^{20}+20p(1-p)^{19}=0.90\). 4. Solving numerically gives \(p\approx 0.0269141\). 5. Therefore, the boundary value is \(p\approx 0.027\) to the nearest thousandth. The acceptance probability is at least \(90\%\) for \(p\le 0.0269141\).

Answer

The boundary value is \(p\approx 0.027\). More precisely, the acceptance probability is at least \(90\%\) when \(p\le 0.0269141\).
52352412
The germination probability for each seed of a certain type of flower is \(25\%\), independently of the other seeds. What is the minimum number of seeds a gardener must plant so that the probability that at least \(3\) seeds germinate is at least \(90\%\)?

Hints

- Identify the known and unknown binomial parameters. - Use the complement of “at least \(3\)” to reduce the number of cases. - Evaluate the cumulative probability \(P(X\le 2)\). - Test consecutive integer values of \(n\) to locate the minimum.

Solution

1. Let \(X\) be the number of seeds that germinate. Then \(X\) has a binomial distribution with success probability \(p=0.25\) and an unknown number of trials \(n\). 2. The condition is \(P(X\ge 3)\ge 0.90\). 3. Using the complement, require \(P(X\le 2)\le 0.10\). 4. Test integer values of \(n\) using \(P(X\le 2)=\sum_{k=0}^{2}\binom{n}{k}(0.25)^k(0.75)^{n-k}\). 5. For \(n=19\), \(P(X\le 2)\approx 0.1113\), so \(P(X\ge 3)\approx 0.8887\). 6. For \(n=20\), \(P(X\le 2)\approx 0.0913\), so \(P(X\ge 3)\approx 0.9087\). 7. Therefore, the minimum number of seeds is \(20\).

Answer

The gardener must plant at least \(20\) seeds.
52352912
An archer has the same probability \(p\) of hitting the bull's-eye on each of \(10\) independent shots. The archer wants at least an \(80\%\) chance of hitting the bull's-eye on at least \(8\) shots. Find the threshold value \(p_0\) such that the goal is met when \(p\ge p_0\). Report \(p_0\) to three decimal places. Then, if \(p\) can be specified only in increments of \(0.01\), state the smallest value that meets the goal.

Hints

- Write the probability of at least \(8\) hits as the sum of the probabilities of exactly \(8\), \(9\), and \(10\) hits. - The unknown probability appears in several powers, so solve the threshold equation numerically. - After finding the threshold, test the adjacent hundredth values rather than rounding without checking the inequality.

Solution

1. Let \(X\sim\operatorname{Bin}(10,p)\). The condition is \(P(X\ge 8)\ge 0.80\). 2. The upper-tail probability is \(P(X\ge 8)=\binom{10}{8}p^8(1-p)^2+\binom{10}{9}p^9(1-p)+p^{10}\). 3. This probability increases as \(p\) increases. Solving \(P(X\ge 8)=0.80\) numerically gives \(p_0\approx 0.842365\), so the threshold rounded to three decimal places is \(0.842\). 4. At \(p=0.84\), \(P(X\ge 8)\approx 0.7936<0.80\), while at \(p=0.85\), \(P(X\ge 8)\approx 0.8202>0.80\). Therefore, the smallest hundredth that meets the goal is \(0.85\).

Answer

The threshold is \(p_0\approx 0.842\). If \(p\) is restricted to increments of \(0.01\), the smallest value that meets the goal is \(0.85\).
52353012
A shipment contains \(20\) electronic components. Each component independently meets quality standards with probability \(p\). The probability that at least \(18\) components meet the standards must be at least \(70\%\). Find the threshold value \(p_0\) such that the requirement is met when \(p\ge p_0\). Report \(p_0\) to three decimal places. Then, if \(p\) can be specified only in increments of \(0.01\), state the smallest value that meets the requirement.

Hints

- Write the probability of at least \(18\) acceptable components as the sum of the probabilities of exactly \(18\), \(19\), and \(20\). - Solve the threshold equation numerically because it cannot be isolated for \(p\) by elementary algebra. - After finding the threshold, test the adjacent hundredth values rather than rounding without checking the inequality.

Solution

1. Let \(X\sim\operatorname{Bin}(20,p)\). The condition is \(P(X\ge 18)\ge 0.70\). 2. The upper-tail probability is \(P(X\ge 18)=\binom{20}{18}p^{18}(1-p)^2+\binom{20}{19}p^{19}(1-p)+p^{20}\). 3. This probability increases as \(p\) increases. Solving \(P(X\ge 18)=0.70\) numerically gives \(p_0\approx 0.904057\), so the threshold rounded to three decimal places is \(0.904\). 4. At \(p=0.90\), \(P(X\ge 18)\approx 0.6769<0.70\), while at \(p=0.91\), \(P(X\ge 18)\approx 0.7334>0.70\). Therefore, the smallest hundredth that meets the requirement is \(0.91\).

Answer

The threshold is \(p_0\approx 0.904\). If \(p\) is restricted to increments of \(0.01\), the smallest value that meets the requirement is \(0.91\).
52353812
A quality-control inspector examines boxes containing \(50\) glass bottles each. Each bottle independently has probability \(p\) of containing a small air bubble. The probability that a box contains at most \(2\) bottles with air bubbles is exactly \(15\%\). Find \(p\). Round your answer to the nearest thousandth.

Hints

- Write the cumulative binomial probability \(P(X\le 2)\). - Substitute \(n=50\) and the possible values \(X=0,1,2\). - Use a numerical solver or a table because the equation is difficult to solve algebraically. - Test nearby values of \(p\) to locate where the cumulative probability is \(0.15\).

Solution

1. Let \(X\) be the number of bottles with air bubbles. Then \(X\) has a binomial distribution with \(n=50\) and probability \(p\). 2. The condition is \(P(X\le 2)=0.15\). 3. Therefore, \(\sum_{k=0}^{2}\binom{50}{k}p^k(1-p)^{50-k}=0.15\). 4. Solve the equation numerically. 5. The solution is \(p\approx 0.0919156\). For comparison, \(P(X\le 2)\approx 0.1550\) when \(p=0.091\) and \(P(X\le 2)\approx 0.1495\) when \(p=0.092\). 6. Rounded to the nearest thousandth, \(p\approx 0.092\).

Answer

The probability that a bottle contains a small air bubble is approximately \(0.092\), or \(9.2\%\).
52353912
Each computer chip produced by a manufacturer independently has a \(4\%\) chance of being defective. A sample of \(20\) chips is selected. a) What is the probability that exactly \(1\) chip in the sample is defective? b) What is the probability that at most \(2\) chips in the sample are defective? c) What is the minimum number of chips that must be checked so that the probability of finding at least \(1\) defective chip is at least \(99\%\)?

Hints

- Identify the binomial parameters for the sample. - For “exactly \(k\),” use the binomial probability formula directly. - Which values of \(X\) are included in “at most \(2\)”? - For “at least \(1\),” use the complement of finding no defective chips. - Use logarithms or systematic trial when the unknown appears in an exponent.

Solution

1. For parts a) and b), let \(X\) be the number of defective chips in a sample of \(20\). Then \(X\) has a binomial distribution with \(n=20\) and \(p=0.04\). 2. \(P(X=1)=\binom{20}{1}(0.04)(0.96)^{19}\approx 0.3683\). 3. \(P(X\le 2)=\sum_{k=0}^{2}\binom{20}{k}(0.04)^k(0.96)^{20-k}\approx 0.9561\). 4. For part c), require \(P(X\ge 1)\ge 0.99\). Using the complement gives \(1-(0.96)^n\ge 0.99\), so \((0.96)^n\le 0.01\). 5. Taking logarithms gives \(n\ge \frac{\ln(0.01)}{\ln(0.96)}\approx 112.811\). Therefore, the minimum is \(n=113\).

Answer

a) The probability is approximately \(36.83\%\). b) The probability is approximately \(95.61\%\). c) At least \(113\) chips must be checked.
52354512
A factory produces LED bulbs, and each bulb independently has a \(4\%\) probability of being defective. a) A package contains \(25\) bulbs. What is the probability that at least \(1\) bulb is defective? b) What is the minimum number of bulbs an inspector must test so that the probability of finding at least \(1\) defective bulb is at least \(90\%\)? c) What is the minimum number of bulbs needed so that the probability of finding at least \(2\) defective bulbs is at least \(90\%\)?

Hints

- For each “at least” event, consider using a complement. - What is the complement of finding at least \(1\) defective bulb? - Use logarithms when the unknown sample size appears in an exponent. - Remember that dividing an inequality by a negative logarithm reverses the inequality sign. - For part c), test integer values of \(n\) using a cumulative binomial probability.

Solution

1. For part a), \(P(X\ge 1)=1-P(X=0)=1-(0.96)^{25}\approx 0.6396\). 2. For part b), require \(1-(0.96)^n\ge 0.90\), so \((0.96)^n\le 0.10\). 3. Taking logarithms gives \(n\ge \frac{\ln(0.10)}{\ln(0.96)}\approx 56.406\). Thus, the minimum is \(n=57\). 4. For part c), require \(P(X\ge 2)\ge 0.90\), or equivalently \(P(X\le 1)\le 0.10\). 5. Test consecutive integer values using \(P(X\le 1)=(0.96)^n+n(0.04)(0.96)^{n-1}\). 6. For \(n=95\), \(P(X\ge 2)\approx 0.8974\). For \(n=96\), \(P(X\ge 2)\approx 0.9007\). 7. Therefore, the minimum is \(n=96\).

Answer

a) The probability is approximately \(63.96\%\). b) At least \(57\) bulbs must be tested. c) At least \(96\) bulbs must be tested.
52361612
During normal operation, \(5\%\) of a factory's components are defective. A random sample of \(20\) components is inspected, and defect outcomes are treated as independent. Let \(X\) be the number of defective components. a) Find the probability that the sample contains at least \(2\) defective components. b) Find the smallest integer \(k\) for which the probability that the sample contains at least \(k\) defective components is less than \(1\%\).

Hints

- Identify the binomial parameters from the situation. - Translate “at least \(2\)” into an inequality for \(X\). - Use a complement to simplify the first probability. - For part b), test consecutive values of \(k\) or use cumulative binomial probabilities.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(20,0.05)\). 2. For part a), use the complement: \(P(X\ge 2)=1-P(X\le 1)\). 3. Since \(P(X=0)=(0.95)^{20}\approx 0.3585\) and \(P(X=1)=20\cdot 0.05\cdot (0.95)^{19}\approx 0.3774\), using unrounded values gives \(P(X\ge 2)=1-P(X=0)-P(X=1)\approx 0.2642\). 4. For part b), seek the smallest \(k\) such that \(P(X\ge k)<0.01\), or equivalently \(P(X\le k-1)>0.99\). 5. Since \(P(X\le 3)\approx 0.9841<0.99\) but \(P(X\le 4)\approx 0.9974>0.99\), the smallest qualifying value is \(k=5\). In fact, \(P(X\ge 5)\approx 0.0026\).

Answer

a) \(P(X\ge 2)\approx 0.2642\), or about \(26.42\%\) b) \(k=5\)
52520812
The fill weight \(X\) of coffee packages is normally distributed with mean \(\mu=500\,\text{g}\) and standard deviation \(\sigma=4\,\text{g}\). A package is underweight if it weighs less than \(495\,\text{g}\). 1. Find the proportion of packages classified as underweight. 2. A supermarket receives \(100\) packages. Find the probability that exactly \(10\) are underweight. 3. What is the minimum number of packages in a delivery for the probability of at least one underweight package to exceed \(99\%\)?

Hints

- Use the population proportion as the binomial success probability. - Apply the formula for exactly \(k\) successes. - For “at least one,” use the complement of zero successes. - Use logarithms to solve an inequality with the variable in an exponent.

Solution

1. The standard score for \(495\) is \(z=\frac{495-500}{4}=-1.25\). Thus, \(p=P(X<495)=\Phi(-1.25)\approx0.10565\). 2. Let \(Y\sim\operatorname{Bin}(100,p)\). Then \(P(Y=10)=\binom{100}{10}p^{10}(1-p)^{90}\approx0.1296\). 3. For a delivery of \(n\) packages, \(P(Y\ge1)=1-(1-p)^n\). We need \(1-(1-p)^n>0.99\), so \((1-p)^n<0.01\). 4. Therefore, \(n>\frac{\ln(0.01)}{\ln(1-p)}\approx41.24\). The minimum integer is \(42\).

Answer

1. Approximately \(0.1056\), or \(10.56\%\) 2. Approximately \(0.1296\), or \(12.96\%\) 3. \(42\) packages
52574912
A spinner has five equal sections labeled \(1\), \(2\), \(3\), \(4\), and \(5\). It is spun three times, producing results \(x_1\), \(x_2\), and \(x_3\). Define the partial sums \(S_1=x_1\), \(S_2=x_1+x_2\), and \(S_3=x_1+x_2+x_3\). a) Find the probability that none of the three partial sums is divisible by \(5\). b) The three-spin game is played \(10\) times. Find the probability that none of the partial sums is divisible by \(5\) in exactly \(5\) of the games.

Hints

- For each spin, determine how many of the five outcomes would make the current partial sum divisible by \(5\). - Does the number of disallowed outcomes on a later spin depend on the previous total? - For the repeated games, identify the binomial parameters \(n\), \(p\), and \(k\).

Solution

1. For \(S_1\) not to be divisible by \(5\), the first spin cannot be \(5\), so the probability is \(\frac45\). 2. If a current partial sum is not divisible by \(5\), exactly one of the five possible next spin values would make the new partial sum divisible by \(5\). Thus, at each of the next two spins, the probability of avoiding a multiple of \(5\) is also \(\frac45\). 3. Therefore, the probability of success in one game is \(p=\left(\frac45\right)^3=\frac{64}{125}=0.512\). 4. For \(10\) independent games, let \(K\) count the successful games. Then \(K\) is binomial with \(n=10\) and \(p=0.512\). 5. \(P(K=5)=\binom{10}{5}(0.512)^5(0.488)^5\approx0.2454\).

Answer

a) \(\frac{64}{125}=0.512\) b) \(P(K=5)\approx0.2454\)
52703412
Let \(X\sim\operatorname{Bin}\left(60,\frac{1}{3}\right)\). a) Find the probability that \(X\) differs from its mean by at most \(2\). b) Find \(P(|X-20|\ge 4)\). c) Find the smallest nonnegative integer \(k\) for which \(P(|X-20|\le k)>0.5\).

Hints

- First find the mean of the binomial distribution. - Translate “differs by at most \(2\)” into an interval for \(X\). - Use a complement for the event in part b). - Test small values of \(k\) in increasing order.

Solution

1. The mean is \(\mu=np=60\cdot\frac{1}{3}=20\). 2. For part a), \(|X-20|\le 2\) means \(18\le X\le 22\). Using a binomial cumulative distribution function, \(P(18\le X\le 22)=P(X\le 22)-P(X\le 17)\approx 0.5061\). 3. For part b), use the complement: \(P(|X-20|\ge 4)=1-P(|X-20|<4)=1-P(17\le X\le 23)\approx 0.3377\). 4. For \(k=1\), \(P(19\le X\le 21)\approx 0.3183\), which is not greater than \(0.5\). For \(k=2\), part a) gives approximately \(0.5061>0.5\). Therefore, the smallest value is \(k=2\).

Answer

a) \(P(|X-20|\le 2)\approx 0.5061\) b) \(P(|X-20|\ge 4)\approx 0.3377\) c) \(k=2\)
52704912
Let \(X\) be a binomial random variable with \(n=14\) and \(p=0.4\). Find every value of \(k\) for which \(P(X=k)\) is greatest. Justify the maximum by analyzing the ratio \(\frac{P(X=k)}{P(X=k-1)}\).

Hints

- A probability is larger than the preceding one when their ratio is greater than \(1\). - Form the ratio of two consecutive binomial probabilities. - Interpret what happens when the ratio equals \(1\). - Simplify the ratio of the consecutive binomial coefficients.

Solution

1. For a binomial random variable, \(\frac{P(X=k)}{P(X=k-1)}=\frac{n-k+1}{k}\frac{p}{1-p}\). 2. Substituting \(n=14\) and \(p=0.4\) gives \(\frac{P(X=k)}{P(X=k-1)}=\frac{15-k}{k}\frac{2}{3}\). 3. The probabilities are nondecreasing while this ratio is at least \(1\): \(\frac{15-k}{k}\frac23\ge1\). 4. Solving gives \(2(15-k)\ge3k\), so \(30\ge5k\), or \(k\le6\). 5. At \(k=6\), the ratio equals \(1\), so \(P(X=6)=P(X=5)\). Before that point the probabilities increase, and after that point they decrease. 6. Therefore, the maximum occurs at both \(k=5\) and \(k=6\).

Answer

\(P(X=k)\) is greatest for \(k=5\) and \(k=6\).
52707012
A certain type of flower seed germinates independently with probability \(0.90\). Each packet contains \(12\) seeds. a) Find the probability that at least \(11\) seeds in a randomly selected packet germinate. b) A gardener buys \(5\) packets. Find the probability that at most \(2\) of the packets each have at least \(11\) seeds germinate.

Hints

- Which values are included in “at least \(11\)”? - Use the unrounded result from part a) as the success probability in part b). - Which packet counts are included in “at most \(2\)”? - Round only the final result.

Solution

1. Let \(X\) be the number of seeds that germinate in one packet. Then \(X\) has a binomial distribution with \(n=12\) and \(p=0.90\). 2. \(P(X\ge 11)=P(X=11)+P(X=12)\). 3. \(P(X\ge 11)=\binom{12}{11}(0.90)^{11}(0.10)+(0.90)^{12}\approx 0.659002\). 4. Let \(Y\) be the number of the \(5\) packets in which at least \(11\) seeds germinate. Then \(Y\) has a binomial distribution with \(n=5\) and \(p\approx 0.659002\). 5. \(P(Y\le 2)=\sum_{k=0}^{2}\binom{5}{k}(0.659002)^k(1-0.659002)^{5-k}\approx 0.2214\).

Answer

a) The probability is approximately \(65.90\%\). b) The probability is approximately \(22.14\%\).
52707412
At a distribution center, an automated conveyor system sorts packages with an error rate of \(4\%\). A sample of \(50\) consecutively sorted packages is inspected. 1. Using a binomial model, find the probability that more than \(3\) packages were sorted incorrectly. 2. Using the same model, find the probability that the number of incorrectly sorted packages equals the expected value. 3. Evaluate the independence assumption if sorting errors are often caused by brief mechanical jams that affect several consecutive packages.

Hints

- Distinguish “more than \(3\)” from “at least \(3\).” - Use \(E(X)=np\) for a binomial random variable. - Ask whether one package’s outcome changes the probability of the next package’s outcome. - Independence is a required condition for the binomial model.

Solution

1. Under a binomial model, let \(X\) be the number of sorting errors. Then \(X\) has parameters \(n=50\) and \(p=0.04\). 2. \(P(X>3)=1-P(X\le 3)\approx 1-0.8609=0.1391\). 3. The expected value is \(E(X)=np=50\cdot0.04=2\). 4. \(P(X=2)=\binom{50}{2}(0.04)^2(0.96)^{48}\approx 0.2762\). 5. If a mechanical jam affects several consecutive packages, the error outcomes are clustered rather than independent. An error on one package makes errors on nearby packages more likely, so a binomial model would not be exact.

Answer

1. The probability is approximately \(13.91\%\). 2. The expected value is \(2\), and \(P(X=2)\approx 27.62\%\). 3. The independence condition is not reasonable when one jam can cause errors for several consecutive packages.
52708412
A multiple-choice test has \(20\) questions. Each question has \(4\) answer choices, exactly one of which is correct. A test taker guesses independently and randomly on every question. Find the probability of each event. a) Only the first \(5\) answers are correct; all remaining answers are incorrect. b) More than \(15\) answers are incorrect. c) The first \(3\) answers are correct, and exactly \(6\) answers are correct in total. d) At least \(4\) of the first \(5\) answers are correct, and at most \(1\) of the last \(5\) answers is correct.

Hints

- What is the probability of guessing one question correctly? - Translate more than \(15\) incorrect answers into a condition on the number correct. - In part c), how many additional correct answers are needed among the remaining \(17\) questions? - Treat the first \(5\) and last \(5\) questions as separate independent blocks in part d).

Solution

1. The probability of a correct answer is \(p=0.25\), and the probability of an incorrect answer is \(0.75\). 2. Part a) requires one fixed sequence, so \(P=(0.25)^5(0.75)^{15}\approx 0.00001305\). 3. More than \(15\) incorrect answers means fewer than \(5\) correct answers. If \(X\) is the total number correct, \(P(X\le 4)\approx 0.4148\). 4. In part c), the first \(3\) answers are already correct, so exactly \(3\) of the remaining \(17\) must be correct. Thus, \(P=\binom{17}{3}(0.25)^6(0.75)^{14}\approx 0.00296\). 5. For part d), let \(Y_1\) be the number correct among the first \(5\), and let \(Y_2\) be the number correct among the last \(5\). 6. \(P(Y_1\ge 4)=\binom{5}{4}(0.25)^4(0.75)+(0.25)^5=0.015625\). 7. \(P(Y_2\le 1)=(0.75)^5+5(0.25)(0.75)^4=0.6328125\). 8. The two blocks are independent, so the required probability is \(0.015625\cdot 0.6328125\approx 0.0099\).

Answer

a) \(P\approx 0.00001305\) b) \(P\approx 0.4148\) c) \(P\approx 0.00296\) d) \(P\approx 0.0099\)
52709612
A medication produces the desired response in \(85\%\) of patients. Assume patient responses are independent. A preliminary study must be large enough that the probability of observing the response in at least \(10\) patients is at least \(90\%\). Find the minimum number \(n\) of patients needed. Show how you used systematic testing or appropriate technology.

Hints

- Rewrite “at least \(10\)” using a complement so that a cumulative probability can be used. - Because \(n\) is unknown, begin with an integer somewhat larger than \(10\) and test consecutive values. - Consider how increasing the sample size affects the chance of observing at least \(10\) responses.

Solution

1. For a study of size \(n\), let \(X\sim\operatorname{Bin}(n,0.85)\). The requirement is \(P(X\ge 10)\ge 0.90\). 2. Using the complement, the condition is \(P(X\le 9)\le 0.10\). 3. Test consecutive sample sizes near the threshold. For \(n=13\), \(P(X\le 9)\approx 0.1180>0.10\), so \(P(X\ge 10)\approx 0.8820<0.90\). 4. For \(n=14\), \(P(X\le 9)\approx 0.0467\le 0.10\), so \(P(X\ge 10)\approx 0.9533\ge 0.90\). 5. Because \(n=13\) fails and \(n=14\) succeeds, the minimum sample size is \(n=14\).

Answer

The study needs at least \(14\) patients.
52710012
A certain flower seed germinates independently with probability \(0.80\). a) Twelve seeds are planted. Find the probabilities of these events: - \(C\): At most \(10\) seeds germinate. - \(D\): Exactly \(2\) seeds do not germinate. b) Explain the meaning of \(1-(0.8)^{12}\) in context. c) Among plants that germinate, \(2\%\) are affected by a particular pest. Assume the pest outcomes are independent. What is the minimum number of seeds that must be planted so that the probability of producing at least \(1\) affected plant is greater than \(95\%\)?

Hints

- For “at most \(10\),” decide whether a complement is shorter. - Exactly \(2\) failures is the same as how many germinations? - What event does \((0.8)^{12}\) represent? - Multiply the germination probability by the conditional pest probability for one seed. - Use logarithms to solve an exponential inequality for \(n\).

Solution

1. For part a), let \(X\) be the number of seeds that germinate. Then \(X\) has a binomial distribution with \(n=12\) and \(p=0.80\). 2. \(P(C)=P(X\le 10)=1-[P(X=11)+P(X=12)]\approx 0.7251\). 3. Event \(D\) means exactly \(10\) seeds germinate, so \(P(D)=\binom{12}{10}(0.80)^{10}(0.20)^2\approx 0.2835\). 4. The value \((0.8)^{12}\) is the probability that all \(12\) seeds germinate. Therefore, \(1-(0.8)^{12}\) is the probability that at least \(1\) seed does not germinate. 5. For one planted seed, the probability that it germinates and the resulting plant is affected is \((0.80)(0.02)=0.016\). 6. Require \(1-(0.984)^n>0.95\), so \((0.984)^n<0.05\). 7. Taking logarithms gives \(n>\frac{\ln(0.05)}{\ln(0.984)}\approx 185.731\). Therefore, the minimum is \(n=186\).

Answer

a) \(C\): \(P(C)\approx 72.51\%\) \(D\): \(P(D)\approx 28.35\%\) b) The expression is the probability that at least \(1\) of the \(12\) seeds does not germinate. c) At least \(186\) seeds must be planted.
52710912
A glass manufacturer finds that each drinking glass independently has an \(8\%\) chance of having small inclusions and being classified as defective. a) Six glasses are inspected in order. Find the probability of each event. \(A\): Only the last \(2\) glasses are defective. \(B\): At most \(2\) glasses are defective. b) A sample of \(50\) glasses is selected. Find the probability that at least \(2\) are defective. c) Three glasses are packed in each box. A box is acceptable if none of its \(3\) glasses is defective. Find the probability that a randomly selected box is acceptable. d) What is the minimum number of three-glass boxes that must be inspected so that the probability of finding at least \(1\) acceptable box is at least \(95\%\)? e) What is the maximum defect probability \(p\) that would make the probability of finding no defective glasses in a box of \(20\) at least \(90\%\)?

Hints

- Distinguish a fixed sequence from any arrangement with the same number of defects. - For “at least” events, consider using a complement. - To find a minimum number of trials, write the probability of no successes and use logarithms. - In part e), the probability of a nondefective glass is \(1-p\).

Solution

1. The defect probability is \(0.08\), and the nondefect probability is \(0.92\). 2. Event \(A\) requires one fixed sequence: \((0.92)^4(0.08)^2\approx 0.004585\). 3. For Event \(B\), let \(X\) be the number of defective glasses among \(6\). Then \(P(X\le 2)\approx 0.991488\). 4. For part b), \(P(X\ge 2)=1-[P(X=0)+P(X=1)]\approx 0.917288\) for \(n=50\) and \(p=0.08\). 5. For part c), all \(3\) glasses must be nondefective, so \(p_B=(0.92)^3=0.778688\). 6. For part d), require \(1-(1-p_B)^n\ge 0.95\). Thus, \((0.221312)^n\le 0.05\), giving \(n\ge \frac{\ln(0.05)}{\ln(0.221312)}\approx 1.986\). The minimum is \(2\) boxes. 7. For part e), require \((1-p)^{20}\ge 0.90\). 8. Solving gives \(p\le 1-(0.90)^{1/20}\approx 0.005254\), or approximately \(0.525\%\).

Answer

a) \(A\): \(P(A)\approx 0.46\%\) \(B\): \(P(B)\approx 99.15\%\) b) \(P(X\ge 2)\approx 91.73\%\) c) \(P(\text{acceptable})\approx 77.87\%\) d) At least \(2\) boxes must be inspected. e) The defect probability must be at most approximately \(0.525\%\).
52711012
A seed company states that a certain variety of vegetable seed has a germination probability of \(75\%\). Assume that the seeds germinate independently. a) A gardener plants \(5\) seeds in a pot. Find the probability that exactly \(4\) seeds germinate. b) Find the probability that at least \(4\) of the \(5\) seeds germinate. c) The gardener plants a row of \(10\) seeds. Find the probability that the first \(3\) seeds germinate and at least \(4\) of the remaining \(7\) seeds germinate. d) What is the minimum number of seeds the gardener must plant so that the probability that at least \(1\) seed germinates is at least \(99\%\)? e) Suppose the germination probability is unknown. What is the minimum value of \(p\) needed so that all \(10\) seeds in a packet germinate with probability at least \(95\%\)?

Hints

- In part c), multiply the probabilities of the two independent conditions. - For the minimum number in part d), use the complement in which no seeds germinate. - When solving for \(p\), write an inequality in which \(p\) is the base of a power.

Solution

1. Let \(X\sim\operatorname{Bin}(5,0.75)\). 2. For part a), \(P(X=4)=\binom{5}{4}(0.75)^4(0.25)\approx 0.395508\). 3. For part b), \(P(X\ge 4)=P(X=4)+P(X=5)\approx 0.395508+0.237305=0.632812\). 4. For part c), the probability that the first \(3\) seeds germinate is \((0.75)^3\). If \(Y\sim\operatorname{Bin}(7,0.75)\), then \(P(Y\ge 4)\approx 0.929443\). 5. By independence, the required probability is \((0.75)^3P(Y\ge 4)\approx 0.392109\). 6. For part d), require \(1-(0.25)^n\ge 0.99\). Thus, \(n\ge \frac{\ln(0.01)}{\ln(0.25)}\approx 3.322\), so the minimum is \(4\). 7. For part e), require \(p^{10}\ge 0.95\). Therefore, \(p\ge (0.95)^{1/10}\approx 0.994884\).

Answer

a) \(P(X=4)\approx 39.55\%\) b) \(P(X\ge 4)\approx 63.28\%\) c) The probability is approximately \(39.21\%\). d) The gardener must plant at least \(4\) seeds. e) The germination probability must be at least approximately \(99.488\%\).
52711412
During basketball practice, a player shoots free throws in sets of \(4\). The player makes each free throw with probability \(0.70\), independently of all other shots. A set is considered successful if the player makes at least \(3\) of the \(4\) free throws. The player completes \(10\) sets. Find the probability that at least \(9\) sets are successful.

Hints

- First find the probability that one set is successful. Include both possible successful outcomes. - Once you know the success probability for one set, model the \(10\) sets as a new sequence of Bernoulli trials. - Which possible numbers of successful sets are included in “at least \(9\)”? - Use the binomial probability formula for those outcomes.

Solution

1. The probability that one set is successful is \(p_S=\binom{4}{3}(0.70)^3(0.30)+(0.70)^4=0.6517\). 2. Let \(Y\) be the number of successful sets among the \(10\) sets. Then \(Y\sim\operatorname{Bin}(10,0.6517)\). 3. The required probability is \(P(Y\ge 9)=\binom{10}{9}(0.6517)^9(0.3483)+(0.6517)^{10}\approx 0.087675\).

Answer

The probability is approximately \(8.77\%\).
52711512
A soldering defect occurs in \(5\%\) of electronic components. A random sample is selected for quality inspection, and defect outcomes are treated as independent. Find the minimum sample size \(n\) needed so that the probability of finding at least \(2\) components with soldering defects is at least \(90\%\).

Hints

- Model the number of defective components with a binomial random variable. - Use the complement event that at most one component has the defect. - Determine when the complement probability becomes small enough. - Test consecutive sample sizes with a calculator or statistical software.

Solution

1. For a sample of size \(n\), let \(X\sim\operatorname{Bin}(n,0.05)\). The requirement is \(P(X\ge 2)\ge 0.90\). 2. Using the complement, the condition becomes \(P(X\le 1)\le 0.10\). 3. Therefore, \((0.95)^n+n\cdot 0.05\cdot (0.95)^{n-1}\le 0.10\). 4. Test consecutive sample sizes near the threshold. For \(n=76\), \(P(X\ge 2)\approx 0.8986<0.90\). For \(n=77\), \(P(X\ge 2)\approx 0.9027\ge 0.90\). 5. Because \(76\) fails and \(77\) succeeds, the minimum sample size is \(77\).

Answer

The minimum sample size is \(77\) components.
52711612
A conservation project estimates that each released juvenile animal of a rare species has a \(15\%\) chance of surviving its first year in the wild. Assume the survival outcomes are independent. What is the minimum number of juvenile animals that must be released so that the probability that at least \(5\) survive the first year is at least \(99\%\)?

Hints

- Identify what counts as a success and its probability. - Express “at least \(5\)” as an inequality for a binomial random variable. - Use the complement event “at most \(4\)” to simplify the calculation. - Test consecutive values of the total number released to locate the minimum.

Solution

1. For \(n\) released animals, let \(X\sim\operatorname{Bin}(n,0.15)\). The requirement is \(P(X\ge 5)\ge 0.99\). 2. Using the complement, the condition becomes \(P(X\le 4)\le 0.01\). 3. Test consecutive values of \(n\) near the threshold. For \(n=73\), \(P(X\le 4)\approx 0.0105>0.01\), so \(P(X\ge 5)\approx 0.9895<0.99\). 4. For \(n=74\), \(P(X\le 4)\approx 0.0094\le 0.01\), so \(P(X\ge 5)\approx 0.9906\ge 0.99\). 5. Therefore, the minimum number is \(74\).

Answer

At least \(74\) juvenile animals must be released.
52713012
A factory produces microchips, and \(8\%\) are defective. A quality-control sample of \(400\) chips is tested sequentially, with outcomes treated as independent. Let \(X\) be the number of defective chips in the sample. a) Find the probability that the twelfth chip tested is the first defective chip. b) Find the probability that \(X\) is within one standard deviation of its mean.

Hints

- What must happen on each of the first \(11\) tests? - Use the binomial formulas for the mean and standard deviation. - Which integer counts lie inside \([\mu-\sigma,\mu+\sigma]\)? - Express the interval probability as a difference of cumulative probabilities.

Solution

1. For the twelfth chip to be the first defective one, the first \(11\) chips must be nondefective and the twelfth must be defective. Thus the probability is \((0.92)^{11}(0.08)\approx 0.0320\). 2. For \(X\sim\operatorname{Bin}(400,0.08)\), the mean is \(\mu=np=400\cdot 0.08=32\), and the standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{400\cdot 0.08\cdot 0.92}\approx 5.43\). 3. The interval \([\mu-\sigma,\mu+\sigma]\) is approximately \([26.57,37.43]\). Since \(X\) is integer-valued, the event is \(27\le X\le 37\). 4. Using the binomial cumulative distribution function, \(P(27\le X\le 37)=P(X\le 37)-P(X\le 26)\approx 0.6898\).

Answer

a) \((0.92)^{11}(0.08)\approx 0.0320\), or about \(3.20\%\) b) \(P(27\le X\le 37)\approx 0.6898\), or about \(68.98\%\)
52715212
Historically, each microchip produced by a manufacturer independently has an \(8\%\) chance of being defective. A new manufacturing process is intended to reduce the defect rate. To evaluate the process, the manufacturer tests \(250\) chips from the new production line. The process is recognized as a significant improvement if the number of defective chips in the sample is at least \(1.5\) original standard deviations below the original expected number. Find the probability that the improvement is not recognized when the new process has actually reduced the defect rate to \(4\%\).

Hints

- First find the mean and standard deviation under the original defect rate. - Write the condition for a recognized improvement as an inequality involving the number of defects. - Convert the cutoff carefully to a whole-number count. - Calculate the final probability using the new defect rate and the event above the recognition cutoff.

Solution

1. Under the original process, \(n=250\) and \(p_0=0.08\). The mean is \(\mu_0=250\cdot0.08=20\), and the standard deviation is \(\sigma_0=\sqrt{250\cdot0.08\cdot0.92}\approx 4.2895\). 2. The improvement is recognized when \(X\le \mu_0-1.5\sigma_0\approx 20-1.5\cdot4.2895=13.5657\). Since \(X\) is an integer, recognition occurs when \(X\le 13\). 3. Under the new process, let \(X\sim\operatorname{Bin}(250,0.04)\). 4. The improvement is not recognized when \(X\ge 14\). Therefore, \(P(X\ge 14)=1-P(X\le 13)\approx 0.131050\).

Answer

The probability is approximately \(13.10\%\).
52876012
In a large shipment of electronic components, each component is defective independently with probability \(0.10\). A random sample of \(20\) components is selected. Let \(X\) be the number of defective components. a) Find the mean of \(X\). b) Find the shortest interval of consecutive integer values that contains the mean and has probability greater than \(90\%\).

Hints

- Use the mean formula for a binomial random variable. - Add point probabilities for consecutive integer counts near the mean. - Start with short intervals containing the mean and increase the number of included values only as needed. - Remember that a count-valued random variable takes only integer values.

Solution

1. The random variable has distribution \(X\sim\operatorname{Bin}(20,0.1)\), so its mean is \(\mu=np=20\cdot 0.1=2\). 2. The relevant point probabilities are \(P(X=0)\approx 0.1216\), \(P(X=1)\approx 0.2702\), \(P(X=2)\approx 0.2852\), \(P(X=3)\approx 0.1901\), \(P(X=4)\approx 0.0898\), and \(P(X=5)\approx 0.0319\). 3. The four-value intervals containing the mean are \([-1,2]\), \([0,3]\), \([1,4]\), and \([2,5]\). Their probabilities are approximately \(0.6770\), \(0.8670\), \(0.8352\), and \(0.5970\), respectively, so none exceeds \(0.90\). Therefore, no interval containing the mean with at most four integer values can exceed \(0.90\). 4. The five-value interval \([0,4]\) has probability \(P(0\le X\le 4)\approx 0.9568>0.90\). 5. Therefore, the shortest qualifying interval is \([0,4]\).

Answer

a) \(\mu=2\) b) \([0,4]\), with probability approximately \(0.9568\), or \(95.68\%\)
53101012
In a microchip manufacturing process, \(5\%\) of chips are defective, independently of one another. Chips are packaged in groups of \(10\). A package is rejected if it contains at least one defective chip. An inspector examines \(20\) packages. 1. Find the probability that one randomly selected package is rejected. 2. Find the probability that exactly \(5\) of the \(20\) packages are rejected. 3. Find the probability that at most \(1\) of the \(20\) packages is rejected.

Hints

- Find the package rejection probability using the complement of no defective chips. - Use that probability as the success probability in a second binomial model. - “At most \(1\)” includes two exact counts.

Solution

1. A package is accepted only if all \(10\) chips are nondefective. Therefore, the rejection probability is \(p_R=1-(0.95)^{10}\approx 0.4013\). 2. Let \(Y\) be the number of rejected packages. Then \(Y\sim\operatorname{Bin}(20,p_R)\), so \(P(Y=5)=\binom{20}{5}p_R^5(1-p_R)^{15}\approx 0.0735\). 3. At most \(1\) rejected package means \(Y=0\) or \(Y=1\). Thus \(P(Y\le 1)=(1-p_R)^{20}+20p_R(1-p_R)^{19}\approx 0.000505\).

Answer

1. \(p_R=1-(0.95)^{10}\approx 0.4013\) 2. \(P(Y=5)\approx 0.0735\) 3. \(P(Y\le 1)\approx 0.000505\)
53109212
Consider a binomial setting with \(n\) trials and success probability \(p=\frac1n\). Let \(X\) be the number of successes. a) Prove that for every \(n>1\), \(P(X=1)=\frac{n}{n-1}P(X=0)\). b) For large \(n\), \(\left(1-\frac1n\right)^n\approx\frac1e\). Rewrite \(P(X=2)\) to show that for large \(n\), \(P(X=2)\approx\frac{1}{2e}\).

Hints

- Substitute \(k=0\), \(k=1\), and \(k=2\) into the binomial formula. - Use exponent rules to simplify ratios with the same base. - Expand \(\binom n2\). - Evaluate the limiting value of each factor separately.

Solution

1. \(P(X=0)=\left(1-\frac1n\right)^n\), while \(P(X=1)=n\left(\frac1n\right)\left(1-\frac1n\right)^{n-1}=\left(1-\frac1n\right)^{n-1}\). 2. Therefore, \(\frac{P(X=1)}{P(X=0)}=\frac{1}{1-\frac1n}=\frac{n}{n-1}\), which proves part a). 3. For two successes, \(P(X=2)=\binom n2\left(\frac1n\right)^2\left(1-\frac1n\right)^{n-2}\). 4. Simplifying gives \(P(X=2)=\frac{n-1}{2n}\left(1-\frac1n\right)^{n-2}\). 5. As \(n\) becomes large, \(\frac{n-1}{2n}\to\frac12\), and \(\left(1-\frac1n\right)^{n-2}\to\frac1e\). Therefore, \(P(X=2)\approx\frac{1}{2e}\).

Answer

a) \(\frac{P(X=1)}{P(X=0)}=\frac{n}{n-1}\), so \(P(X=1)=\frac{n}{n-1}P(X=0)\). b) \(P(X=2)=\frac{n-1}{2n}\left(1-\frac1n\right)^{n-2}\approx\frac12\cdot\frac1e=\frac{1}{2e}\).
53109412
For quality control, a glass panel is divided into \(80\) equal sections. The panel contains \(60\) tiny air bubbles that are distributed independently and uniformly across the panel. a) Find the probability that one specified section contains more than \(2\) air bubbles. b) Find the expected number of the \(80\) sections that contain no air bubbles.

Hints

- What is the probability that one air bubble lands in a specified section? - Use a complement for “more than \(2\).” - Find the probability that one specified section is empty, then use it to find the expected number of empty sections.

Solution

1. For one specified section, each air bubble lands there with probability \(\frac{1}{80}=0.0125\). Thus the number \(X\) of air bubbles in that section has distribution \(X\sim\operatorname{Bin}\left(60,\frac{1}{80}\right)\). 2. Use the complement for part a): \(P(X>2)=1-[P(X=0)+P(X=1)+P(X=2)]\approx 0.0395\). 3. A specified section is empty with probability \(P(X=0)=\left(\frac{79}{80}\right)^{60}\approx 0.4701\). 4. By linearity of expectation, the expected number of empty sections is \(80P(X=0)=80\cdot \left(\frac{79}{80}\right)^{60}\approx 37.61\).

Answer

a) \(P(X>2)\approx 0.0395\), or about \(3.95\%\) b) About \(37.6\) sections
53111012
A company has \(20\) electric vehicles available through an employee car-sharing program. At a randomly selected time, each participating employee independently needs a vehicle with probability \(0.10\). What is the greatest number of employees who can participate so that the probability that every employee who needs a vehicle can get one is at least \(95\%\)?

Hints

- The success probability for needing a vehicle is given. - Find the largest \(n\) for which the cumulative probability through \(20\) remains at least \(0.95\). - Test nearby values of \(n\) using binomial technology. - Equivalently, the probability of needing more than \(20\) vehicles must remain at most \(5\%\).

Solution

1. If \(n\) employees participate, let \(X\sim\operatorname{Bin}(n,0.10)\) be the number who need a vehicle at the selected time. 2. The requirement is \(P(X\le 20)\ge 0.95\). 3. Evaluating nearby values gives \(P(X\le 20)\approx 0.952401\) for \(n=143\) and \(P(X\le 20)\approx 0.949330\) for \(n=144\). 4. For fixed success probability \(0.10\), \(P(X\le 20)\) decreases as \(n\) increases. Therefore, the greatest allowable number of employees is \(143\).

Answer

At most \(143\) employees can participate.
53113212
An urn contains \(25\%\) red balls and \(75\%\) blue balls. A ball is drawn with replacement \(192\) times. Let \(X\) be the number of red balls drawn and \(Y\) the number of blue balls drawn. 1. Find the standard deviation of each random variable. 2. Use the binomial distribution to find the probability that \(X\) lies within two standard deviations of its mean. 3. Without another probability calculation, explain why the corresponding two-standard-deviation probability for \(Y\) is exactly the same.

Hints

- Compare the success probabilities for red and blue. - Notice what happens to \(np(1-p)\) when \(p\) and \(1-p\) are interchanged. - Relate “\(k\) red” to “\(192-k\) blue.” - Transform the interval for \(X\) into an interval for \(Y\).

Solution

1. For \(X\), \(\sigma_X=\sqrt{192\cdot0.25\cdot0.75}=\sqrt{36}=6\). For \(Y\), \(\sigma_Y=\sqrt{192\cdot0.75\cdot0.25}=6\). 2. The mean of \(X\) is \(\mu_X=192\cdot0.25=48\), so the two-standard-deviation interval is \([48-12, 48+12]=[36, 60]\). 3. Therefore, \(P(36\le X\le60)=P(X\le60)-P(X\le35)\approx0.9632\). 4. Since \(Y=192-X\), the event \(36\le X\le60\) is exactly equivalent to \(132\le Y\le156\). Because \(\mu_Y=144\) and \(\sigma_Y=6\), this is the two-standard-deviation interval for \(Y\). The two events therefore have the same probability.

Answer

1. \(\sigma_X=6\) and \(\sigma_Y=6\) 2. \(P(36\le X\le60)\approx0.9632\) 3. The probabilities are equal because \(Y=192-X\), so the two intervals describe the same outcomes.
53116812
A multiple-choice test has \(25\) questions. A well-prepared student answers each question correctly with probability \(0.75\), independently of the other questions. Let \(X\) be the number of correct answers. a) Briefly explain why cumulative probabilities are useful for finding an interval probability such as \(P(15\le X\le 20)\). b) Find \(P(15\le X\le 20)\) by defining \(Y\) as the number of incorrect answers and using \(p=0.25\).

Hints

- Compare the work required to add several point probabilities with the work required to subtract two cumulative probabilities. - Translate the interval for correct answers into an interval for incorrect answers. Remember that a larger value of \(X\) gives a smaller value of \(Y\). - Express an inclusive interval probability using two cumulative probabilities.

Solution

1. Cumulative probabilities allow an interval probability to be found by subtracting two cumulative values instead of adding many individual probabilities. 2. Let \(Y\) be the number of incorrect answers. Then \(Y\sim\operatorname{Bin}(25,0.25)\) and \(Y=25-X\). 3. The condition \(15\le X\le 20\) is equivalent to \(5\le Y\le 10\). 4. Therefore, \(P(15\le X\le 20)=P(5\le Y\le 10)=P(Y\le 10)-P(Y\le 4)\approx 0.9703-0.2137=0.7566\).

Answer

a) An interval probability can be found from two cumulative probabilities, \(P(X\le b)-P(X\le a-1)\), instead of summing many point probabilities. b) \(P(15\le X\le 20)=P(5\le Y\le 10)\approx 0.7566\), or about \(75.66\%\)

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