Free math worksheets

- 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 \(n\).

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 \(k=5\).

a) \(n=20\), \(p=0.75\), and \(k=11\) b) \(n=50\), \(p=0.05\), and \(k=2\) c) \(n=9\), \(p=\frac14\), and \(k=5\)

- 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 \(n\).

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.

a) \(P(X=4)=\binom{16}{4}\left(\frac13\right)^4\left(\frac23\right)^{12}\) b) The exponent is \(12\), found from \(n-k\).

- Check whether the number of trials is fixed in advance. - Ask whether each trial has two relevant outcomes with the same success probability. - Check whether the trial outcomes can reasonably be treated as independent.

1. Situation A has a fixed number of trials, two relevant outcomes on each trial, independent trials, and a constant success probability. 2. Therefore, \(X\sim\operatorname{Bin}(10,0.65).\) 3. Situation B is a waiting-time setting, situation C changes probabilities because sampling is without replacement, and situation D does not describe a fixed number of Bernoulli trials.

Situation A. The parameters are \(n=10\) and \(p=0.65\).

- 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 \(1\).

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, \(P(X=6)=\binom{15}{6}(0.25)^6(0.75)^9\approx0.0917\).

\(k=6\), \(p=0.25\), \(m=9\), and \(P(X=6)\approx0.0917\)

- 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 \(1\).

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 \(\square=10\).

a) \(\square=3\), \(\triangle=3\), and \(\bigcirc=7\) b) \(\square=5\) and \(\triangle=\frac23\), with \(n=12\) c) \(\square=10\)

- 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.

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, \(P(X=4)\approx0.1936\).

\(P(X=4)\approx0.1936\)

- Identify \(n\) and \(p\). - Use the binomial probability formula for each fixed number of hits. - Scientific notation is useful for the very small zero-hit probability.

1. Let \(X\sim\operatorname{Bin}(12,0.80)\). 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. \(P(X=0)=(0.2)^{12}=4.096\times10^{-9}.\)

a) \(\binom{12}{10}(0.8)^{10}(0.2)^2\approx0.2835\). b) \(\binom{12}{8}(0.8)^8(0.2)^4\approx0.1329\). c) \(4.096\times10^{-9}\).

- 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.

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.

Machine B. Its probability is \(0.36015\), compared with \(0.32805\) for Machine A.

- 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.

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.

Leah hitting the center on all \(3\) shots is more likely: \(0.421875\) compared with \(0.375\).

- 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.

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 \(10\cdot0.216\cdot0.16=0.3456\).

\(P(X=3)=0.3456\), or \(34.56\%\)

- Define a success as selecting a defective component. - Identify \(n\), \(p\), and \(k\). - Use the binomial formula for exactly \(2\) successes.

1. Let \(X\sim\operatorname{Bin}(10,0.2)\). 2. \(P(X=2)=\binom{10}{2}(0.2)^2(0.8)^8.\) 3. Since \(\binom{10}{2}=45\), \(P(X=2)=0.301989888.\)

\(P(X=2)=0.301989888\).

- Model the count with a binomial distribution. - “At most one” includes \(0\) and \(1\). - Keep exact expressions until the final rounding.

1. Let \(X\sim\operatorname{Bin}(30,0.05)\). 2. \(P(X=0)=0.95^{30}\approx0.2146.\) 3. \(P(X\le1)=0.95^{30}+30(0.05)(0.95)^{29}\approx0.5535.\)

a) \(0.95^{30}\approx0.2146\). b) \(0.95^{30}+30(0.05)(0.95)^{29}\approx0.5535\).

- Identify \(n\) and \(p\). - “At least \(4\)” includes both \(4\) and \(5\) hits. - Keep exact expressions until the final rounding.

1. Let \(X\sim\operatorname{Bin}(5,0.85)\). 2. \(P(X=4)=\binom54(0.85)^4(0.15)\approx0.3915.\) 3. \(P(X\ge4)=\binom54(0.85)^4(0.15)+(0.85)^5\approx0.8352.\)

a) \(\binom54(0.85)^4(0.15)\approx0.3915\). b) \(\binom54(0.85)^4(0.15)+(0.85)^5\approx0.8352\).

- 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.

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 \(P(X=2)=\binom42(0.6)^2(0.4)^2=6\cdot0.36\cdot0.16=0.3456\).

\(P(X=2)=0.3456\), or \(34.56\%\)

- 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.

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, \(P(X=5)=21\cdot0.0131072=0.2752512\).

\(P(X=5)=0.2752512\), or about \(27.53\%\)

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

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), \(P(X=0)=\binom40(0.5)^0(0.5)^4=0.0625\).

a) \(0.2304\) b) \(0.0984\) c) \(0.0625\)

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

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.

\(X=2\) is more likely: about \(30.2\%\), compared with about \(26.8\%\) for \(X=1\).

- 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.

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.

Random experiment: Roll a fair six-sided die \(12\) times. Event \(A\): Exactly two of the rolls show a \(6\).

- 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.

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), \(P(X=2)=\binom{50}{2}(0.05)^2(0.95)^{48}\approx0.2611\).

a) \(0.2501\) b) \(0.1593\) c) \(0.2611\)

- 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.

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), \(P(X=0)=\binom{10}{0}\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^{10}\approx0.1615\).

a) \(0.2907\) b) \(0.0130\) c) \(0.1615\)

- Find the probability of one specified face on a fair four-sided die. - Identify \(n\), \(k\), and \(p\). - Include the probability of the six rolls that are not \(3\)s.

1. Since the four outcomes are equally likely, \(p=\frac14=0.25\), and \(1-p=0.75\). 2. \(P(X=2)=\binom82(0.25)^2(0.75)^6.\) 3. Since \(\binom82=28\), \(P(X=2)=0.31146240234375.\)

a) \(p=0.25\). b) \(P(X=2)=0.31146240234375\).

- 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.

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 \(P(X=60)\approx0.0474\).

\(P(X=60)\approx0.0474\)

- 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 \(1\)?

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 \(P(C)=1-P(B)=1-0.9011=0.0989\).

Event A: \(P(A)=0.6553\) Event B: \(P(B)=0.9011\) Event C: \(P(C)=0.0989\)

- 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.

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. \(P(X=3)=\binom{3}{3}(0.4)^3(0.6)^0=0.064\).

<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>

- 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.

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.

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) \(0.32768+0.4096+0.2048+0.0512+0.0064+0.00032=1\).

- 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.

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: \(P(X\ge1)=1-P(X=0)=1-\frac{9}{64}=\frac{55}{64}\).

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) \(P(X\ge1)=\frac{55}{64}\approx85.9\%\)

- 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.

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), \(P(X=0)=(0.75)^6=\frac{729}{4096}\approx0.1780\).

a) \(\frac{81}{4096}\approx0.0198\) b) \(\frac{1215}{4096}\approx0.2966\) c) \(\frac{729}{4096}\approx0.1780\)

- The success probability is \(1/4\). - Identify \(n=15\) and \(k=5\). - Apply the binomial point-probability formula.

1. \(X\sim\operatorname{Bin}(15,0.25)\). 2. \(P(X=5)=\binom{15}{5}(0.25)^5(0.75)^{10}\). 3. Therefore, \(P(X=5)\approx0.1651\).

\(\binom{15}{5}(0.25)^5(0.75)^{10}\approx0.1651\).

- 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 \(1\).

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 \(P(Y=6)=\binom86\left(\frac23\right)^6\left(\frac13\right)^2=\frac{1792}{6561}\approx0.2731\).

a) \(\frac{1792}{6561}\approx0.2731\), or about \(27.31\%\) b) \(\frac{1792}{6561}\approx0.2731\), or about \(27.31\%\)

- 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.

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 \(0.8\cdot0.2\cdot0.2\cdot0.8=(0.8)^2\cdot(0.2)^2=0.0256\).

a) \(0.1536\), or \(15.36\%\) b) \(0.0256\), or \(2.56\%\)

- 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?

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 \(p=0.5\).

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 \(p=0.5\).

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

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 \(n=15\).

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 \(p=\frac{1}{3}\).

- 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?

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 \(P(X=2)=\binom62\left(\frac13\right)^2\left(\frac23\right)^4=\frac{80}{243}\approx0.3292\).

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, \(P(X=2)=\frac{80}{243}\approx0.3292\).

- 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.

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 \((0.9)^3=0.729\).

1. \(0.059049\), or about \(5.90\%\) 2. \(0.354294\), or about \(35.43\%\) 3. \(0.729\), or \(72.9\%\)

- 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.

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 \(0.7130\).

1. \(P(X=3)\approx 0.2363\) 2. \(P(X\le 4)\approx 0.9102\) 3. \(P(2\le X\le 5)\approx 0.7130\)

- 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.

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, \(F(k)-F(k-1)=P(X=k)\).

Subtracting \(F(k-1)\) from \(F(k)\) removes all probabilities for values below \(k\), leaving exactly \(P(X=k)\).

- 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.

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 \(1-P(X\le 1)=P(X\ge 2)\approx 0.6630\).

a) At most \(2\) successes; \(P\approx 0.8159\) b) At least \(2\) successes; \(P\approx 0.6630\)

- 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.

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.

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

- 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.

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, \(P(X>2)=\binom43(0.6)^3(0.4)+\binom44(0.6)^4=0.4752\).

a) \((0.6)^2(0.4)^2=0.0576\) b) \(\binom42(0.6)^2(0.4)^2=0.3456\) c) \(\binom43(0.6)^3(0.4)+\binom44(0.6)^4=0.4752\)

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

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.

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.

- 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.

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.

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.

- Determine the smallest count that is a majority of \(15\). - Define a binomial count for regular bicycle commuters. - Use a cumulative probability or its complement.

1. Let \(X\) be the number of people in the group who regularly commute by bicycle. Then \(X\sim\operatorname{Bin}(15,0.60)\). 2. A majority of \(15\) people means at least \(8\) people. 3. Therefore, \(P(X\ge8)=1-P(X\le7)\approx0.7869.\)

\(P(X\ge8)\approx0.7869\).

- Recall all four binomial conditions, including the fixed number of trials. - Define a success as a misrouted package. - Match the sample size and single-package probability to the binomial parameters.

1. A binomial model requires a fixed number of trials, two relevant outcomes on each trial, independent trials, and a constant success probability. 2. Here the fixed number is \(40\), the two outcomes are misrouted or not misrouted, independence must be reasonable for the sampled packages, and the misrouting probability must remain \(0.05\). 3. Let \(X\) be the number of misrouted packages among the \(40\) sampled packages. Then \(n=40\) and \(p=0.05\). 4. \(P(X=2)=\binom{40}{2}(0.05)^2(0.95)^{38}\approx0.2777.\)

a) Fixed \(n=40\), two relevant outcomes, independence, and constant \(p=0.05\). b) \(X\) is the number of misrouted packages; \(n=40\), \(p=0.05\). c) \(P(X=2)\approx0.2777\).

- Identify the binomial parameters from the context. - Express the inclusive interval in \(F\) as a difference of cumulative probabilities. - Find the complement of the event \(X\ge175\).

1. \(X\sim\operatorname{Bin}(200,0.80)\). 2. \(P(E)=P(X=165)=\binom{200}{165}(0.80)^{165}(0.20)^{35}\approx0.0495.\) 3. \(P(F)=P(155\le X\le170)=P(X\le170)-P(X\le154)\approx0.8066.\) 4. The complement of \(X\ge175\) is \(X\le174\). Therefore, \(1-P(X\ge175)\) is the probability that at most \(174\) of the \(200\) inspected trees are pine trees.

a) \(P(E)\approx0.0495\) and \(P(F)\approx0.8066\). b) It is the probability that at most \(174\) of the \(200\) inspected trees are pine trees.

- 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.

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: \(P(X\ge 3)=1-\sum_{k=0}^{2}\binom{15}{k}(0.25)^k(0.75)^{15-k}\).

a) \(\binom{15}{6}(0.25)^6(0.75)^9\) b) \((0.75)^{15}+15(0.25)(0.75)^{14}\) c) \(1-\sum_{k=0}^{2}\binom{15}{k}(0.25)^k(0.75)^{15-k}\)

- Convert “1 in 20” to a probability. - Use \(np\) for the expected count. - Use a point probability for exactly one defect and a cumulative probability for at most two.

1. The defect probability is \(p=\frac{1}{20}=0.05\). 2. For part a), the expected number is \(E(X)=1400(0.05)=70.\) 3. For parts b) and c), let \(X\sim\operatorname{Bin}(25,0.05)\). 4. \(P(X=1)=\binom{25}{1}(0.05)(0.95)^{24}\approx0.3650.\) 5. \(P(X\le2)=\sum_{k=0}^{2}\binom{25}{k}(0.05)^k(0.95)^{25-k}\approx0.8729.\)

a) \(70\) defective diodes. b) \(0.3650\). c) \(0.8729\).

- State all four binomial conditions, including the fixed number of trials. - Identify \(n\) and \(p\). - Use the binomial-mode rule involving \((n+1)p\).

1. A binomial model requires a fixed number of trials, two relevant outcomes on each trial, independent trials, and a constant success probability. 2. Here, the fixed number is \(n=45\), the two outcomes are make or miss, the shots must be independent, and the make probability must remain \(p=0.75\). 3. Since \((n+1)p=46(0.75)=34.5\) is not an integer, the unique mode is \(\lfloor34.5\rfloor=34.\)

a) Fixed \(n=45\), two relevant outcomes, independence, and constant \(p=0.75\). b) \(34\) made free throws.

- 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.

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.

The greatest probability occurs at \(k=36\).

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

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.

a) Binomial with \(n=12\) and \(p=0.80\). b) Not binomial because the ace probability changes without replacement.

- 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?

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.

a) Binomial with \(n=20\) and \(p=0.5\). b) Not binomial because the recorded outcome has more than two categories.

- Use one binomial model for all three events in part a). - At least four and at most three are complementary events. - Independence means previous outcomes do not change the next-trial probability.

1. Let \(X\sim\operatorname{Bin}(50,0.04)\). 2. \(P(X=2)\approx0.2762.\) 3. \(P(X\ge4)\approx0.1391.\) 4. \(P(X\le3)\approx0.8609.\) 5. Independence means the first \(100\) results do not change the next-component defect probability, so \(P(\text{next defective})=0.04.\)

a) (1) \(0.2762\); (2) \(0.1391\); (3) \(0.8609\). b) \(0.04\).

- Translate the stated range into an inequality for the binomial count. - Use a difference of cumulative probabilities. - Subtract through \(11\) so that \(12\) remains included.

1. Let \(X\sim\operatorname{Bin}(120,0.15)\). 2. The required event is \(12\le X\le24\). 3. \(P(12\le X\le24)=P(X\le24)-P(X\le11)\approx0.9052.\)

\(0.9052\).

- 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?

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.

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.

- 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.

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), \(P(X\le 1)=P(X=0)+P(X=1)=(0.20)^6+6(0.80)(0.20)^5=0.0016\).

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 \(0.16\%\).

- State all four Bernoulli/binomial conditions. - Use the binomial point formula for exactly one success. - Use a complement for at least two successes.

1. Each bottle must be classified into two relevant outcomes, the sample size must be fixed at \(10\), the air-bubble probability must remain \(0.04\), and the outcomes must be independent. 2. Let \(X\sim\operatorname{Bin}(10,0.04)\). Then \(P(X=1)=\binom{10}{1}(0.04)(0.96)^9\approx0.2770.\) 3. \(P(X\ge2)=1-\left[(0.96)^{10}+10(0.04)(0.96)^9\right]\approx0.0582.\)

a) Fixed \(n=10\), two relevant outcomes, constant \(p=0.04\), and independence. b) \(0.2770\). c) \(0.0582\).

- 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 \(90\%\).

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.

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 \(90\%\).

- 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 \(1\).

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 \(P(X=5)=\frac{1}{243}\approx0.0041\).

<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>

- 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.

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, \(P(X\ge 5)=\binom{6}{5}\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)+\left(\frac{1}{3}\right)^6=\frac{13}{729}\approx 0.017833\).

1. The probability is \(\frac{80}{243}\approx 32.92\%\). 2. The probability is \(\frac{13}{729}\approx 1.78\%\).

- Use \(X\sim\operatorname{Bin}(6,0.70)\). - “At most \(1\)” includes the counts \(0\) and \(1\). - “At least \(5\)” includes the counts \(5\) and \(6\).

1. Let \(X\sim\operatorname{Bin}(6,0.70)\). 2. \(P(X=4)=\binom64(0.70)^4(0.30)^2=0.324135\approx0.3241\). 3. \(P(X=0)=(0.30)^6=0.000729\approx0.0007\). 4. \(P(X\le1)=P(X=0)+P(X=1)=0.010935\approx0.0109\). 5. \(P(X\ge5)=P(X=5)+P(X=6)=0.420175\approx0.4202\).

a) \(0.3241\). b) \(0.0007\). c) \(0.0109\). d) \(0.4202\).

- Let success mean that a bulb is defective. - Use the binomial point-probability formula for exactly two defects. - For “at least \(1\),” use the complement of no defects.

1. Let \(X\sim\operatorname{Bin}(8,0.05)\), where \(X\) is the number of defective bulbs. 2. \(P(X=0)=(0.95)^8\approx0.6634\). 3. \(P(X=2)=\binom82(0.05)^2(0.95)^6\approx0.0515\). 4. \(P(X\ge1)=1-P(X=0)\approx0.3366\).

a) \(0.6634\). b) \(0.0515\). c) \(0.3366\).

- The success probability on one roll is \(1/6\). - “At most \(1\)” includes \(0\) and \(1\). - Use a complement for “at least \(3\).”

1. Let \(X\sim\operatorname{Bin}(10,\frac16)\). 2. \(P(X=2)=\binom{10}{2}(\frac16)^2(\frac56)^8\approx0.2907\). 3. \(P(X\le1)=(\frac56)^{10}+10(\frac16)(\frac56)^9\approx0.4845\). 4. \(P(X\ge3)=1-P(X\le2)\approx0.2248\).

a) \(0.2907\). b) \(0.4845\). c) \(0.2248\).

- Sampling with replacement keeps the success probability constant and supports independence. - “No defects or exactly one defect” is the event \(X\le1\). - Add the two binomial point probabilities.

1. Let \(X\sim\operatorname{Bin}(15,0.10)\). 2. The required event is \(X\le1\), so \(P(X\le1)=(0.90)^{15}+15(0.10)(0.90)^{14}.\) 3. Therefore, \(P(X\le1)\approx0.5490.\)

\(0.5490\).

- Determine the one-spin prize probability from the equal sections. - The complement of “at least \(3\)” contains \(0\), \(1\), and \(2\) successes. - Use the binomial distribution for the ten independent spins.

1. The success probability is \(p=\frac15=0.20\), so \(X\sim\operatorname{Bin}(10,0.20)\). 2. Use the complement: \(P(X\ge3)=1-P(X\le2)\). 3. Thus, \(P(X\ge3)=1-\sum_{k=0}^{2}\binom{10}{k}(0.20)^k(0.80)^{10-k}\approx0.3222\).

\(P(X\ge3)\approx0.3222\).

- Sampling with replacement keeps the red probability constant from draw to draw. - Translate “at most \(2\)” into the included success counts. - Add the corresponding binomial point probabilities.

1. Let \(X\sim\operatorname{Bin}(15,0.20)\). 2. “At most \(2\)” means \(X=0,1,2\). 3. Therefore, \(P(X\le2)=\sum_{k=0}^{2}\binom{15}{k}(0.20)^k(0.80)^{15-k}\approx0.3980\).

\(P(X\le2)\approx0.3980\).

- Determine the minimum number of games Sarah must win. - Model the number of wins with a binomial random variable. - Add the probabilities of all winning counts.

1. Let \(X\sim\operatorname{Bin}(7,0.55)\), where \(X\) is the number of games Sarah wins. 2. Sarah wins the match when \(X\ge4\). 3. \(P(X\ge4)=\sum_{k=4}^{7}\binom{7}{k}(0.55)^k(0.45)^{7-k}\approx0.6083.\)

\(0.6083\).

- Determine the smallest number of wins that is more than half of ten games. - Model Marc's number of wins with a binomial random variable. - Add the probabilities of all series-winning counts.

1. Let \(X\sim\operatorname{Bin}(10,0.40)\), where \(X\) is the number of games Marc wins. 2. Marc is the overall winner exactly when \(X\ge6\). 3. Therefore, \(P(X\ge6)=\sum_{k=6}^{10}\binom{10}{k}(0.40)^k(0.60)^{10-k}\approx0.1662\).

\(P(\text{Marc wins the series})\approx0.1662\).

- Identify \(n=15\) and \(p=0.40\). - Use one point probability for part a). - Translate the “at most” and “at least” phrases into the included integer counts.

1. Let \(X\sim\operatorname{Bin}(15,0.40)\). 2. \(P(X=6)=\binom{15}{6}(0.40)^6(0.60)^9\approx0.2066\). 3. \(P(X\le2)=\sum_{k=0}^{2}\binom{15}{k}(0.40)^k(0.60)^{15-k}\approx0.0271\). 4. \(P(X\ge13)=\sum_{k=13}^{15}\binom{15}{k}(0.40)^k(0.60)^{15-k}\approx0.0003\).

a) \(0.2066\). b) \(0.0271\). c) \(0.0003\).

- Let success mean that a component is defective. - A complement is efficient for “at least \(2\).” - “No components are defective” means \(X=0\).

1. Let \(X\sim\operatorname{Bin}(15,0.04)\). 2. \(P(X=1)=\binom{15}{1}(0.04)(0.96)^{14}\approx0.3388\). 3. \(P(X\ge2)=1-[P(X=0)+P(X=1)]\approx0.1191\). 4. \(P(X=0)=(0.96)^{15}\approx0.5421\).

a) \(0.3388\). b) \(0.1191\). c) \(0.5421\).

- Use the given success probability to define a binomial count. - A complement is efficient for “at least \(2\).” - List the included counts for the inclusive interval before calculating.

1. \(X\sim\operatorname{Bin}(10,0.40)\). 2. \(P(X=4)=\binom{10}{4}(0.40)^4(0.60)^6\approx0.2508\). 3. \(P(X\ge2)=1-P(X\le1)\approx0.9536\). 4. \(P(3\le X\le5)=P(X=3)+P(X=4)+P(X=5)\approx0.6665\).

a) \(0.2508\). b) \(0.9536\). c) \(0.6665\).

- Identify the binomial parameters from the sample size and defect rate. - Use a complement for “more than \(2\).” - For part c), list the included integer counts before calculating.

1. \(X\sim\operatorname{Bin}(15,0.05)\). 2. \(P(X=1)=\binom{15}{1}(0.05)(0.95)^{14}\approx0.3658\). 3. \(P(X>2)=1-P(X\le2)\approx0.0362\). 4. \(P(1\le X\le3)=P(X=1)+P(X=2)+P(X=3)\approx0.5312\).

a) \(0.3658\). b) \(0.0362\). c) \(0.5312\).

- Translate the verbal condition into the included win counts. - Use the binomial formula for each exact count. - Add the three disjoint point probabilities.

1. The number of wins has distribution \(X\sim\operatorname{Bin}(10,0.20).\) 2. The required event is \(1\le X\le3.\) 3. Therefore, \(P(1\le X\le3) =\sum_{k=1}^{3}\binom{10}{k}(0.20)^k(0.80)^{10-k} \approx0.7718.\)

\(P(1\le X\le3) =\sum_{k=1}^{3}\binom{10}{k}(0.20)^k(0.80)^{10-k} \approx0.7718.\)

- Identify the three included integer success counts. - Use the binomial point-probability formula for each count. - Add the probabilities because the exact-count events are disjoint.

1. The number made has distribution \(X\sim\operatorname{Bin}(15,0.60).\) 2. The required event is \(8\le X\le10.\) 3. Therefore, \(P(8\le X\le10) =\sum_{k=8}^{10}\binom{15}{k}(0.60)^k(0.40)^{15-k} \approx0.5696.\)

\(P(8\le X\le10) =\sum_{k=8}^{10}\binom{15}{k}(0.60)^k(0.40)^{15-k} \approx0.5696.\)

- Convert the no-show probability to an attendance probability. - Determine which attendance counts exceed the available seats. - Add the corresponding binomial point probabilities.

1. Each person attends with probability \(0.90\), so \(X\sim\operatorname{Bin}(15,0.90)\). 2. The theater is over capacity when \(X>13\), so \(X=14\) or \(X=15\). 3. Therefore, \(P(X>13)=\binom{15}{14}(0.90)^{14}(0.10)+(0.90)^{15}\approx0.5490\).

\(P(\text{over capacity})\approx0.5490\).

- Convert the cancellation probability to the probability that a reservation is used. - Identify the attendance counts that would create a shortage. - Use the complement of the shortage event.

1. Each reservation is used with probability \(0.75\), so \(X\sim\operatorname{Bin}(12,0.75)\). 2. There are enough tables when \(X\le10\). 3. Using the complement, \(P(X\le10)=1-[P(X=11)+P(X=12)]\approx0.8416\).

\(P(\text{enough tables})\approx0.8416\).

- 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.

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 \(P(X=3)\approx0.2188\).

\(2\) vehicles with a paint defect

- 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.

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, \(P(X=11)=P(X=12)\approx0.2252\).

\(11\) and \(12\) germinating seeds

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

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, \(P(X=1)=3\cdot0.2\cdot0.64=0.384\).

\(P(X=1)=0.384\), or \(38.4\%\)

- Model the number of made free throws with a binomial random variable. - Use the point-probability formula for exactly three makes. - “At least \(4\)” means \(4\) or \(5\) makes.

1. Let \(X\sim\operatorname{Bin}(5,0.70)\). 2. \(P(X=3)=\binom53(0.70)^3(0.30)^2=0.3087\). 3. \(P(X\ge4)=P(X=4)+P(X=5)\). 4. \(P(X=4)=\binom54(0.70)^4(0.30)=0.36015\), and \(P(X=5)=(0.70)^5=0.16807\). 5. Therefore, \(P(X\ge4)=0.52822\).

a) \(0.3087\). b) \(0.52822\).

- The probability of blue on one spin is \(1/5\). - “At most \(1\)” includes \(0\) and \(1\). - Use the complement of zero blue results for part c).

1. \(X\sim\operatorname{Bin}(6,0.20)\). 2. \(P(X=2)=\binom62(0.20)^2(0.80)^4=0.24576\). 3. \(P(X\le1)=P(X=0)+P(X=1)=0.65536\). 4. \(P(X\ge1)=1-P(X=0)=1-(0.80)^6=0.737856\).

a) \(0.24576\). b) \(0.65536\). c) \(0.737856\).

- Use the sample size and defect probability as the binomial parameters. - For “more than \(2\),” use the complement through \(2\). - Keep unrounded values through the final subtraction.

1. \(X\sim\operatorname{Bin}(10,0.05)\). 2. \(P(X=1)=\binom{10}{1}(0.05)(0.95)^9\approx0.3151\). 3. \(P(X=0)=(0.95)^{10}\approx0.5987\). 4. \(P(X>2)=1-P(X\le2)\approx0.0115\), using unrounded probabilities in the subtraction.

a) \(0.3151\). b) \(0.5987\). c) \(0.0115\).

- Use the binomial model with \(n=8\) and \(p=1/4\). - “More than \(6\)” includes only the counts \(7\) and \(8\). - Keep an exact form for the very small upper-tail probability before rounding.

1. Since a random guess is correct with probability \(0.25\), \(X\sim\operatorname{Bin}(8,0.25)\). 2. \(P(X=3)=\binom83(0.25)^3(0.75)^5\approx0.2076\). 3. More than \(6\) means \(7\) or \(8\) correct answers. 4. Thus, \(P(X>6)=P(X=7)+P(X=8)=\frac{25}{65536}\approx0.000381\).

a) \(0.2076\). b) \(\frac{25}{65536}\approx0.000381\).

- Identify the number of trials and the sprouting probability. - Exactly eleven sprouts means one failure among twelve positions. - Use the binomial point-probability formula.

1. Let \(X\sim\operatorname{Bin}(12,0.95)\). 2. \(P(X=11)=\binom{12}{11}(0.95)^{11}(0.05)\). 3. Therefore, \(P(X=11)\approx0.3413\).

\(\binom{12}{11}(0.95)^{11}(0.05)\approx0.3413\).

- Let success mean a made free throw. - Exactly seven makes also means exactly three misses. - Use the binomial point-probability formula.

1. Let \(X\sim\operatorname{Bin}(10,0.70)\). 2. \(P(X=7)=\binom{10}{7}(0.70)^7(0.30)^3\). 3. Therefore, \(P(X=7)\approx0.2668\).

\(\binom{10}{7}(0.70)^7(0.30)^3\approx0.2668\).

- Define success as a correct guess. - State the fixed number of questions and the one-question success probability. - “No correct answers” corresponds to the binomial count \(X=0\).

1. Let \(X\) be the number of correct answers. Each guess is correct with probability \(1/4\), so \(X\sim\operatorname{Bin}(15,0.25)\). 2. No correct answers means \(X=0\). 3. Thus, \(P(X=0)=\binom{15}{0}(0.25)^0(0.75)^{15}=(0.75)^{15}\approx0.01336\).

a) \(X\sim\operatorname{Bin}(15,0.25)\). b) \((0.75)^{15}\approx0.01336\).

- Define success as arrival within \(24\) hours. - State \(n\) and \(p\) before calculating. - “All \(20\)” corresponds to the endpoint count \(X=20\).

1. Let \(X\) be the number of packages that arrive within \(24\) hours. Then \(X\sim\operatorname{Bin}(20,0.95)\). 2. All packages arrive within \(24\) hours exactly when \(X=20\). 3. Therefore, \(P(X=20)=\binom{20}{20}(0.95)^{20}(0.05)^0=(0.95)^{20}\approx0.3585\).

a) \(X\sim\operatorname{Bin}(20,0.95)\). b) \((0.95)^{20}\approx0.3585\).

- A random guess is correct with probability \(1/4\). - “Exactly half” of ten questions means five successes. - Use \(np\) for the expected binomial count.

1. Let \(X\) be the number of correct answers. Then \(X\sim\operatorname{Bin}(10,0.25)\). 2. Exactly half means \(X=5\), so \(P(X=5)=\binom{10}{5}(0.25)^5(0.75)^5\approx0.0584\). 3. The expected number is \(E(X)=np=10(0.25)=2.5\).

a) \(X\) is the number of correct answers, with \(X\sim\operatorname{Bin}(10,0.25)\). b) \(0.0584\). c) \(2.5\) correct answers.

- Use the binomial point formula for exactly three coupons. - For at least one coupon, use the complement of zero coupons. - Keep the same \(n\) and \(p\) in both parts.

1. \(X\sim\operatorname{Bin}(12,0.20)\). 2. \(P(X=3)=\binom{12}{3}(0.20)^3(0.80)^9\approx0.2362\). 3. \(P(X\ge1)=1-P(X=0)=1-(0.80)^{12}\approx0.9313\).

a) \(0.2362\). b) \(0.9313\).

- Use the same binomial model in both parts. - “At least \(5\)” includes exactly five and exactly six makes. - Apply the point-probability formula to the included counts.

1. \(X\sim\operatorname{Bin}(6,0.75)\). 2. \(P(X=4)=\binom64(0.75)^4(0.25)^2\approx0.2966\). 3. \(P(X\ge5)=\binom65(0.75)^5(0.25)+(0.75)^6\approx0.5339\).

a) \(0.2966\). b) \(0.5339\).

- 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?

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 \(p'=\frac{2}{8}=0.25\).

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 \(p'=0.25\).

- 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 \(1\).

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 \(P(X=4)=0.00020736\approx0.0002\).

<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>

- 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.

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 \(P(Y=5)\approx0.0017\).

<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>

- State all four binomial conditions. - Identify the fixed trial count and the two relevant outcomes. - Apply the binomial formula to every possible value of \(X\). - Check that the probabilities add to \(1\).

1. A binomial model requires a fixed number of trials, two relevant outcomes on each trial, independent trials, and a constant success probability. 2. Here, \(n=4\), the relevant outcomes are center hit or not, the shots must be independent, and \(p=0.70\) must remain constant. 3. Let \(X\sim\operatorname{Bin}(4,0.70)\). 4. The probabilities are \(P(X=0)=0.0081,\quad P(X=1)=0.0756,\quad P(X=2)=0.2646,\quad P(X=3)=0.4116,\quad P(X=4)=0.2401.\)

a) Fixed \(n=4\), two relevant outcomes, independence, and constant \(p=0.70\). b) \(0.0081,\ 0.0756,\ 0.2646,\ 0.4116,\ 0.2401\) for \(X=0,1,2,3,4\), respectively.

- Use \(Y\sim\operatorname{Bin}(12,0.20)\). - Complements are efficient for upper or endpoint events. - In part d), use the probability already found in part a) rather than repeating the calculation.

1. \(Y\sim\operatorname{Bin}(12,0.20)\). 2. \(P(Y\ge1)=1-P(Y=0)=1-(0.80)^{12}\approx0.9313\). 3. \(P(Y>3)=1-P(Y\le3)\approx0.2054\). 4. \(P(Y\le10)=1-[P(Y=11)+P(Y=12)]\approx0.9999998\). 5. The event \(Y=0\) is the complement of \(Y\ge1\), so \(P(Y=0)=1-P(Y\ge1)=(0.80)^{12}\approx0.0687\).

a) \(0.9313\). b) \(0.2054\). c) \(1-[P(Y=11)+P(Y=12)]\approx0.9999998\). d) \(0.0687\); “no red results” and “at least one red result” are complementary events.

- Use \(n=12\) and \(p=0.15\). - “At most \(2\)” is a lower-tail event. - Use a complement for “more than \(4\).”

1. Let \(X\sim\operatorname{Bin}(12,0.15)\). 2. \(P(X=3)=\binom{12}{3}(0.15)^3(0.85)^9\approx0.1720\). 3. \(P(X\le2)\approx0.7358\). 4. \(P(X>4)=1-P(X\le4)\approx0.0239\).

a) \(0.1720\). b) \(0.7358\). c) \(0.0239\).

- Use \(n=15\) and \(p=0.40\). - Translate “fewer than \(5\)” into an integer inequality. - Use a complement for “at least \(8\).”

1. Let \(X\sim\operatorname{Bin}(15,0.40)\). 2. \(P(X=6)=\binom{15}{6}(0.40)^6(0.60)^9\approx0.2066\). 3. Fewer than five means \(X\le4\), so \(P(X<5)\approx0.2173\). 4. \(P(X\ge8)=1-P(X\le7)\approx0.2131\).

a) \(0.2066\). b) \(0.2173\). c) \(0.2131\).

- Use cumulative binomial probabilities. - Express the upper tail with a complement. - For the interval, subtract the cumulative probability through the value just below the lower endpoint.

1. \(P(X\le4)\approx0.2552\). 2. \(P(X>8)=1-P(X\le8)\approx0.1287\). 3. \(P(5\le X\le10)=P(X\le10)-P(X\le4)\approx0.7192\).

a) \(0.2552\). b) \(0.1287\). c) \(0.7192\).

- Pay particular attention to the strict inequalities in part b). - Convert the open interval to its attainable integer values. - Use a difference of cumulative probabilities.

1. \(P(X\le20)\approx0.0368.\) 2. Because \(X\) is integer-valued, \(22<X<32\) means \(23\le X\le31\). 3. \(P(22<X<32)=P(X\le31)-P(X\le22)\approx0.6980.\)

a) \(0.0368\). b) \(0.6980\).

- Use one binomial model for all three parts. - For the inclusive interval, subtract the cumulative probability through \(7\). - Use a complement for the upper tail.

1. \(X\sim\operatorname{Bin}(200,0.05)\). 2. \(P(X\le10)\approx0.5831\). 3. \(P(8\le X\le12)=P(X\le12)-P(X\le7)\approx0.5832\). 4. \(P(X>15)=1-P(X\le15)\approx0.0444\).

a) \(0.5831\). b) \(0.5832\). c) \(0.0444\).

- Use a complement for the upper-tail event. - “Fewer than \(10\)” ends at \(9\). - Use a difference of cumulative probabilities for the inclusive interval.

1. \(X\sim\operatorname{Bin}(50,0.25)\). 2. \(P(X\ge15)=1-P(X\le14)\approx0.2519\). 3. \(P(X<10)=P(X\le9)\approx0.1637\). 4. \(P(10\le X\le15)=P(X\le15)-P(X\le9)\approx0.6732\).

a) \(0.2519\). b) \(0.1637\). c) \(0.6732\).

- Use complements for upper-tail events. - Use cumulative subtraction for the inclusive interval. - Convert the condition on incorrect answers to a condition on the number correct.

1. \(X\sim\operatorname{Bin}(60,0.25)\). 2. \(P(X>12)=1-P(X\le12)\approx0.7684\). 3. \(P(10\le X\le20)=P(X\le20)-P(X\le9)\approx0.9007\). 4. Fewer than \(40\) incorrect answers means at least \(21\) correct answers, so \(P(X\ge21)=1-P(X\le20)\approx0.0541\).

a) \(0.7684\). b) \(0.9007\). c) \(0.0541\).

- Use one binomial model for all three parts. - Use a complement for “more than \(50\).” - For the inclusive interval, subtract the cumulative probability through \(41\).

1. \(X\sim\operatorname{Bin}(60,0.75)\). 2. \(P(X\le40)\approx0.0925\). 3. \(P(X>50)=1-P(X\le50)\approx0.0452\). 4. \(P(42\le X\le48)=P(X\le48)-P(X\le41)\approx0.7038\).

a) \(0.0925\). b) \(0.0452\). c) \(0.7038\).

- Use a point probability for exactly five defects. - Translate “fewer than \(8\)” to an upper endpoint of \(7\). - Use a complement for the upper tail.

1. \(X\sim\operatorname{Bin}(100,0.05)\). 2. \(P(X=5)=\binom{100}{5}(0.05)^5(0.95)^{95}\approx0.1800\). 3. At least \(3\) but fewer than \(8\) means \(3\le X\le7\), so \(P(3\le X\le7)=P(X\le7)-P(X\le2)\approx0.7538\). 4. \(P(X>10)=1-P(X\le10)\approx0.0115\).

a) \(0.1800\). b) \(0.7538\). c) \(0.0115\).

- Express the “at least” event using a complement. - Use a difference of cumulative probabilities for the interval. - Check the value immediately below the lower endpoint.

1. \(P(X\le15)\approx0.8369\). 2. \(P(X\ge10)=1-P(X\le9)\approx0.8363\). 3. \(P(8\le X\le14)=P(X\le14)-P(X\le7)\approx0.7028\).

a) \(0.8369\). b) \(0.8363\). c) \(0.7028\).

- Use a complement for the upper-tail probability. - Convert the open interval to the included integer values before using a CDF difference. - Subtract the cumulative probability through the value just below the lower included count.

1. \(P(X\le75)\approx0.1314\). 2. \(P(X>85)=1-P(X\le85)\approx0.0804\). 3. Because \(X\) is integer-valued, \(78<X<88\) means \(79\le X\le87\). 4. Therefore, \(P(78<X<88)=P(X\le87)-P(X\le78)\approx0.6287\).

a) \(0.1314\). b) \(0.0804\). c) \(0.6287\).

- Use one binomial model for all three events. - “Every” means the endpoint count \(X=20\). - Use a complement for “more than \(15\).”

1. Let \(X\sim\operatorname{Bin}(20,0.75)\). 2. \(P(X=20)=(0.75)^{20}\approx0.0032\). 3. \(P(X=15)=\binom{20}{15}(0.75)^{15}(0.25)^5\approx0.2023\). 4. \(P(X>15)=1-P(X\le15)\approx0.4148\).

a) \(0.0032\). b) \(0.2023\). c) \(0.4148\).

- Let success mean a defective component. - Use a point probability for exactly four defects. - Translate “fewer than \(4\)” into the largest included integer count.

1. Let \(X\sim\operatorname{Bin}(80,0.05)\). 2. \(P(X=0)=(0.95)^{80}\approx0.0165\). 3. \(P(X=4)=\binom{80}{4}(0.05)^4(0.95)^{76}\approx0.2004\). 4. Fewer than four means \(X\le3\), so \(P(X<4)\approx0.4284\).

a) \(0.0165\). b) \(0.2004\). c) \(0.4284\).

- Convert the no-show probability to an attendance probability. - Determine the largest attendance count that fits on the bus. - Use a cumulative binomial probability for all acceptable attendance counts.

1. Each customer arrives with probability \(0.90\), so \(X\sim\operatorname{Bin}(54,0.90)\). 2. Everyone has a seat exactly when \(X\le50\). 3. Therefore, \(P(X\le50)\approx0.8015\).

\(P(\text{everyone has a seat})\approx0.8015\).

- Use \(n=150\) and \(p=0.06\). - Use a point probability for exactly nine defects. - Use a complement for the upper tail.

1. \(X\sim\operatorname{Bin}(150,0.06)\). 2. \(P(X=9)=\binom{150}{9}(0.06)^9(0.94)^{141}\approx0.1359\). 3. \(P(X\le5)\approx0.1083\). 4. \(P(X\ge12)=1-P(X\le11)\approx0.1909\).

a) \(0.1359\). b) \(0.1083\). c) \(0.1909\).

- Distinguish “more than” from “at least.” - “Fewer than \(15\)” ends at \(14\). - Use a cumulative difference for the inclusive interval.

1. \(X\sim\operatorname{Bin}(80,0.25)\). 2. \(P(X>20)=1-P(X\le20)\approx0.4403\). 3. \(P(X<15)=P(X\le14)\approx0.0740\). 4. \(P(18\le X\le22)=P(X\le22)-P(X\le17)\approx0.4811\).

a) \(0.4403\). b) \(0.0740\). c) \(0.4811\).

- Translate each verbal event into integer bounds for \(X\). - Use a complement for the upper tail. - Use a CDF difference for the inclusive interval.

1. \(X\sim\operatorname{Bin}(60,0.80)\). 2. \(P(X=50)=\binom{60}{50}(0.80)^{50}(0.20)^{10}\approx0.1102\). 3. \(P(X<45)=P(X\le44)\approx0.1306\). 4. \(P(X\ge52)=1-P(X\le51)\approx0.1268\). 5. \(P(46\le X\le54)=P(X\le54)-P(X\le45)\approx0.7814\).

a) \(0.1102\). b) \(0.1306\). c) \(0.1268\). d) \(0.7814\).

- Use one binomial model for all four parts. - Use a complement for “more than \(20\).” - Use a cumulative difference for the inclusive interval.

1. \(X\sim\operatorname{Bin}(150,0.12)\). 2. \(P(X=18)\approx0.0998\). 3. \(P(X\le15)\approx0.2715\). 4. \(P(X>20)=1-P(X\le20)\approx0.2587\). 5. \(P(10\le X\le25)=P(X\le25)-P(X\le9)\approx0.9543\).

a) \(0.0998\). b) \(0.2715\). c) \(0.2587\). d) \(0.9543\).

- Count failures rather than successes if that makes the event simpler. - Translate “at least \(65\) sprout” into a condition on the number that fail. - Use the corresponding binomial cumulative probability.

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, \(P(Y\le15)\approx0.8625\).

\(P(\text{at least 65 flowers})\approx0.8625\).

- Use \(n=80\) and \(p=0.025\). - “At most \(3\)” is a lower-tail event. - Use the complement of zero ticket holders for part c).

1. \(X\sim\operatorname{Bin}(80,0.025)\). 2. \(P(X=2)=\binom{80}{2}(0.025)^2(0.975)^{78}\approx0.2741\). 3. \(P(X\le3)\approx0.8594\). 4. \(P(X\ge1)=1-P(X=0)=1-(0.975)^{80}\approx0.8681\).

a) \(0.2741\). b) \(0.8594\). c) \(0.8681\).

- Use the supplied one-vehicle occupancy probability. - Translate “more than half of six” into a whole-number demand threshold. - Use a complement for the upper-tail probability.

1. Let \(X\) be the number of vehicles needing a charger at the selected time. Then \(X\sim\operatorname{Bin}(24,1/6)\). 2. More than half of six stations means at least four stations are needed. 3. Therefore, \(P(X\ge4)=1-P(X\le3)\approx0.5845\).

\(P(X\ge4)\approx0.5845\).

- Translate “at least \(80\)” into an upper-tail event. - Use the complement through \(79\). - Keep the endpoint in the complement consistent with the inclusive threshold.

1. \(X\sim\operatorname{Bin}(450,0.15)\). 2. \(P(X\ge80)=1-P(X\le79)\approx0.0590\).

\(P(X\ge80)\approx0.0590\).

- Translate “outside” the inclusive interval into two disjoint tails. - The lower tail ends at \(164\), and the upper tail begins at \(196\). - Add the two tail probabilities.

1. \(X\sim\operatorname{Bin}(600,0.30)\). 2. Being outside \([165,195]\) means \(X\le164\) or \(X\ge196\). 3. The two events are disjoint, so \(P(X\le164\text{ or }X\ge196)=P(X\le164)+P(X\ge196)\approx0.1672\).

\(0.1672\).

- Use the binomial formulas for center and spread. - Translate “within \(5\)” into an inclusive integer interval centered at the mean. - Use a difference of cumulative probabilities.

1. \(\mu=np=60\), and \(\sigma=\sqrt{np(1-p)}=\sqrt{36}=6\). 2. Being within five of the mean means \(55\le X\le65\). 3. Therefore, \(P(55\le X\le65)=P(X\le65)-P(X\le54)\approx0.6407\).

a) \(\mu=60\), \(\sigma=6\). b) \(0.6407\).

- Express the inclusive interval as a difference of cumulative probabilities. - To keep \(80\) included, subtract the cumulative probability through \(79\). - Evaluate with binomial CDF technology.

1. \(P(80\le X\le110)=P(X\le110)-P(X\le79)\). 2. Therefore, \(P(80\le X\le110)\approx0.9068\).

\(0.9068\).

- Find \(np\) first. - Compare each endpoint with the mean separately. - Use the cumulative probability through \(71\) to keep \(72\) included.

1. \(\mu=np=80\). 2. The lower endpoint \(72\) is \(8\) below the mean, while \(90\) is \(10\) above it, so the interval is not symmetric about the mean. 3. \(P(72\le X\le90)=P(X\le90)-P(X\le71)\approx0.8251\).

a) \(\mu=80\). b) \(72\) is \(8\) below the mean and \(90\) is \(10\) above it; the interval is not symmetric. c) \(0.8251\).

- Use the binomial model with the supplied parameters. - Express the inclusive interval as a CDF difference. - Subtract through the integer immediately below the lower endpoint.

1. \(X\sim\operatorname{Bin}(150,0.85)\). 2. \(P(122\le X\le132)=P(X\le132)-P(X\le121)\approx0.7875\).

\(0.7875\).

- Use the supplied binomial model for all three events. - Use a complement for the upper tail. - Use a difference of cumulative probabilities for the inclusive interval.

1. \(X\sim\operatorname{Bin}(400,0.02)\). 2. \(P(X\le5)\approx0.1885\). 3. \(P(X>10)=1-P(X\le10)\approx0.1821\). 4. \(P(6\le X\le12)=P(X\le12)-P(X\le5)\approx0.7497\), using unrounded cumulative values.

a) \(0.1885\). b) \(0.1821\). c) \(0.7497\).

- Translate the strict lower-tail boundary carefully. - Use a complement for the upper tail. - Find the binomial mean before translating “within \(5\).”

1. \(X\sim\operatorname{Bin}(100,0.40)\). 2. \(P(X<35)=P(X\le34)\approx0.1303\). 3. \(P(X\ge45)=1-P(X\le44)\approx0.1789\). 4. The mean is \(\mu=np=40\), so being within five of the mean means \(35\le X\le45\). 5. Thus, \(P(35\le X\le45)=P(X\le45)-P(X\le34)\approx0.7386\).

a) \(0.1303\). b) \(0.1789\). c) \(0.7386\).

- Use a lower-tail CDF for part a). - Use a complement for “at least \(25\).” - Use a cumulative difference for the inclusive interval.

1. \(X\sim\operatorname{Bin}(500,0.04)\). 2. \(P(X\le15)\approx0.1513\). 3. \(P(X\ge25)=1-P(X\le24)\approx0.1522\). 4. \(P(18\le X\le22)=P(X\le22)-P(X\le17)\approx0.4314\).

a) \(0.1513\). b) \(0.1522\). c) \(0.4314\).

- The red probability is \(1/4\). - Translate each verbal boundary into an integer inequality. - Use a complement or cumulative difference where efficient.

1. \(X\sim\operatorname{Bin}(40,0.25)\). 2. \(P(X>12)=1-P(X\le12)\approx0.1791\). 3. \(P(8\le X\le12)=P(X\le12)-P(X\le7)\approx0.6389\). 4. \(P(X<5)=P(X\le4)\approx0.0160\).

a) \(0.1791\). b) \(0.6389\). c) \(0.0160\).

- Use the per-holder attendance probability in the binomial model. - The first over-capacity attendance count is \(1201\). - Evaluate the upper tail using a complement.

1. Let \(X\) be the number of ticket holders who attend. Then \(X\sim\operatorname{Bin}(1400,0.85)\). 2. The section is over capacity when \(X>1200\). 3. Therefore, \(P(X>1200)=1-P(X\le1200)\approx0.2169\).

\(P(\text{over capacity})\approx0.2169\).

- Counting defective drives makes the threshold simple. - At least forty-two working drives allows at most eight defects. - Use the corresponding cumulative binomial probability.

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, \(P(X\le8)\approx0.9421\).

\(0.9421\).

- Translate the strict inequalities to their included integer counts. - Use a complement for the upper tail. - Use a cumulative difference for the open-ended interval in part c).

1. \(X\sim\operatorname{Bin}(200,0.15)\). 2. \(P(X\le25)\approx0.1876\). 3. \(P(X>35)=1-P(X\le35)\approx0.1387\). 4. More than \(25\) but fewer than \(40\) means \(26\le X\le39\), so \(P(26\le X\le39)=P(X\le39)-P(X\le25)\approx0.7788\).

a) \(0.1876\). b) \(0.1387\). c) \(0.7788\).

- Translate “fewer than \(10\)” to an endpoint of \(9\). - Use a complement for “at least \(15\).” - Use a cumulative difference for the inclusive interval.

1. \(X\sim\operatorname{Bin}(120,0.12)\). 2. \(P(X<10)=P(X\le9)\approx0.0782\). 3. \(P(X\ge15)=1-P(X\le14)\approx0.4745\). 4. \(P(12\le X\le20)=P(X\le20)-P(X\le11)\approx0.7403\).

a) \(0.0782\). b) \(0.4745\). c) \(0.7403\).

- Match the information in the problem to \(n\), \(p\), and \(k\). - Count the arrangements containing exactly two wins. - State all four binomial conditions, including the fixed number of trials.

1. The binomial parameters are \(n=6\), \(p=0.30\), and \(k=2\). 2. \(P(X=2)=\binom62(0.30)^2(0.70)^4=0.324135.\) 3. A binomial model requires a fixed number of trials, two relevant outcomes per trial, independent trials, and a constant success probability.

a) \(P(X=2)=0.324135\). b) Fixed \(n=6\), two relevant outcomes, independence, and constant \(p=0.30\).

- Use a point probability for exactly fifteen returns. - “Fewer than \(10\)” ends at \(9\). - Use a complement for “more than \(20\).”

1. Let \(X\sim\operatorname{Bin}(50,0.30)\). 2. \(P(X=15)=\binom{50}{15}(0.30)^{15}(0.70)^{35}\approx0.1223\). 3. \(P(X<10)=P(X\le9)\approx0.0402\). 4. \(P(X>20)=1-P(X\le20)\approx0.0478\).

a) \(0.1223\). b) \(0.0402\). c) \(0.0478\).

- Use a cumulative probability for part a) and a CDF difference for part b). - Find the binomial mean before translating “within \(2\).” - Keep all interval endpoints inclusive.

1. \(X\sim\operatorname{Bin}(200,0.05)\). 2. \(P(X\le12)\approx0.7965\). 3. \(P(5\le X\le15)=P(X\le15)-P(X\le4)\approx0.9292\). 4. The mean is \(\mu=np=10\), so being within two of the mean means \(8\le X\le12\). 5. Thus, \(P(8\le X\le12)=P(X\le12)-P(X\le7)\approx0.5832\).

a) \(0.7965\). b) \(0.9292\). c) \(0.5832\).

- Count the total balls and the green balls in the displayed urn. - Replacement preserves the same success probability on every draw. - Use the binomial formula for exactly two green results.

1. The urn shows \(10\) balls total and \(3\) green balls, so \(p=3/10=0.30\). 2. Replacement makes the six draws independent with the same success probability, so \(X\sim\operatorname{Bin}(6,0.30)\). 3. \(P(X=2)=\binom62(0.30)^2(0.70)^4=0.324135\). 4. As a percentage rounded to one decimal place, the probability is \(32.4\%\).

\(32.4\%\).

- Count the yellow and blue sections in the displayed spinner. - For each part, treat the named color as success and all other colors as failure. - Apply the binomial point-probability formula.

1. From the spinner, \(P(\text{yellow})=3/6=1/2\) and \(P(\text{blue})=2/6=1/3\). 2. \(P(\text{exactly 4 yellow})=\binom64(1/2)^4(1/2)^2=15/64=23.4375\%\approx23.44\%\). 3. \(P(\text{exactly 2 blue})=\binom62(1/3)^2(2/3)^4=80/243\approx32.92\%\).

a) \(23.44\%\). b) \(32.92\%\).

- 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.

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 \(56\cdot(0.4)^3(0.6)^5=0.27869184\approx0.279\).

\(0.279\)

- Count the blue sections in the displayed spinner. - “At most \(1\)” includes \(0\) and \(1\). - “At least \(3\)” includes \(3,4,5\).

1. The spinner shows \(2\) blue sections out of \(5\), so \(p=2/5=0.40\) and \(X\sim\operatorname{Bin}(5,0.40)\). 2. \(P(X\le1)=P(X=0)+P(X=1)=(0.60)^5+5(0.40)(0.60)^4=0.33696\). 3. \(P(X\ge3)=P(X=3)+P(X=4)+P(X=5)=0.31744\).

a) \(0.33696\), or \(33.696\%\). b) \(0.31744\), or \(31.744\%\).

- 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.

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 \(24.16\%\).

a) \(24.75\%\) b) \(24.16\%\)

- 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.

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), \(P(Z=1)=\binom41(0.125)(0.875)^3\approx0.3350\).

a) \(0.3433\), or about \(34.33\%\) b) \(0.2734\), or about \(27.34\%\) c) \(0.3350\), or about \(33.50\%\)

- Count the blue and red sections in the displayed spinner. - Use a separate binomial model for each named color. - A complement is efficient for “at least \(2\).”

1. The spinner shows \(5\) blue sections out of \(10\), so \(p_B=0.50\). Thus, \(P(\text{at most 4 blue})\approx0.1938\). 2. The spinner shows \(2\) red sections out of \(10\), so \(p_R=0.20\). Thus, \(P(\text{at least 2 red})=1-P(\text{0 or 1 red})\approx0.7251\).

a) \(0.1938\). b) \(0.7251\).

- 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.

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 \(20\cdot0.008\cdot0.512=0.08192\).

\(P(X=3)=0.08192\), or \(8.192\%\)

- Count the yellow sections in the displayed spinner. - Model the number of yellow results with a binomial random variable. - “At most \(10\)” is a cumulative probability.

1. The spinner shows \(2\) yellow sections out of \(6\), so \(p=1/3\). 2. Let \(X\sim\operatorname{Bin}(50,1/3)\). 3. \(P(X\le10)=\sum_{k=0}^{10}\binom{50}{k}(1/3)^k(2/3)^{50-k}\approx0.028440\). 4. Rounded to three decimal places, the probability is \(0.028\).

\(0.028\).

- Identify the binomial parameters. - Use the point-probability formula for each possible value of \(X\). - Add the exact fractions to check the distribution total.

1. \(X\sim\operatorname{Bin}(3,1/6)\). 2. \(P(X=0)=(5/6)^3=125/216\). 3. \(P(X=1)=3(1/6)(5/6)^2=75/216\). 4. \(P(X=2)=3(1/6)^2(5/6)=15/216\). 5. \(P(X=3)=(1/6)^3=1/216\). 6. Their sum is \((125+75+15+1)/216=1\).

a) \(P(X=0)=125/216\), \(P(X=1)=75/216\), \(P(X=2)=15/216\), and \(P(X=3)=1/216\). b) The probabilities sum to \(1\).

- Ask whether the number of trials is fixed before the experiment begins. - Ask whether the random variable counts successes or counts waiting time. - Both models can use the same independent Bernoulli trials but summarize them differently.

1. \(A\) counts successes in a fixed number of \(10\) independent attempts, so \(A\sim\operatorname{Bin}(10,0.80).\) 2. \(B\) counts trials through the first success, so \(B\sim\operatorname{Geom}(0.80).\) 3. A binomial random variable counts how many successes occur in a fixed number of trials. A geometric random variable counts how long the process waits until the first success.

a) \(A\sim\operatorname{Bin}(10,0.80)\). b) \(B\sim\operatorname{Geom}(0.80)\). c) \(A\) counts successes in a fixed number of trials; \(B\) counts trials until the first success.

- First find the probability that all \(12\) glasses in one box are bubble-free. - Use that box-level probability as the success probability for the second binomial model. - “At least \(5\)” includes \(5\) and \(6\).

1. One glass is bubble-free with probability \(0.97\). 2. Thus, the probability one box is entirely bubble-free is \(p_B=0.97^{12}\approx0.6938.\) 3. Let \(Y\sim\operatorname{Bin}(6,p_B)\). 4. Then \(P(Y\ge5)=\binom65p_B^5(1-p_B)+p_B^6\approx0.4070.\)

a) \(0.97^{12}\approx0.6938\). b) \(\binom65(0.97^{12})^5(1-0.97^{12})+(0.97^{12})^6\approx0.4070\).

- 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.

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.

Event 1 is more likely. Its probability is \(94.77\%\), compared with approximately \(31.10\%\) for Event 2.

- 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.

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.

Strategy B has the greater success probability: approximately \(24.61\%\), compared with approximately \(10.35\%\) for Strategy A.

- 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 \(1-p\).

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 \(0.0026\).

The probability that one tablet is defective is approximately \(0.0026\), or \(0.26\%\).

- Use \(p=1/5\) for red. - For part b), use the complement of \(0\) or \(1\) red. - For part c), use the complement of no red results.

1. The red probability is \(p=0.2\). 2. \(P(X=3)=\binom83(0.2)^3(0.8)^5\approx0.1468.\) 3. \(P(X\ge2)=1-\left[(0.8)^8+8(0.2)(0.8)^7\right]\approx0.4967.\) 4. For part c), \(1-(0.8)^n>0.90\), so \(n>10.319\ldots\). The minimum integer is \(11\).

a) \(\binom83(0.2)^3(0.8)^5\approx0.1468\). b) \(1-[(0.8)^8+8(0.2)(0.8)^7]\approx0.4967\). c) \(11\) spins.

- The success probability is \(1/4\). - Use a complement for at least one correct answer. - Test upper-tail probabilities for candidate passing scores.

1. \(X\sim\operatorname{Bin}(10,0.25)\). 2. \(P(X=3)=\binom{10}{3}(0.25)^3(0.75)^7\approx0.2503\). 3. \(P(X\ge1)=1-(0.75)^{10}\approx0.9437\). 4. \(P(X\ge4)\approx0.2241>0.15\), while \(P(X\ge5)\approx0.0781<0.15\). 5. Therefore, the minimum passing score is \(5\).

a) \(\binom{10}{3}(0.25)^3(0.75)^7\approx0.2503\). b) \(1-(0.75)^{10}\approx0.9437\). c) \(5\) correct answers.

- Model the number of arriving guests with a binomial distribution. - Overbooking occurs when the arrival count exceeds \(45\). - For the maximum in part b), compare consecutive reservation counts around the threshold.

1. With \(52\) reservations, let \(X\sim\operatorname{Bin}(52,0.80)\). 2. Overbooking means \(X>45\), so \(P(X>45)=P(X\ge46)\approx0.0823.\) 3. To maximize reservations subject to the \(10\%\) limit, compare consecutive reservation counts. 4. With \(52\) reservations, the overbooking probability is about \(0.0823\le0.10\). 5. With \(53\) reservations, the overbooking probability is about \(0.1420>0.10\). 6. Therefore, \(52\) is the maximum allowed number of reservations.

a) \(P(X>45)\approx0.0823\). b) \(52\) reservations.

- Use the binomial model with \(n=20\) and \(p=0.05\). - “At most \(2\)” includes \(0,1,2\). - Use the complement of no defective components for “at least \(1\).”

1. \(X\sim\operatorname{Bin}(20,0.05)\). 2. \(P(X=1)=\binom{20}{1}(0.05)(0.95)^{19}\approx0.3774.\) 3. \(P(X\le2)=\sum_{k=0}^{2}\binom{20}{k}(0.05)^k(0.95)^{20-k}\approx0.9245.\) 4. \(P(X\ge1)=1-(0.95)^{20}\approx0.6415.\)

a) \(\binom{20}{1}(0.05)(0.95)^{19}\approx0.3774\). b) \(\sum_{k=0}^{2}\binom{20}{k}(0.05)^k(0.95)^{20-k}\approx0.9245\). c) \(1-(0.95)^{20}\approx0.6415\).

- Translate each verbal condition into a set of integer hit counts. - Use a binomial tail or finite sum. - Keep the endpoint inequalities strict or inclusive as stated.

1. \(X\sim\operatorname{Bin}(10,0.7)\). 2. \(P(X\ge8)=\sum_{k=8}^{10}\binom{10}{k}(0.7)^k(0.3)^{10-k}\approx0.3828.\) 3. The event \(5<X<9\) is \(X\in\{6,7,8\}\), so \(P(5<X<9)=\sum_{k=6}^{8}\binom{10}{k}(0.7)^k(0.3)^{10-k}\approx0.7004.\)

a) \(\sum_{k=8}^{10}\binom{10}{k}(0.7)^k(0.3)^{10-k}\approx0.3828\). b) \(\sum_{k=6}^{8}\binom{10}{k}(0.7)^k(0.3)^{10-k}\approx0.7004\).

- Model the number of red results first. - Use a complement for “at least \(2\).” - Translate “at least \(9\) nonred” into a condition on the red count.

1. For red, \(X\sim\operatorname{Bin}(10,0.25)\). 2. \(P(X=2)=\binom{10}{2}(0.25)^2(0.75)^8\approx0.2816.\) 3. \(P(X\le1)=(0.75)^{10}+10(0.25)(0.75)^9\approx0.2440.\) 4. \(P(X\ge2)=1-P(X\le1)\approx0.7560.\) 5. At least \(9\) nonred results means at most \(1\) red result, so the probability is the same as in part b), about \(0.2440\).

a) \(\binom{10}{2}(0.25)^2(0.75)^8\approx0.2816\). b) \((0.75)^{10}+10(0.25)(0.75)^9\approx0.2440\). c) \(1-[(0.75)^{10}+10(0.25)(0.75)^9]\approx0.7560\). d) \((0.75)^{10}+10(0.25)(0.75)^9\approx0.2440\).

- Count sixes as successes. - Use complements for the cumulative events. - Translate the number of non-sixes into the corresponding number of sixes.

1. Let \(X\) be the number of sixes. Then \(X\sim\operatorname{Bin}(6,\frac16)\). 2. \(P(X=1)=6\left(\frac16\right)\left(\frac56\right)^5 =\left(\frac56\right)^5\approx0.4019.\) 3. \(P(X=0)=\left(\frac56\right)^6\approx0.3349.\) 4. \(P(X\ge2)=1-P(X=0)-P(X=1)\approx0.2632.\) 5. “A number other than \(6\) at most \(5\) times” is equivalent to at least one \(6\), so \(P(X\ge1)=1-\left(\frac56\right)^6\approx0.6651.\)

a) \((5/6)^5\approx0.4019\). b) \((5/6)^6\approx0.3349\). c) \(1-(5/6)^6-(5/6)^5\approx0.2632\). d) \(1-(5/6)^6\approx0.6651\).

- Recall the four conditions for a binomial setting. - Use a complement or binomial upper tail for part b). - When using the interval around the mean, remember that \(X\) can take only integer values.

1. A binomial model requires a fixed number of trials, two relevant outcomes per trial, independent trials, and a constant success probability. Here, \(n=40\) and \(p=0.75\). 2. \(P(X\ge35)=0.04327398\ldots\approx0.0433\). 3. The mean is \(\mu=np=30\), and the standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{7.5}\approx2.7386\). 4. The interval is approximately \([27.2614,32.7386]\). Since \(X\) is an integer count, this corresponds to \(28\le X\le32\). 5. Therefore, \(P(28\le X\le32)=0.63891160\ldots\approx0.6389\).

a) Fixed \(n\), two relevant outcomes, independence, and constant \(p\); \(n=40\), \(p=0.75\). b) \(P(X\ge35)\approx0.0433\). c) \(\mu=30\) and \(\sigma=\sqrt{7.5}\approx2.7386\). d) \(P(28\le X\le32)\approx0.6389\).

- Use the fact that \(E(X)=np\) is an integer to list the possible values of \(p\). - Use the row entry for \(P(X=0)\) to identify which possible value of \(p\) fits. - Add the table entries that correspond to the requested interval. - For “at least one,” consider the complement.

1. Because \(E(X)=np=10p\) is an integer, \(p\) must be one of \(0,0.1,0.2,\ldots,1\). 2. The table gives \(P(X=0)=(1-p)^{10}\approx0.3487\). Testing the possible values shows that \(p=0.1\), since \((0.9)^{10}\approx0.3487\). 3. Add the listed probabilities: \(P(1\le X\le3)=0.3874+0.1937+0.0574=0.6385\). 4. Use the complement: \(P(X\ge1)=1-P(X=0)=1-0.3487=0.6513\).

a) \(p=0.1\) b) \(P(1\le X\le3)=0.6385\) c) \(P(X\ge1)=0.6513\)

- 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.

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.

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.

- 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.

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.

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 \(0.80\).

- 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.

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.

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 \(P(X=0)+P(X=1)=P(X\le 1)\).

- Use \(p=1/4\) for a correct guess. - Part a is an upper-tail probability. - For part b, compare consecutive integer cutoffs around the \(10\%\) threshold.

1. \(X\sim\operatorname{Bin}(12,0.25)\). 2. \(P(X\ge5)=\sum_{j=5}^{12}\binom{12}{j}(0.25)^j(0.75)^{12-j}\approx0.1576\). 3. For the cutoff, compare successive upper-tail probabilities. 4. \(P(X\ge5)\approx0.1576>0.10\), while \(P(X\ge6)\approx0.0544<0.10\). 5. Therefore, the smallest qualifying value is \(k=6\).

a) \(P(X\ge5)\approx0.1576\). b) \(k=6\).

- 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.

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.

There are \(5\) blue marbles in the container.

- 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.

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, \(k=25\cdot0.2=5\).

The shipment contains \(5\) defective components.

- Use the binomial point formula for part a). - Use cumulative probabilities for the tails and interval. - Check the endpoint inclusion carefully.

1. \(P(X=6)=\binom{20}{6}(0.3)^6(0.7)^{14}\approx0.1916.\) 2. \(P(X\le4)=\sum_{k=0}^{4}\binom{20}{k}(0.3)^k(0.7)^{20-k}\approx0.2375.\) 3. \(P(X>7)=1-P(X\le7)\approx0.2277.\) 4. \(P(5\le X\le9)=P(X\le9)-P(X\le4)\approx0.7145.\)

a) \(0.1916\). b) \(0.2375\). c) \(0.2277\). d) \(0.7145\).

- Standardize the lifetime cutoff first. - Use the one-bulb probability as the success probability in the binomial count. - Keep the unrounded normal probability when evaluating the binomial probability.

1. The standard score is \(z=\frac{13{,}500-15{,}000}{1200}=-1.25.\) Therefore, \(p=P(X<13{,}500)=\Phi(-1.25)\approx0.1056.\) 2. Let \(Y\sim\operatorname{Bin}(200,p)\), where \(p=\Phi(-1.25)\). 3. \(P(Y\le30)=\sum_{k=0}^{30}\binom{200}{k}p^k(1-p)^{200-k}\approx0.9805.\)

1. \(\Phi(-1.25)\approx0.1056\). 2. \(P(Y\le30)\approx0.9805\).

- Standardize the one-cup cutoff. - Use the resulting probability as the binomial success probability. - Keep the unrounded one-cup probability in the second calculation.

1. The standardized cutoff is \(z=\frac{195-200}{5}=-1.\) Thus, \(p=P(X<195)=\Phi(-1)\approx0.1587.\) 2. Let \(Y\sim\operatorname{Bin}(50,p)\). 3. \(P(Y\le10)=\sum_{k=0}^{10}\binom{50}{k}p^k(1-p)^{50-k}\approx0.8406.\)

a) \(\Phi(-1)\approx0.1587\). b) \(P(Y\le10)\approx0.8406\).

- Convert the specification endpoints to z-scores. - Use the one-bolt fit probability as the binomial success probability. - Keep the unrounded normal probability for the shipment calculation.

1. The endpoints have standard scores \(-1\) and \(1\), so \(p=P(9.95\le X\le10.05)=\Phi(1)-\Phi(-1)\approx0.6827.\) 2. Let \(Y\sim\operatorname{Bin}(200,p)\). 3. \(P(Y\ge140)=1-P(Y\le139)\approx0.3291.\)

a) \(\Phi(1)-\Phi(-1)\approx0.6827\). b) \(P(Y\ge140)\approx0.3291\).

- Standardize the one-LED cutoff. - Use the resulting lifetime probability as the binomial success probability. - Keep the unrounded value of \(p\) for the second calculation.

1. The standardized cutoff is \(z=\frac{1410-1500}{60}=-1.5.\) Thus, \(p=P(X\ge1410)=\Phi(1.5)\approx0.9332.\) 2. Let \(Y\sim\operatorname{Bin}(40,p)\). 3. \(P(Y\ge38)=P(Y=38)+P(Y=39)+P(Y=40)\approx0.4947.\)

a) \(\Phi(1.5)\approx0.9332\). b) \(P(Y\ge38)\approx0.4947\).

- Distinguish estimating parameters from checking model shape. - The specification endpoints are one standard deviation from the stated mean. - Convert \(65\%\) of \(150\) to an attainable integer count.

1. The manufacturer could take a sufficiently large random sample of pins. The sample mean estimates \(\mu\), and the sample standard deviation estimates \(\sigma\). A histogram or normal probability plot can be used to assess whether a normal model is plausible. 2. For one pin, \(p=P(119.5\le X\le120.5)=\Phi(1)-\Phi(-1)\approx0.6827.\) 3. Let \(Y\sim\operatorname{Bin}(150,p)\). 4. At least \(65\%\) of \(150\) means at least \(98\) pins because \(0.65(150)=97.5\). 5. \(P(Y\ge98)=\sum_{k=98}^{150}\binom{150}{k}p^k(1-p)^{150-k}\approx0.8059.\)

a) Estimate \(\mu\) with the sample mean and \(\sigma\) with the sample standard deviation; inspect a histogram or normal probability plot for normal-model plausibility. b) \(P(Y\ge98)\approx0.8059\).

- Standardize the \(505\,\text{g}\) cutoff. - Convert \(10\%\) of \(400\) to a count. - Use the unrounded one-bag probability in the binomial calculation.

1. The standardized cutoff is \(z=\frac{505-500}{4}=1.25.\) Thus, \(p=P(W>505)=1-\Phi(1.25)\approx0.1056.\) 2. Let \(X\sim\operatorname{Bin}(400,p)\). 3. At most \(10\%\) of \(400\) means \(X\le40\). 4. \(P(X\le40)=\sum_{k=0}^{40}\binom{400}{k}p^k(1-p)^{400-k}\approx0.3950.\)

a) \(1-\Phi(1.25)\approx0.1056\). b) \(P(X\le40)\approx0.3950\).

- Standardize the one-bottle cutoff. - Use the resulting probability as the binomial success probability. - Use a complement and logarithms for the minimum inspection count.

1. The standardized cutoff is \(z=\frac{750-754}{4}=-1,\) so \(p=P(X<750)=\Phi(-1)\approx0.1587.\) 2. Let \(Y\sim\operatorname{Bin}(12,p)\). Then \(P(Y\le1)=(1-p)^{12}+12p(1-p)^{11}\approx0.4105.\) 3. For \(n\) inspected bottles, \(P(\text{at least one underfilled})=1-(1-p)^n.\) Requiring this to be at least \(0.95\) gives \(n\ge\frac{\ln(0.05)}{\ln(1-p)}\approx17.34.\) The minimum integer is \(18\).

a) \(\Phi(-1)\approx0.1587\). b) \(P(Y\le1)\approx0.4105\). c) \(18\) bottles.

- Relate \(0.10\,\text{mm}\) to the given standard deviation. - Use exactly the instructed approximation \(p=0.05\) in parts b) and c). - For the large-counts check, solve both inequalities and use the stronger requirement.

1. 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. 2. Using the instructed approximation \(p=0.05\), let \(X\sim\operatorname{Bin}(50,0.05)\). 3. \(P(X=2)=\binom{50}{2}(0.05)^2(0.95)^{48}\approx0.2611.\) 4. The large-counts conditions are \(0.05n\ge10\) and \(0.95n\ge10\). 5. These require \(n\ge200\) and \(n\ge10.53\ldots\), respectively, so the minimum integer is \(200\). 6. At \(n=200\), the two counts are \(10\) and \(190\), so both conditions hold.

a) About \(5\%\). b) \(P(X=2)\approx0.2611\). c) \(n=200\); the large-counts values are \(10\) and \(190\).

- 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.

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 \(119\).

a) Standard: approximately \(95\%\); Overweight: approximately \(2.5\%\) b) \(119\) bags

- Standardize each fill cutoff. - Use the below-\(492\) probability as the binomial success probability in part c). - Keep the unrounded normal probability when evaluating the binomial probability.

1. The z-scores for \(495\) and \(505\) are \(-1.25\) and \(1.25\), so \(P(495\le X\le505)=\Phi(1.25)-\Phi(-1.25)\approx0.7887.\) 2. For \(492\), \(z=-2\), so \(p=P(X<492)=\Phi(-2)\approx0.0228.\) 3. Let \(Y\sim\operatorname{Bin}(12,p)\). Then \(P(Y\le1)=(1-p)^{12}+12p(1-p)^{11}\approx0.9706.\)

a) \(0.7887\). b) \(0.0228\). c) \(0.9706\).

- Standardize the interval endpoints and the \(16\)-hour cutoff. - Use the long-life probability as the success probability for the count. - Keep the unrounded value of \(p\) in the binomial calculation.

1. The z-scores for \(12\) and \(15\) are \(-4/3\) and \(2/3\), so \(P(12\le X\le15)\approx0.6563.\) 2. For \(16\), \(z=4/3\), so \(p=P(X>16)\approx0.0912.\) 3. Let \(Y\sim\operatorname{Bin}(50,p)\). Then \(P(Y\ge5)=1-\sum_{k=0}^{4}\binom{50}{k}p^k(1-p)^{50-k}\approx0.4844.\)

a) \(0.6563\). b) \(0.0912\). c) \(0.4844\).

- Use the binomial mean formula to find the sample size. - Then use the binomial standard-deviation formula. - Substitute the values into both parts of the success-failure condition.

1. Since \(E(X)=np=20\) and \(p=0.05\), \(n=\frac{20}{0.05}=400\). 2. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{400\cdot0.05\cdot0.95}=\sqrt{19}\approx4.36\). 3. For the success-failure condition, \(np=400\cdot0.05=20\) and \(n(1-p)=400\cdot0.95=380\). 4. Both values are at least \(10\), so a normal approximation is appropriate.

a) \(n=400\) b) \(\sigma=\sqrt{19}\approx4.36\) c) Yes. Since \(np=20\) and \(n(1-p)=380\), both parts of the success-failure condition are satisfied.

- 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 \(1\).

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.

a) Mila answers none of the \(6\) questions correctly. b) Mila answers at least \(1\) question correctly. c) Mila answers exactly \(1\) of the \(6\) questions correctly. d) Mila answers at least \(2\) questions correctly.

- Choose the appropriate success definition for each part. - Use a complement for “at least one.” - “More than \(13\)” means \(14\) or \(15\).

1. For a specified day, the success probability is \(1/7\). 2. \(P(X=2)=\binom{15}{2}\left(\frac17\right)^2\left(\frac67\right)^{13}\approx0.2889.\) 3. \(P(X\ge1)=1-\left(\frac67\right)^{15}\approx0.9010.\) 4. For Monday through Friday, \(p=5/7\), so \(P(X>13)=P(X=14)+P(X=15)\approx0.0450.\)

a) \(0.2889\). b) \(0.9010\). c) \(0.0450\).

- Use combinations to count bit positions in part a). - For part b), add the zero-error and one-error probabilities. - State all four binomial conditions in context.

1. The number of blocks with exactly two ones is \(\binom{20}{2}=190\). 2. If \(X\) is the number of bit errors, then \(X\sim\operatorname{Bin}(20,0.01)\). 3. \(P(X\le1)=(0.99)^{20}+20(0.01)(0.99)^{19}\approx0.9831.\) 4. A binomial model requires a fixed number of bits, two relevant outcomes per bit, a constant error probability, and independent error outcomes.

a) \(190\). b) \((0.99)^{20}+20(0.01)(0.99)^{19}\approx0.9831\). c) Fixed \(n\), two relevant outcomes, constant \(p\), and independence.

- Count one-letter and two-letter prefixes separately. - Count the special prefixes beginning with A. - Use the resulting probability in the binomial formula.

1. There are \((26+26^2)(99)=69{,}498\) possible codes. 2. There are \((1+26)(99)=2673\) special codes, so \(p=\frac{2673}{69498}=\frac1{26}.\) 3. If \(X\) counts special codes, then \(X\sim\operatorname{Bin}(20,1/26)\), so \(P(X=3)=\binom{20}{3}\left(\frac1{26}\right)^3\left(\frac{25}{26}\right)^{17}\approx0.0333.\)

1. \(69{,}498\). 2. \(p=\frac1{26}\). 3. \(\binom{20}{3}(1/26)^3(25/26)^{17}\approx0.0333\).

- Count all possible numeric suffixes and those beginning with \(9\). - Model the ten generated codes with a binomial random variable. - Use a complement for “more than \(2\).”

1. There are \(5(1000)=5000\) possible codes and \(5(100)=500\) priority codes, so \(p=500/5000=0.1\). 2. If \(Y\sim\operatorname{Bin}(10,0.1)\), then \(P(Y>2)=1-\sum_{k=0}^{2}\binom{10}{k}(0.1)^k(0.9)^{10-k}\approx0.0702.\)

1. \(p=0.1\). 2. \(P(Y>2)\approx0.0702\).

- Treat Ava and Ben as one block when counting favorable seatings. - Use the one-night probability as the success probability for ten nights. - The probability in part b) is an exact finite decimal.

1. Put five distinct name tiles in a bag and draw all five without replacement; the draw order represents the seating order. 2. There are \(5!=120\) seating orders. Treat Ava and Ben as a block: \(2(4!)=48\) orders place them together. 3. Thus, \(p=48/120=0.4\). 4. Across \(10\) independent nights, let \(X\sim\operatorname{Bin}(10,0.4)\). 5. \(P(X=4)=\binom{10}{4}(0.4)^4(0.6)^6=0.250822656.\)

a) Use five name tiles drawn without replacement; \(p=0.4\). b) \(P(X=4)=0.250822656\).

- Use the binomial model with \(n=25\). - Use a complement for at least two returns. - Expected count is \(np\).

1. For part a), \(X\sim\operatorname{Bin}(25,0.15)\). 2. \(P(X=3)=\binom{25}{3}(0.15)^3(0.85)^{22}\approx0.2174.\) 3. \(P(X=0)=(0.85)^{25}\approx0.0172.\) 4. \(P(X\ge2)=1-P(X=0)-P(X=1)\approx0.9069.\) 5. At \(p=0.15\), \(E(X)=25(0.15)=3.75\). At \(p=0.20\), \(E(X)=25(0.20)=5\), an increase of \(1.25\).

a) (1) \(0.2174\); (2) \(0.0172\); (3) \(0.9069\). b) The expected count increases from \(3.75\) to \(5\), an increase of \(1.25\).

- Count the possible positions of the three correct answers. - Use the binomial coefficient when any three positions may be correct. - Do not use a binomial coefficient for one specified order.

1. The number of patterns is \(\binom{10}{3}=120\). 2. \(P(X=3)=\binom{10}{3}(0.2)^3(0.8)^7=0.201326592\). 3. For the single specified order, \((0.2)^3(0.8)^7=0.0016777216\).

a) \(120\). b) \(0.201326592\). c) \(0.0016777216\).

- Use a combination to count the positions of the four special-handling packages. - Use the same combination inside the exact binomial probability. - “At most \(2\)” includes \(0,1,2\).

1. The number of sequences is \(\binom{20}{4}=4845\). 2. \(P(X=4)=\binom{20}{4}(0.15)^4(0.85)^{16}\approx0.1821.\) 3. \(P(X\le2)=\sum_{k=0}^{2}\binom{20}{k}(0.15)^k(0.85)^{20-k}\approx0.4049.\)

a) \(4845\). b) \(\binom{20}{4}(0.15)^4(0.85)^{16}\approx0.1821\). c) \(\sum_{k=0}^{2}\binom{20}{k}(0.15)^k(0.85)^{20-k}\approx0.4049\).

- Compare the number of rare sixes in the two fixed sequences. - A fixed sequence does not use a binomial coefficient. - An exact success count includes every arrangement with that count.

1. Sequence B is more likely because it contains only three occurrences of the less likely result \(6\). 2. \(P(A)=(1/6)^6=1/46656\). 3. \(P(B)=(1/6)^3(5/6)^3=125/46656\). 4. Exactly three sixes can occur in \(\binom63=20\) positions, so \(P(X=3)=\binom63(1/6)^3(5/6)^3=2500/46656\). 5. Sequence B is only one of the \(20\) ordered patterns included in \(X=3\).

a) Sequence B. b) \(P(A)=1/46656\), \(P(B)=125/46656\). c) \(P(X=3)=2500/46656\); it includes \(20\) possible arrangements, not just Sequence B.

- Compare the frequency of the rare outcome \(D\). - A specified record fixes the defect position. - An exact defect count allows every possible defect position.

1. Record (3) is least likely because every position contains the rare \(D\) outcome. 2. \(P(1)=(0.95)^8\), while \(P(2)=0.05(0.95)^7\). 3. Exactly one defective chip can occur in any of eight positions, so \(P(X=1)=8(0.05)(0.95)^7\). 4. Therefore, the claim is false: the exact-one event has eight times the probability of the one specified record.

a) Record (3). b) \(P(1)=(0.95)^8\), \(P(2)=0.05(0.95)^7\). c) The claim is false; \(P(X=1)=8(0.05)(0.95)^7\).

- Decide whether the positions of the two nongerminating seeds are specified. - Use a binomial coefficient only when any arrangement with the given count is allowed. - Compare one fixed sequence with all sequences having the same count.

1. For part a), any \(13\) of the \(15\) positions may be the germinating seeds, so \(P(X=13)=\binom{15}{13}(0.85)^{13}(0.15)^2.\) 2. Part b specifies one particular ordered pattern, so \(P=(0.85)^{13}(0.15)^2.\) 3. The binomial exact-count probability is \(\binom{15}{13}\) times the probability of that one specified pattern.

a) \(\binom{15}{13}(0.85)^{13}(0.15)^2\). b) \((0.85)^{13}(0.15)^2\).

- State all four binomial conditions. - Identify \(n\), \(p\), and \(k\). - Use the binomial point-probability formula.

1. The model has a fixed \(12\) trials, two relevant outcomes on each shot, constant make probability \(0.75\), and independent shot outcomes. 2. Thus, \(X\sim\operatorname{Bin}(12,0.75)\). 3. \(P(X=9)=\binom{12}{9}(0.75)^9(0.25)^3.\)

Fixed \(n=12\), two relevant outcomes, constant \(p=0.75\), and independence. The probability is \(\binom{12}{9}(0.75)^9(0.25)^3\).

- 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 \(1-p\).

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, \(P(X=1)=\binom{5}{1}(0.15)^1(0.85)^4\approx 0.3915\).

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.

- Distinguish a fixed ordered pattern from a count event. - Multiply success and failure probabilities for a fixed sequence. - Use a binomial coefficient only when the failure position is not specified.

1. Part a specifies one sequence with five successes and one failure: \((0.85)^5(0.15).\) 2. Part b is \((0.85)^6.\) 3. Part c specifies four successes followed by two failures: \((0.85)^4(0.15)^2.\) 4. Part d allows the one failure in any of six positions: \(\binom65(0.85)^5(0.15).\)

a) \((0.85)^5(0.15)\). b) \((0.85)^6\). c) \((0.85)^4(0.15)^2\). d) \(\binom65(0.85)^5(0.15)\).

- Translate “more than \(6\)” into the correct binomial tail. - Find the binomial mean and standard deviation before forming the interval. - Use the decimal interval endpoints to select only attainable integer counts.

1. \(Y\sim\operatorname{Bin}(80,0.05)\). 2. \(P(Y>6)=0.10528525\ldots\approx0.1053\). 3. The mean is \(\mu=4\), and the standard deviation is \(\sigma=\sqrt{3.8}\approx1.9494\). 4. The interval \([\mu-\sigma,\mu+\sigma]\) is approximately \([2.0506,5.9494]\), so the integer values are \(3,4,5\). 5. Therefore, \(P(3\le Y\le5)=0.55860417\ldots\approx0.5586\).

1. \(P(Y>6)\approx0.1053\). 2. The integer values are \(3,4,5\), and \(P(3\le Y\le5)\approx0.5586\).

- Read the lower and upper summation limits as possible values of \(Y\). - A leading \(1-\) indicates a complement. - Translate the expression before evaluating it numerically.

1. Part a is \(P(Y\le10)\), the probability of at most \(10\) defective components, and it is approximately \(0.9885\). 2. Part b is \(P(Y\ge3)\), the probability of at least \(3\) defective components, and it is approximately \(0.8817\). 3. Part c is \(P(4\le Y\le7)\), the probability of from \(4\) through \(7\) defective components, and it is approximately \(0.6142\).

a) At most \(10\) defective components; \(0.9885\). b) At least \(3\) defective components; \(0.8817\). c) From \(4\) through \(7\) defective components; \(0.6142\).

- 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.

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 \(P(a<X\le b)=P(X\le b)-P(X\le a)=F(b)-F(a)\).

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, \(P(a<X\le b)=F(b)-F(a)\).

- 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 \(1\).

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. \(P(80\le X<110)=\sum_{i=80}^{109}\binom{150}{i}(0.6)^i(0.4)^{150-i}\approx 0.9587\).

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) \(\sum_{i=80}^{109}\binom{150}{i}(0.6)^i(0.4)^{150-i}\approx 0.9587\)

- Rewrite each absolute-value condition as an interval. - Intersect the interval with the integer support of \(X\). - Use a complement for part b)(2).

1. \(|X-20|\le3\) gives \(17\le X\le23\), so the possible values are \(17\) through \(23\). 2. \(P(17\le X\le23)\approx0.6340\). 3. The complement is \(P(|X-20|>3)\approx0.3660\). 4. \(|X-18|<2\) gives \(16<X<20\), so \(X\in\{17,18,19\}\), with probability approximately \(0.2731\).

a) \(\{17,18,19,20,21,22,23\}\). b) (1) \(0.6340\); (2) \(0.3660\); (3) \(0.2731\).

- Find the binomial mean first. - Translate each absolute-value condition into integer bounds. - For part c, test small integer radii in increasing order.

1. The mean is \(60(1/3)=20\). 2. Part a is \(P(18\le X\le22)\approx0.5061\). 3. Part b is \(P(X\le16)+P(X\ge24)\approx0.3377\). 4. For part c, \(P(|X-20|\le1)<0.5\), while \(P(|X-20|\le2)\approx0.5061>0.5\). 5. Thus, the smallest value is \(k=2\).

a) \(0.5061\). b) \(0.3377\). c) \(k=2\).

- Use \(n=8\) and \(p=0.70\). - “At least \(7\)” includes \(7\) and \(8\). - Part c is the complement of at least \(6\).

1. Let \(X\sim\operatorname{Bin}(8,0.70)\). 2. \(P(X=6)\approx0.2965\). 3. \(P(X\ge7)\approx0.2553\). 4. \(P(X\le5)\approx0.4482\).

a) \(0.2965\). b) \(0.2553\). c) \(0.4482\).

- Read the summation limits as allowed success counts. - Add the corresponding binomial point probabilities. - Preserve the exact fraction where requested.

1. Part a represents \(3\), \(4\), or \(5\) successes: \(P(3\le X\le5)\approx0.3172.\) 2. Part b represents \(4\), \(5\), or \(6\) successes: \(P(4\le Y\le6)=\frac{2211}{4096}.\)

a) \(3\), \(4\), or \(5\) successes; \(0.3172\). b) \(4\), \(5\), or \(6\) successes; \(\frac{2211}{4096}\).

- First determine the one-spin success probability for each event. - Use a binomial model when only the number of successes matters. - Do not use a binomial coefficient for one specified order.

1. Four of the eight labels are prime, so \(p=1/2\). Thus, \(P(X=4)=\binom{10}{4}(1/2)^{10}\approx0.2051.\) 2. \(P(X\le3)=\sum_{k=0}^{3}\binom{10}{k}(1/2)^{10}\approx0.1719.\) 3. Two labels exceed \(6\), so \(p=1/4\). Therefore, \(P(Y\ge8)=\sum_{k=8}^{10}\binom{10}{k}(1/4)^k(3/4)^{10-k}\approx0.0004.\) 4. The order is fixed, so \((1/2)^3(1/2)^7=(1/2)^{10}\approx0.0010.\)

1. \(0.2051\). 2. \(0.1719\). 3. \(0.0004\). 4. \(0.0010\).

- 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.

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.

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.

- 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.

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.

The most likely number is \(k=4\). The least likely number is \(k=20\).

- Use a binomial model with \(n=20\) and \(p=0.85\). - Translate the verbal boundaries into integer counts. - Use cumulative probabilities or complements for the tails.

1. \(X\sim\operatorname{Bin}(20,0.85)\). 2. \(P(X=18)\approx0.2293\). 3. \(P(X\le16)\approx0.3523\). 4. \(P(X\ge15)\approx0.9327\). 5. \(P(17\le X\le19)\approx0.6090\).

a) \(0.2293\). b) \(0.3523\). c) \(0.9327\). d) \(0.6090\).

- “At least \(2\)” is easier through a complement. - Include both zero and one defect in the complementary event. - Keep the exact expression until the final rounding.

1. \(X\sim\operatorname{Bin}(15,0.04)\). 2. Use the complement of zero or one defect: \(P(X\ge2)=1-(0.96)^{15}-15(0.04)(0.96)^{14}.\) 3. Therefore, \(P(X\ge2)\approx0.1191.\)

\(1-(0.96)^{15}-15(0.04)(0.96)^{14}\approx0.1191\).

- 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.

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 \(k\in\{0, 1, \ldots, 14\}\cup\{23, 24, \ldots, 30\}\).

a) \(k\in\{16, 17, 18, 19, 20\}\) b) \(k\in\{0, 1, \ldots, 14\}\cup\{23, 24, \ldots, 30\}\)

- 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.

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 \(k=35\).

a) \(k=32\) b) \(k=35\)

- Translate “more than \(3\)” into an integer tail. - A complement of \(0,1,2,3\) defects is efficient. - Convert the final probability to a percentage only after calculating it.

1. Let \(X\sim\operatorname{Bin}(20,0.10)\). 2. A package is rejected when \(X\ge4\), so \(P(X\ge4)=1-\sum_{k=0}^{3}\binom{20}{k}(0.10)^k(0.90)^{20-k}\approx0.1330.\) 3. As a percentage, this is approximately \(13.3\%\).

Approximately \(13.3\%\).

- Translate the word “fewer” carefully. - Use the binomial cumulative probability through \(7\). - Round only the final result.

1. Let \(X\sim\operatorname{Bin}(12,0.75)\). 2. “Fewer than \(8\)” means \(X\le7\). 3. \(P(X\le7)=\sum_{k=0}^{7}\binom{12}{k}(0.75)^k(0.25)^{12-k}\approx0.1576.\)

\(P(X\le7)\approx0.1576\).

- 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.

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, \(P(X\ge4)=P(X\le2)\).

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 \(P(X\le2)\).

- First find the probability that all \(20\) LEDs in one strip work. - Use that strip-level probability in a second binomial model. - Keep the unrounded strip probability in part b).

1. A strip is completely functional with probability \(p=(0.98)^{20}\approx0.6676.\) 2. If \(Y\) counts functional strips among \(15\), then \(Y\sim\operatorname{Bin}(15,p)\). 3. \(P(Y=12)=\binom{15}{12}p^{12}(1-p)^3\approx0.1310.\)

a) \((0.98)^{20}\approx0.6676\). b) \(\binom{15}{12}[(0.98)^{20}]^{12}[1-(0.98)^{20}]^3\approx0.1310\).

- First calculate the one-packet success probability. - Then treat each packet as one Bernoulli trial. - Round only after the second-stage calculation is complete.

1. For one packet, \(p=P(X\ge11)=\binom{12}{11}(0.9)^{11}(0.1)+(0.9)^{12}\approx0.6590.\) 2. Let \(Y\sim\operatorname{Bin}(5,p)\). 3. \(P(Y\le2)=\sum_{k=0}^{2}\binom5k p^k(1-p)^{5-k}\approx0.2214.\)

a) \(0.6590\). b) \(0.2214\).

- Use the same binomial model for all three parts. - Part b is a lower-tail cumulative probability. - Part c is an inclusive integer interval.

1. \(X\sim\operatorname{Bin}(200,0.04)\). 2. \(P(X=8)\approx0.1425\). 3. \(P(X\le10)\approx0.8200\). 4. \(P(5\le X\le12)\approx0.8451\).

a) \(0.1425\). b) \(0.8200\). c) \(0.8451\).

- Translate each verbal condition to integer bounds. - Use a complement for the upper tail if convenient. - “Fewer than \(10\)” stops at \(9\).

1. \(X\sim\operatorname{Bin}(30,0.25)\). 2. \(P(X>10)\approx0.1057\). 3. The second event is \(5\le X\le9\), with probability approximately \(0.7055\).

a) \(0.1057\). b) \(0.7055\).

- Use the stated binomial parameters. - Translate “at most” and “at least” into cumulative tails. - State all four binomial conditions in context.

1. Let \(X\sim\operatorname{Bin}(30,0.85)\). 2. \(P(X\le22)\approx0.0698\). 3. \(P(X\ge28)\approx0.1514\). 4. A binomial model requires fixed \(n=30\), two relevant outcomes, independent trials, and constant \(p=0.85\).

1. \(0.0698\). 2. \(0.1514\). 3. Fixed \(n\), two outcomes, independence, and constant \(p\).

- Use the requested binomial model for the numerical parts. - The expected count is \(np\). - In part 3, ask whether one package's error status can affect nearby packages.

1. Under the binomial model, \(X\sim\operatorname{Bin}(50,0.04)\), and \(P(X>3)\approx0.1391\). 2. \(E(X)=50(0.04)=2\), and \(P(X=2)\approx0.2762\). 3. Independence is questionable because one jam can create several nearby errors, making consecutive package outcomes dependent.

1. \(0.1391\). 2. Expected count \(2\); \(P(X=2)\approx0.2762\). 3. Independence is not reasonable if a single jam affects several consecutive packages.

- Use one binomial model for all three events. - A strict upper tail can be found by complement. - Part 3 includes both endpoints.

1. \(X\sim\operatorname{Bin}(250,0.04)\). 2. \(P(X\le5)\approx0.0633\). 3. \(P(X>12)\approx0.2045\). 4. \(P(8\le X\le15)\approx0.7400\).

1. \(0.0633\). 2. \(0.2045\). 3. \(0.7400\).

- Translate “fewer than” to the correct integer endpoint. - Use cumulative binomial probabilities for the tails. - Use a point probability for exactly \(12\).

1. Let \(X\sim\operatorname{Bin}(80,0.15)\). 2. \(P(X<10)=P(X\le9)\approx0.2211\). 3. \(P(X\ge15)\approx0.2126\). 4. \(P(X=12)\approx0.1240\).

1. \(0.2211\). 2. \(0.2126\). 3. \(0.1240\).

- Convert the percentage threshold to a defect count. - “More than \(5\%\)” means more than \(4\) defects. - Use a binomial point probability for A and a tail for B.

1. Let \(X\sim\operatorname{Bin}(80,0.025)\). 2. \(P(A)=P(X=2)\approx0.2741\). 3. Five percent of \(80\) is \(4\), so event B is \(X>4\). 4. \(P(B)=P(X\ge5)\approx0.0504\).

A: \(0.2741\). B: \(0.0504\).

- Use the same binomial model for both events. - “At most \(5\)” is a lower tail. - Event D includes both endpoints.

1. \(X\sim\operatorname{Bin}(60,0.15)\). 2. \(P(C)=P(X\le5)\approx0.0968\). 3. \(P(D)=P(8\le X\le12)\approx0.5891\).

C: \(0.0968\). D: \(0.5891\).

- Use separate binomial models for the two sample sizes. - “At least \(14\)” includes \(14\) and \(15\). - A probability less than \(1\) cannot support a guarantee.

1. For part a), \(X\sim\operatorname{Bin}(15,0.90)\), so \(P(X\ge14)\approx0.5490\). 2. For part b), \(Y\sim\operatorname{Bin}(11,0.90)\), and \(P(Y\ge10)\approx0.6974\). 3. Since this probability is below \(1\), buying \(10\%\) extra does not guarantee \(10\) plants.

a) \(0.5490\). b) \(0.6974\); the guarantee claim is false.

- Use one binomial model for the whole problem. - Compare the two point probabilities directly in part b). - Do not infer likelihood only from closeness to the mean without calculation.

1. Let \(X\sim\operatorname{Bin}(50,0.05)\). 2. \(P(X\le3)\approx0.7604\). 3. \(P(X=5)\approx0.0658\), while \(P(X=0)\approx0.0769\). 4. Since \(0.0658<0.0769\), the claim is false.

a) \(0.7604\). b) \(P(X=5)\approx0.0658\), \(P(X=0)\approx0.0769\); the claim is false.

- 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?

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.

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.

- “At least \(4\)” is efficiently handled with a complement. - The draws are not exactly independent because sampling is without replacement. - Ask when removing one sampled component changes the composition of the remaining population only negligibly.

1. Let \(X\) be the number of selected components with a quality seal. 2. Because the sample is small relative to the very large production lot, the success probability changes negligibly from draw to draw, so use \(X\approx\operatorname{Bin}(12,0.15)\). 3. Use a complement: \(P(X\ge4)=1-\sum_{k=0}^{3}\binom{12}{k}(0.15)^k(0.85)^{12-k}.\) 4. This gives \(P(X\ge4)\approx0.0922.\) 5. The approximation is reasonable when the sample is only a small fraction of the population, so sampling without replacement creates negligible dependence and the success probability remains nearly constant.

\(P(X\ge4)\approx0.0922\). The binomial approximation is reasonable because the sample is very small relative to the population, making the draws approximately independent with nearly constant success probability.

- Convert the blue central angle into a fraction of \(360^\circ\). - “At most \(1\)” includes \(0\) and \(1\). - Keep the exact expressions until the final rounding.

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 trials is \(n=20\). 3. Let \(X\) be the number of blue results. Then \(P(X=0)=(0.7)^{20}=0.0007979\ldots\approx0.0008.\) 4. \(P(X\le1)=P(X=0)+P(X=1) =(0.7)^{20}+20(0.3)(0.7)^{19} =0.0076373\ldots\approx0.0076.\)

a) \(n=20\), the number of spins, and \(p=0.3\), the probability of blue on each spin. b) \(P(X=0)\approx0.0008\). c) \(P(X\le1)\approx0.0076\).

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

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. \(P(E)=\binom43(0.75)^3(0.25)+(0.75)^4 =0.73828125.\)

a) \(A\to(4)\), \(B\to(1)\), \(C\to(2)\), \(D\to(3)\), \(E\to(5)\). b) \(P(E)=0.73828125\).

- A first success in a specified position requires failures before that position and a success at that position. - “At most \(1\)” includes \(0\) and \(1\). - In part c), compare the factors in the two fixed-sequence probabilities.

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. Therefore, \(P=(0.96)^2(0.04)=0.036864.\) 3. For part b), let \(X\sim\operatorname{Bin}(10,0.04)\). 4. \(P(X\le1)=(0.96)^{10}+10(0.04)(0.96)^9\approx0.9419.\) 5. Each event in part c) has exactly one defective outcome and nine nondefective outcomes in a specified order. 6. Both probabilities are \((0.04)(0.96)^9\). Changing the position does not change the product.

a) \(0.036864\). b) \(P(X\le1)\approx0.9419\). c) Both events have probability \((0.04)(0.96)^9\), so they are equally likely.

- Model the number germinating with a binomial random variable. - Use the binomial-mode rule involving \((n+1)p\). - “At least \(24\)” includes \(24\) through \(30\).

1. Let \(X\sim\operatorname{Bin}(30,0.85)\). 2. Since \((n+1)p=31(0.85)=26.35\) is not an integer, the unique binomial mode is \(\lfloor26.35\rfloor=26.\) 3. \(P(X\ge24)=\sum_{k=24}^{30}\binom{30}{k}(0.85)^k(0.15)^{30-k}\approx0.8474.\)

a) \(26\) seeds. b) \(P(X\ge24)\approx0.8474\).

- Translate “fewer than \(50\)” into attainable integer values. - Use cumulative probabilities to calculate the interval. - The prompt already supplies fixed \(n\) and two relevant outcomes; identify the two remaining binomial conditions.

1. The binomial parameters are \(n=50\) and \(p=0.94\). 2. \(P(X=48)=\binom{50}{48}(0.94)^{48}(0.06)^2\approx0.2262.\) 3. \(P(X\le45)\approx0.1794.\) 4. \(P(46\le X<50)=P(46\le X\le49)\approx0.7753.\) 5. The additional modeling conditions are independent battery outcomes and the same success probability \(0.94\) for every battery.

a) (1) \(0.2262\); (2) \(0.1794\); (3) \(0.7753\). b) Independence and a constant success probability \(p=0.94\).

- Translate “enough with probability at least \(95\%\)” into a cumulative-probability condition. - Compare consecutive integer brochure counts to prove minimality. - Determine exactly when at least one of \(18\) brochures remains.

1. \(X\sim\operatorname{Bin}(80,0.20)\). 2. For part a), seek the smallest integer \(k\) such that \(P(X\le k)\ge0.95\). 3. Since \(P(X\le21)\approx0.9340<0.95\) and \(P(X\le22)\approx0.9612\ge0.95\), the minimum is \(22\) brochures. 4. With \(18\) brochures, at least one remains exactly when fewer than \(18\) visitors want one, so \(P(X\le17)\approx0.6708.\)

a) \(22\) brochures. b) \(P(X\le17)\approx0.6708\).

- Use \(p=1/6\) for a winner. - Translate the required confidence level into a cumulative-probability condition. - For part b), use the complement of having at most \(30\) winners.

1. Let \(X\sim\operatorname{Bin}(150,\frac16)\). 2. For part a), seek the smallest integer \(k\) such that \(P(X\le k)\ge0.90\). 3. Since \(P(X\le30)\approx0.8842<0.90\) and \(P(X\le31)\approx0.9197\ge0.90\), the minimum is \(31\) coupons. 4. With \(30\) coupons, a shortage occurs when \(X>30\), so \(P(X>30)=1-P(X\le30)\approx0.1158.\)

a) \(31\) coupons. b) \(P(X>30)\approx0.1158\).

- Determine the largest attendance count that fits. - Counting no-shows can make the event easier to express. - Use a cumulative binomial probability or its complement.

1. Let \(X\) be the number of people who attend. Then \(X\sim\operatorname{Bin}(50,0.90)\). 2. Everyone has a seat when \(X\le45\). 3. Equivalently, let \(Y\) be the number of no-shows. Then \(Y\sim\operatorname{Bin}(50,0.10)\), and everyone has a seat when \(Y\ge5\). 4. \(P(Y\ge5)=1-P(Y\le4)\approx0.5688.\)

\(P(\text{everyone has a seat})\approx0.5688\).

- 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.

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.

a) \(8\) questions b) \(14\) questions

- Translate “more than \(1\)” into a complement of \(0\) or \(1\). - Use the appropriate defect probability in each part. - Keep the sample size fixed at \(20\).

1. Let \(X\) be the number of cracked eggs in the sample. Rejection occurs when \(X>1\). 2. For part a), \(X\sim\operatorname{Bin}(20,0.02)\), so \(P(X>1)=1-\left[(0.98)^{20}+20(0.02)(0.98)^{19}\right]\approx0.0599.\) 3. For part b), \(X\sim\operatorname{Bin}(20,0.01)\), so \(P(X>1)=1-\left[(0.99)^{20}+20(0.01)(0.99)^{19}\right]\approx0.0169.\)

a) \(0.0599\). b) \(0.0169\).

- 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.

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.

a) Plant \(20\) seeds; exactly \(18\) germinate. b) Plant \(10\) seeds; at least \(2\) germinate. c) Plant \(20\) seeds; at least \(17\) germinate.

- In part a), convert the percentage threshold to the first qualifying whole-number defect count. - In part b), interpret one term \(q^kp\) before interpreting the sum. - In part c), express the probability that all \(30\) chips are nondefective.

1. For part a), let \(X\sim\operatorname{Bin}(60,0.05)\). More than \(10\%\) of \(60\) means at least \(7\) defective chips. 2. \(P(X\ge7)\approx0.0297.\) 3. In part b), \((0.95)^k(0.05)\) 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 nondefective proportion. The requirement is \(q^{30}\ge0.80.\) Hence, \(q\ge(0.80)^{1/30}\approx0.992589,\) or approximately \(99.259\%\).

a) \(0.0297\). b) The first defective chip is found during one of the first \(5\) tests. c) Approximately \(99.259\%\) must be nondefective.

- Part a is a cumulative binomial probability. - In part b), interpret \(0.88\) as the no-rare-item probability. - For part c), use the complement of receiving no rare items.

1. For part a), let \(X\sim\operatorname{Bin}(40,0.12)\). 2. \(P(X\le3)=\sum_{k=0}^{3}\binom{40}{k}(0.12)^k(0.88)^{40-k}\approx0.2768.\) 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 one rare item. 4. For part c), require \(1-(0.88)^n\ge0.99.\) Thus, \(n\ge\frac{\ln(0.01)}{\ln(0.88)}\approx36.03,\) so the minimum integer is \(37\).

a) \(0.2768\). b) At least one rare item is found in \(15\) chests. c) \(37\) chests.

- First find the probability that one figurine is flawless. - Use independence within a set to find the all-flawless set probability. - Treat each set as one Bernoulli trial in the second-stage model.

1. The probability that one figurine is flawless is \(0.60\). 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. \(P(X\le2)=\sum_{k=0}^{2}\binom{20}{k}(0.07776)^k(0.92224)^{20-k}\approx0.7997.\)

\(P(X\le2)\approx0.7997\).

- First find the probability that one four-shot set is successful. - Treat each set as one Bernoulli trial after finding that set-level success probability. - “At least \(9\)” includes \(9\) and \(10\) successful sets.

1. The probability that one set is successful is \(p_S=\binom43(0.70)^3(0.30)+(0.70)^4=0.6517.\) 2. Let \(Y\) be the number of successful sets among the \(10\) sets. Then \(Y\sim\operatorname{Bin}(10,0.6517).\) 3. Therefore, \(P(Y\ge9) =\binom{10}{9}(0.6517)^9(0.3483)+(0.6517)^{10} \approx0.0877.\)

\(P(Y\ge9)\approx0.0877\).

- Use \(np\) for the expected count. - Apply the binomial point formula for exactly \(18\). - “At least \(19\)” includes \(19\) and \(20\).

1. The germination probability is \(p=0.80\). 2. For \(450\) seeds, the expected number is \(450(0.80)=360\). 3. For \(20\) seeds, let \(X\sim\operatorname{Bin}(20,0.80)\). 4. \(P(X=18)=\binom{20}{18}(0.80)^{18}(0.20)^2\approx0.1369.\) 5. \(P(X\ge19)=\binom{20}{19}(0.80)^{19}(0.20)+(0.80)^{20}\approx0.0692.\)

a) \(360\) seedlings. b) \(0.1369\). c) \(0.0692\).

- Identify the binomial parameters from the survey. - Express “at least” using a complement. - Find an inclusive interval probability by subtracting cumulative probabilities with the correct endpoints.

1. Model the count with \(X\sim\operatorname{Bin}(150,0.35)\). 2. \(P(X\ge60)=1-P(X\le59)\approx0.1160.\) 3. \(P(45\le X\le60)=P(X\le60)-P(X\le44)\approx0.8295.\)

a) \(0.1160\). b) \(0.8295\).

- For part a), identify the required outcome of each of the first twelve tests. - Use the binomial mean and standard-deviation formulas in part b). - Convert the continuous one-standard-deviation interval to attainable integer counts.

1. For the twelfth chip to be the first defective chip, the first \(11\) chips must be nondefective and the twelfth must be defective: \(P=(0.92)^{11}(0.08)\approx0.0320.\) 2. Since \(X\sim\operatorname{Bin}(400,0.08),\) its mean is \(\mu=400(0.08)=32\) and its standard deviation is \(\sigma=\sqrt{400(0.08)(0.92)}=\sqrt{29.44}\approx5.43.\) 3. The interval \([\mu-\sigma,\mu+\sigma]\) is approximately \([26.57,37.43]\), so the included integer values are \(27\) through \(37\). 4. Therefore, \(P(27\le X\le37)\approx0.6898.\)

a) \((0.92)^{11}(0.08)\approx0.0320\). b) \(P(27\le X\le37)\approx0.6898\).

- Use the binomial formula for one exact value. - The mode is near the mean but is determined by the binomial-mode rule. - Evaluate \((n+1)p\).

1. \(X\sim\operatorname{Bin}(120,0.92)\). 2. \(P(X=110)=\binom{120}{110}(0.92)^{110}(0.08)^{10}\approx0.1295.\) 3. Since \((n+1)p=121(0.92)=111.32\) is not an integer, the unique mode is \(111\).

a) \(0.1295\). b) \(111\) seeds.

- Interpret \(np\) as a long-run average count. - Keep the exact square-root form of the standard deviation until the final approximation. - Convert the one-standard-deviation interval to attainable integer counts.

1. Since \(X\sim\operatorname{Bin}(500,0.02)\), \(\mu=np=10.\) Over many groups of \(500\) healthy people, the average number of false-positive results would be about \(10\) per group. 2. \(\sigma=\sqrt{np(1-p)}=\sqrt{9.8}\approx3.13.\) 3. The interval \([\mu-\sigma,\mu+\sigma]\) is approximately \([6.87,13.13]\), so the integer values are \(7\) through \(13\). 4. \(P(7\le X\le13)\approx0.7391.\)

a) \(\mu=10\) and \(\sigma=\sqrt{9.8}\approx3.13\). The mean is the long-run average false-positive count per group of \(500\) healthy people. b) \(0.7391\).

- Use the binomial mean and standard-deviation formulas. - Translate the two-standard-deviation condition into an interval. - Keep only attainable integer values inside that interval.

1. Since \(Y\sim\operatorname{Bin}(300,0.10)\), \(\mu=30.\) 2. \(\sigma=\sqrt{300(0.10)(0.90)}=\sqrt{27}=3\sqrt3\approx5.20.\) 3. The interval \([\mu-2\sigma,\mu+2\sigma]\) is approximately \([19.61,40.39]\), so the integer values are \(20\) through \(40\). 4. \(P(20\le Y\le40)\approx0.9575.\)

a) \(\mu=30\) and \(\sigma=3\sqrt3\approx5.20\). b) \(0.9575\).

- 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.

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 \(P(Y=12)=P(Y=13)\).

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.

- 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.

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 \(0.5<p<1\).

\(0.5<p<1\), or \(p\in(0.5, 1)\).

- 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 \(p\).

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 \(0<p\le0.25\).

\(0<p\le0.25\), or \(p\in(0, 0.25]\).

- 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.

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 \([16,32]\).

\([16,32]\)

- 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.

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 \([22,38]\).

\([22,38]\)

- Compute the recognition cutoff using the original process parameters. - Convert the continuous cutoff to the correct integer-count event. - For the final probability, use the explicitly stated independent new-process outcomes with delayed-call probability \(0.08\). - The requested event is the complement of the recognition event.

1. Under the original rate, \(n=120\) and \(p_0=0.15\). The mean is \(\mu_0=18\), and the standard deviation is \(\sigma_0=\sqrt{120(0.15)(0.85)}\approx3.9115\). 2. Success requires \(X<\mu_0-\sigma_0\approx14.0885\). Since \(X\) is an integer, success means \(X\le14\). 3. Under the new staffing tool, the outcomes are independent with delayed-call probability \(0.08\), so \(X\sim\operatorname{Bin}(120,0.08)\). 4. Failure to recognize the improvement means \(X\ge15\), so \(P(X\ge15)\approx0.0564\).

\(0.0564\).

- Build the recognition cutoff using the original process. - Translate the cutoff to the correct integer event. - For the final probability, use the explicitly stated independent new-process outcomes with defect probability \(0.04\). - The requested event is the complement of recognizing the improvement.

1. Under the original process, \(\mu_0=250(0.08)=20\), and \(\sigma_0=\sqrt{250(0.08)(0.92)}\approx4.2895\). 2. Recognition requires \(X\le20-1.5(4.2895)\approx13.5657\). Since \(X\) is integer-valued, recognition occurs for \(X\le13\). 3. Under the new production process, the outcomes are independent with defect probability \(0.04\), so \(X\sim\operatorname{Bin}(250,0.04)\). 4. Therefore, \(P(\text{not recognized})=P(X\ge14)\approx0.1310\).

\(0.1310\).

- Use a weighted mean for the frequency table. - Relate the observed mean to \(np\). - Use the fitted \(p\) in the binomial point probability. - Compare like quantities: model probability and observed relative frequency.

1. \(\bar{k}=\frac{0(4)+1(21)+2(50)+3(48)+4(22)+5(5)}{150}=2.52.\) 2. Since the binomial mean is \(5p\), set \(5p=2.52\), giving \(p=0.504.\) 3. \(P(X=2)=\binom52(0.504)^2(0.496)^3\approx0.3100.\) 4. The observed relative frequency is \(\frac{50}{150}=\frac13\approx0.3333,\) about \(0.0234\) higher than the model probability.

1. \(\bar{k}=2.52\). 2. \(p=0.504\). 3. Model: \(0.3100\); observed: \(0.3333\).

- Convert the grouped counts into a total number of defective components first. - Use the complement of no defects in a group. - Compare the fitted-model probability with the observed fraction of affected groups.

1. The total number of defective components is \(0(130)+1(55)+2(12)+3(3)=88.\) Out of \(600\) tested components, \(p=\frac{88}{600}=\frac{11}{75}\approx0.1467.\) 2. \(P(X\ge1)=1-\left(\frac{64}{75}\right)^3\approx0.3786.\) 3. The observed relative frequency is \(\frac{55+12+3}{200}=0.3500.\) The fitted model is about \(0.0286\) higher.

1. \(p=\frac{11}{75}\approx0.1467\). 2. \(0.3786\). 3. Model: \(0.3786\); observed: \(0.3500\).

- 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 \(n\).

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.

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 \(k=\frac{n}{2}\).

- 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).

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, \(P(X<10)=P(X\le 10)-P(X=10)\approx 0.5-0.1682=0.3318\).

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) \(P(X=10)\approx 0.1682\) \(P(X<10)\approx 0.3318\)

- Divide the number of observed defects by the total sample size. - Distinguish a probability estimated from data from one derived from a specified model. - Use the binomial point-probability formula in part 2.

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 \(P(X=1)=\binom{20}{1}(0.02)(0.98)^{19}\approx0.2725.\) 3. This second calculation uses a theoretical, model-based probability because it begins with a specified probability model and parameter.

1. \(0.03\), or \(3\%\); experimental or empirical probability. 2. \(\binom{20}{1}(0.02)(0.98)^{19}\approx0.2725\); theoretical, model-based probability.

- Determine a different one-spin success probability for each event. - Use complements when they shorten the cumulative calculation. - Independence means past spins do not change the probability of the next spin.

1. For part a)(1), the multiples of \(3\) are \(3,6,9,12\), so \(p=1/3\). Thus, \(P(X\ge5)\approx0.9538.\) 2. For part a)(2), \(p=1/12\), so \(P(Y>2)\approx0.3467.\) 3. For part a)(3), the prime labels are \(2,3,5,7,11\), so \(p=5/12\), and \(P(Z\le10)\approx0.5180.\) 4. The spins are independent, so previous results do not change the next-spin probability. Therefore, \(P(\text{12 on spin 26})=\frac1{12}.\)

a) (1) \(0.9538\); (2) \(0.3467\); (3) \(0.5180\). b) \(\frac1{12}\).

- 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.

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 \(74\).

At least \(74\) components must be selected.

- Use the binomial formulas for center and spread. - Translate the one-standard-deviation interval into attainable integer counts. - Use cumulative probabilities for the inclusive interval.

1. Since \(X\sim\operatorname{Bin}(100,0.60)\), \(\mu=60,\qquad \sigma=\sqrt{100(0.60)(0.40)}=\sqrt{24}=2\sqrt6\approx4.90.\) 2. The interval \([\mu-\sigma,\mu+\sigma]\) is approximately \([55.10,64.90]\), so the included integer values are \(56\) through \(64\). 3. \(P(56\le X\le64)\approx0.6416.\)

1. \(\mu=60\) and \(\sigma=2\sqrt6\approx4.90\). 2. \(0.6416\).

- Start from the binomial model and find its mean and standard deviation. - The interval endpoints need not be integers, but the random variable can take only integer counts. - After identifying the included counts, evaluate the corresponding binomial interval probability.

1. Since \(X\sim\operatorname{Bin}(80,0.2)\), its mean and standard deviation are \(\mu=80(0.2)=16\) and \(\sigma=\sqrt{80(0.2)(0.8)}=\sqrt{12.8}\approx3.578\). 2. Thus, \(2\sigma\approx7.155\), so the continuous interval is approximately \([8.845,23.155]\). 3. Because \(X\) is an integer-valued count, the values within the interval are \(9,10,\ldots,23\). 4. Therefore, \(P(|X-\mu|\le2\sigma)=P(9\le X\le23)=P(X\le23)-P(X\le8)\approx0.9652\).

a) Approximately \([8.845,23.155]\), containing the integer counts \(9\) through \(23\). b) \(P(9\le X\le23)\approx0.9652\).

- Translate nondefective counts into defective counts for \(E_3\). - “At most two” is a cumulative probability. - Use the complement of no defects for the inverse sample-size condition.

1. \(P(E_1)=P(X_{20}=0)=(0.92)^{20}\approx0.1887.\) 2. \(P(E_2)=P(X_{20}\le2)\approx0.7879.\) 3. Exactly \(17\) or \(18\) nondefective chips means exactly \(3\) or \(2\) defective chips, so \(P(E_3)=P(X_{20}=2)+P(X_{20}=3)\approx0.4125.\) 4. For part b), require \(1-(0.92)^n\ge0.99.\) Hence, \(n\ge\frac{\ln(0.01)}{\ln(0.92)}\approx55.230,\) so the minimum integer is \(56\).

a) \(P(E_1)\approx0.1887\), \(P(E_2)\approx0.7879\), \(P(E_3)\approx0.4125\). b) \(56\).

- Decide whether each event specifies an order or only a total number of successes. - Break “at least” into the qualifying exact counts. - Use no binomial coefficient for a fully specified sequence.

1. Let \(X\sim\operatorname{Bin}(12,0.85)\). 2. \(P(X=9)=\binom{12}{9}(0.85)^9(0.15)^3\approx0.1720.\) 3. \(P(X\ge11)=P(X=11)+P(X=12)\approx0.4435.\) 4. The third event specifies one particular ordered sequence, so \(P=(0.15)^3(0.85)^9\approx0.000782.\)

(1) \(0.1720\). (2) \(0.4435\). (3) \((0.15)^3(0.85)^9\approx0.000782\).

- Find the expected number of positive responses. - Translate “differs by at most \(4\)” into an inclusive integer interval. - Evaluate the interval with cumulative binomial probabilities.

1. Since \(X\sim\operatorname{Bin}(50,0.40)\), \(\mu=np=20.\) 2. A difference of at most \(4\) means \(16\le X\le24.\) 3. Therefore, \(P(16\le X\le24)\approx0.8067.\)

\(0.8067\).

- 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.

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.

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.

- 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.

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 \(P(X=3)=\binom43(0.8)^3(0.2)=0.4096\).

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) \(P(X=3)=0.4096\)

- 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.

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 \(13.9108\%\).

\(p\ge 1-\sqrt[20]{0.05}\). The threshold is approximately \(0.139108\), or \(13.9108\%\).

- 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?

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.

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.

- Use the binomial formulas for the mean and standard deviation. - Find the continuous one-standard-deviation interval before identifying the integer counts it contains. - Sum the binomial probabilities over those attainable counts.

1. Since \(X\sim\operatorname{Bin}(400,0.05)\), \(\mu=np=20.\) 2. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{19}\approx4.36.\) 3. The continuous interval is \([20-\sqrt{19},20+\sqrt{19}]\approx[15.64,24.36].\) Therefore, the attainable integer values in the interval are \(16\) through \(24\). 4. Hence, \(P(\mu-\sigma\le X\le\mu+\sigma) =P(16\le X\le24) \approx0.6990.\)

a) \(\mu=20\), \(\sigma=\sqrt{19}\approx4.36\). b) \(P(16\le X\le24)\approx0.6990\).

- Use the binomial mean and standard-deviation formulas. - Translate the one-standard-deviation interval into attainable integer counts. - The requested event is outside that interval.

1. Since \(X\sim\operatorname{Bin}(150,0.64)\), \(\mu=150(0.64)=96.\) 2. \(\sigma=\sqrt{150(0.64)(0.36)}=\sqrt{34.56}=\frac{12\sqrt6}{5}\approx5.88.\) 3. Values within one standard deviation satisfy approximately \(90.12\le X\le101.88\), so the included integers are \(91\) through \(101\). 4. Therefore, the event more than one standard deviation from the mean is \(X\le90\) or \(X\ge102\), with probability \(1-P(91\le X\le101)\approx0.3495.\)

a) \(\mu=96\) and \(\sigma=\frac{12\sqrt6}{5}\approx5.88\). b) \(0.3495\).

- Use \(p=1/4\) for a correct guess. - “At most \(1\)” is a cumulative probability. - Use a complement to obtain \(P(X>2)\), then compare it directly with \(P(X=2)\).

1. Let \(X\sim\operatorname{Bin}(8,0.25)\). 2. \(P(X=2)\approx0.3115\). 3. \(P(X\le1)\approx0.3671\). 4. \(P(X>2)=1-P(X\le2)\approx0.3215\). 5. Since \(0.3215>0.3115\), more than two correct answers is slightly more likely.

a) \(0.3115\). b) \(0.3671\). c) More than \(2\) correct answers is more likely: \(0.3215\) versus \(0.3115\).

- Model the defective count with one binomial random variable. - A complement is efficient for “at least \(2\).” - In part c), calculate the two event probabilities before comparing them.

1. Let \(X\sim\operatorname{Bin}(12,0.10)\). 2. For part a), \(P(X=0)=(0.90)^{12}\approx0.2824\). 3. For part b), \(P(X\ge2)=1-[P(X=0)+P(X=1)]\approx0.3410\). 4. For part c), \(P(X=1)=12(0.10)(0.90)^{11}\approx0.3766\). 5. Also, \(P(X>2)=1-P(X\le2)\approx0.1109\). Since \(0.3766>0.1109\), exactly one defective figurine is more likely.

a) \(0.2824\). b) \(0.3410\). c) Exactly \(1\) defective figurine is more likely: \(0.3766\) versus \(0.1109\).

- 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.

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, \(P(X\ge 4)=0.234375+0.09375+0.015625=0.34375\).

(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 \(34.375\%\).

- Use the same binomial structure with two different success probabilities. - “At least \(3\)” includes \(3\) and \(4\). - Compare the probabilities after changing only \(p\).

1. For \(p=0.70\), \(P(X=2)=\binom42(0.70)^2(0.30)^2=0.2646,\) and \(P(X\ge3)=\binom43(0.70)^3(0.30)+(0.70)^4=0.6517.\) 2. For \(p=0.75\), \(P(X=2)=\binom42(0.75)^2(0.25)^2=0.2109375\approx0.2109,\) and \(P(X\ge3)=\binom43(0.75)^3(0.25)+(0.75)^4=0.73828125\approx0.7383.\) 3. Raising the success probability shifts probability toward larger success counts: exactly two makes becomes less likely, while at least three makes becomes more likely.

a) \(P(X=2)=0.2646\); \(P(X\ge3)=0.6517\). b) \(P(X=2)\approx0.2109\); \(P(X\ge3)\approx0.7383\). Exactly two makes becomes less likely, while at least three makes becomes more likely.

- Translate each verbal condition into a condition on the number germinating. - “More than \(10\)” means \(11\) or \(12\). - At most two failures is equivalent to at least ten germinations.

1. Let \(X\sim\operatorname{Bin}(12,0.80)\), where \(X\) is the number of seeds that germinate. 2. \(P(X=10)\approx0.2835.\) 3. \(P(X>10)=P(X=11)+P(X=12)\approx0.2749.\) 4. At most \(2\) seeds fail to germinate means at least \(10\) germinate, so \(P(X\ge10)\approx0.5583.\)

(1) \(0.2835\). (2) \(0.2749\). (3) \(0.5583\).

- Identify the two exact success counts included in the event. - The events cannot occur together, so their probabilities can be added. - Use the binomial point-probability formula for each count.

1. Let \(X\sim\operatorname{Bin}(10,0.25)\). 2. The two events are mutually exclusive, so \(P(X=3\text{ or }X=4)=P(X=3)+P(X=4).\) 3. \(P(X=3)=\binom{10}{3}(0.25)^3(0.75)^7,\) and \(P(X=4)=\binom{10}{4}(0.25)^4(0.75)^6.\) 4. Therefore, \(P(X=3\text{ or }X=4)\approx0.3963.\)

\(0.3963\).

- First derive the success probability from the two-dice sample space. - Translate “fewer than \(2\)” into the qualifying binomial counts. - Keep the exact fractions until the final rounding.

1. A sum of \(11\) occurs with \((5,6)\) or \((6,5)\), so the one-trial success probability is \(p=\frac{2}{36}=\frac1{18}.\) 2. Let \(X\sim\operatorname{Bin}(8,\frac1{18})\). 3. Fewer than \(2\) occurrences means \(X=0\) or \(X=1\). 4. \(P(X<2)=\left(\frac{17}{18}\right)^8+8\left(\frac1{18}\right)\left(\frac{17}{18}\right)^7\approx0.9309.\)

\(0.9309\).

- Use the complement of \(0\), \(1\), or \(2\) defects. - Evaluate the probability under the claimed defect rate. - Compare the result with the supplied \(5\%\) decision rule.

1. Under the claim, let \(X\sim\operatorname{Bin}(100,0.01)\). 2. Use the complement: \(P(X\ge3)=1-\sum_{k=0}^{2}\binom{100}{k}(0.01)^k(0.99)^{100-k}.\) 3. Thus, \(P(X\ge3)\approx0.0794.\) 4. Since \(0.0794>0.05\), the claim is not rejected under the stated rule.

\(P(X\ge3)\approx0.0794\). Because this exceeds \(0.05\), the claim is not rejected under the stated rule.

- Use the binomial mode rule involving \((n+1)p\). - The two ranges in part b) do not overlap. - Evaluate the lower tail and upper tail separately, then add them.

1. \(X\sim\operatorname{Bin}(20,0.60)\). 2. Since \((n+1)p=21(0.60)=12.6\) is not an integer, the unique mode is \(\lfloor12.6\rfloor=12\). 3. The events \(X\le8\) and \(X\ge16\) are disjoint, so their probabilities add. 4. Therefore, \(P(X\le8\text{ or }X\ge16)=P(X\le8)+P(X\ge16)\approx0.1075\).

a) \(12\). b) \(0.1075\).

- Use the binomial mode rule involving \((n+1)p\). - The two tail events are disjoint. - Use a complement for the upper tail and add the two tail probabilities.

1. \(X\sim\operatorname{Bin}(20,0.20)\). 2. Since \((n+1)p=21(0.20)=4.2\) is not an integer, the unique mode is \(\lfloor4.2\rfloor=4\). 3. The events in part b) are disjoint. \(P(X\le1)=(0.8)^{20}+\binom{20}{1}(0.2)(0.8)^{19}\approx0.0692\). 4. Also, \(P(X\ge8)=1-P(X\le7)\approx0.0321\). 5. Therefore, \(P(X\le1\text{ or }X\ge8)\approx0.1013\).

a) \(4\). b) \(0.1013\).

- Under fairness, the one-spin probability of \(3\) is \(1/3\). - “At most \(4\)” is a lower-tail binomial probability. - Compare the calculated tail probability with the supplied \(5\%\) criterion.

1. Under a fair spinner, let \(X\sim\operatorname{Bin}(30,\frac13)\). 2. \(P(X\le4)=\sum_{k=0}^{4}\binom{30}{k}\left(\frac13\right)^k\left(\frac23\right)^{30-k}\approx0.0122.\) 3. Since \(0.0122<0.05\), the result is significantly rare under the stated rule and supports the suspicion that \(3\) occurs less often than expected.

1. \(0.0122\). 2. Yes. The observation supports the suspicion under the stated criterion because \(0.0122<0.05\).

- 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.

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, \(P(X=10)>P(X=11)\).

(1) False (2) False (3) True (4) True

- 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 \(0\).

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.

(1) True (2) True (3) False (4) True

- 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.

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.

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 \(0.51>0.42\).

- 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.

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 \(12\): \(P(X\ge 10)=\frac{66+12+1}{4096}=\frac{79}{4096}\approx 0.0193\).

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 \(1.93\%\)

- Count ordered first-and-second-place predictions. - Use that one-prediction probability as the success probability for the group of sixty. - Apply the binomial formula for exactly one correct prediction.

1. There are \(10\cdot9=90\) possible ordered pairs for first and second place. 2. Exactly one ordered pair is correct, so \(p=\frac1{90}.\) 3. Let \(X\sim\operatorname{Bin}(60,\frac1{90})\). 4. \(P(X=1)=\binom{60}{1}\left(\frac1{90}\right)\left(\frac{89}{90}\right)^{59}\approx0.3448.\)

a) \(p=\frac1{90}\). b) \(0.3448\).

- Use the binomial formula for one exact value. - “At most one” includes zero and one. - Interpret the binomial coefficient as a count of possible success-position choices.

1. \(X\sim\operatorname{Bin}(20,0.05)\). 2. \(P(X=2)=\binom{20}{2}(0.05)^2(0.95)^{18}\approx0.1887.\) 3. \(P(X\le1)=(0.95)^{20}+20(0.05)(0.95)^{19}\approx0.7358.\) 4. \(\binom{20}{2}=190.\) It counts the possible pairs of sample positions occupied by the two defective components.

a) \(0.1887\). b) \(0.7358\). c) \(190\); it counts the possible pairs of defect positions among the twenty sampled components.

- Use the binomial point formula in part a). - In part b), focus on choosing success positions rather than calculating a probability. - Use the complement of zero correct answers in part c).

1. \(X\sim\operatorname{Bin}(10,0.25)\). 2. \(P(X=3)=\binom{10}{3}(0.25)^3(0.75)^7\approx0.2503\). 3. To obtain exactly two correct answers, choose the two correct-answer positions among the ten questions. The number of patterns is \(\binom{10}{2}=45\). 4. \(P(X\ge1)=1-P(X=0)=1-(0.75)^{10}\approx0.9437\).

a) \(0.2503\). b) \(\binom{10}{2}=45\) patterns; the coefficient chooses the two positions containing correct answers. c) \(0.9437\).

- Write both point probabilities using the binomial formula. - Compare \(\binom{10}{2}\) with \(\binom{10}{8}\). - Use the fact that success and failure have the same probability when \(p=0.5\).

1. \(P(X=2)=\binom{10}{2}(0.5)^{10}=\frac{45}{1024}\approx0.0439\). 2. \(P(X=8)=\binom{10}{8}(0.5)^{10}=\frac{45}{1024}\approx0.0439\). 3. The identity \(\binom{10}{2}=\binom{10}{8}\), together with equal success and failure probabilities, makes the two binomial probability terms identical.

a) \(P(X=2)=P(X=8)=\frac{45}{1024}\approx0.0439\). b) The probabilities are equal because \(\binom{10}{2}=\binom{10}{8}\) and both probability products equal \((0.5)^{10}\).

- 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.

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 \(1-\left(\frac{3059}{3060}\right)^{1000}\approx 0.278809\).

The probability is approximately \(27.88\%\).

- Treat the named color as success and all other colors as failure. - The success probability is different for blue, yellow, and red. - Use a complement for more than two red results.

1. For blue, the one-spin success probability is \(10/20=0.5\), so \(P(\text{exactly }5\text{ blue})=\binom{15}{5}(0.5)^{15}\approx0.0916.\) 2. The probability of not landing on yellow is \(0.8\), so \(P(\text{no yellow})=(0.8)^{15}\approx0.0352.\) 3. For red, the one-spin success probability is \(6/20=0.3\). If \(X\sim\operatorname{Bin}(15,0.3)\), then \(P(X>2)=1-P(X\le2)\approx0.8732.\)

a) \(0.0916\). b) \(0.0352\). c) \(0.8732\).

- 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 \(X\).

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. \(P(X\ge3)=P(X=3)+P(X=4)=\frac{108}{256}+\frac{81}{256}=\frac{189}{256}\approx0.7383\).

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) \(P(X\ge3)=\frac{189}{256}\approx0.7383\)

- State the binomial model using the supplied independence and constant-probability assumptions. - List every possible defect count from \(0\) through \(5\). - Use the direct binomial expected-value formula.

1. \(Y\sim\operatorname{Bin}(5,0.10)\). 2. For \(k=0,1,\ldots,5\), \(P(Y=k)=\binom5k(0.10)^k(0.90)^{5-k}.\) 3. This gives probabilities \(0.59049\), \(0.32805\), \(0.0729\), \(0.0081\), \(0.00045\), and \(0.00001\), respectively. 4. \(E(Y)=np=5(0.10)=0.5.\)

a) <table> <tr><th>\(k\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th><th>\(5\)</th></tr> <tr><th>\(P(Y=k)\)</th><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) \(E(Y)=0.5\).

- Relate the number of losses to the number of wins in four spins. - Changing the focus from wins to losses swaps the success and failure probabilities. - Use the identity between complementary counts.

1. \(X\sim\operatorname{Bin}(4,\frac14)\), so for \(k=0,1,2,3,4\), \(P(X=k)=\frac{81}{256},\frac{108}{256},\frac{54}{256},\frac{12}{256},\frac{1}{256}.\) 2. \(Y\sim\operatorname{Bin}(4,\frac34)\), giving the same probabilities in reverse order: \(\frac{1}{256},\frac{12}{256},\frac{54}{256},\frac{108}{256},\frac{81}{256}.\) 3. Every outcome with \(k\) wins has exactly \(4-k\) losses, so \(P(X=k)=P(Y=4-k).\)

a) For \(k=0,1,2,3,4\): \(\frac{81}{256},\frac{108}{256},\frac{54}{256},\frac{12}{256},\frac{1}{256}\). b) For \(k=0,1,2,3,4\): \(\frac{1}{256},\frac{12}{256},\frac{54}{256},\frac{108}{256},\frac{81}{256}\). c) \(P(X=k)=P(Y=4-k)\); the distributions are reversals of each other.

- Express each event in terms of the number of failures. - Germinations plus failures always total ten. - Compare the two event descriptions before doing any arithmetic.

1. \(P(X\ge9)=\binom{10}{9}(0.9)^9(0.1)+(0.9)^{10}.\) 2. \(P(Y\le1)=(0.9)^{10}+10(0.1)(0.9)^9.\) 3. The two expressions are identical because the events are identical: at least \(9\) germinations means at most \(1\) failure.

a) \(\binom{10}{9}(0.9)^9(0.1)+(0.9)^{10}\). b) \((0.9)^{10}+10(0.1)(0.9)^9\). c) The events are the same event described in complementary counts.

- 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.

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 \(p=0.90\cdot 0.60=0.54\).

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 \(p=0.54\).

- Use complements for events near the endpoints of the support. - “At most \(9\)” is the complement of exactly \(10\). - Preserve exact expressions when the probability is extremely close to \(0\) or \(1\).

1. \(X\sim\operatorname{Bin}(10,\frac18)\). 2. \(P(X\ge1)=1-\left(\frac78\right)^{10}\approx0.7369.\) 3. \(P(X>2)=1-P(X\le2)\approx0.1195.\) 4. \(P(X\le9)=1-P(X=10)=1-\left(\frac18\right)^{10}.\) 5. \(P(X=10)=\left(\frac18\right)^{10}.\)

(1) \(0.7369\). (2) \(0.1195\). (3) \(1-(1/8)^{10}\). (4) \((1/8)^{10}\).

- Use the binomial point-probability formula to build the table. - “At most \(2\)” includes \(0,1,2\). - Use the complement of zero defects for “at least \(1\).” - Keep unrounded probabilities through cumulative calculations.

1. \(X\sim\operatorname{Bin}(6,0.20)\), so \(P(X=k)=\binom6k(0.20)^k(0.80)^{6-k}\). 2. For \(k=0,1,\ldots,6\), the probabilities rounded to four decimals are \(0.2621,0.3932,0.2458,0.0819,0.0154,0.0015,0.0001\). 3. Using unrounded point probabilities, \(P(X\le2)=\sum_{k=0}^{2}\binom6k(0.20)^k(0.80)^{6-k}=0.90112\approx0.9011\). 4. Also, \(P(X\ge1)=1-P(X=0)=1-(0.80)^6=0.737856\approx0.7379\).

a) <table> <tr><th>\(k\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th><th>\(5\)</th><th>\(6\)</th></tr> <tr><th>\(P(X=k)\)</th><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. \(0.9011\). 2. \(0.7379\).

- With \(p=0.5\), all length-ten answer patterns are equally likely. - Use symmetry about \(5\) in part c). - Preserve exact fractions instead of introducing unnecessary rounding.

1. \(X\sim\operatorname{Bin}(10,0.5)\). 2. \(P(X=5)=\binom{10}{5}(0.5)^{10}=\frac{252}{1024}=\frac{63}{256}\). 3. \(P(X\ge8)=\frac{\binom{10}{8}+\binom{10}{9}+\binom{10}{10}}{2^{10}}=\frac{56}{1024}=\frac{7}{128}\). 4. Because the distribution is symmetric about \(5\), \(P(X<5)=P(X>5)\). Hence \(P(X>5)=\frac{1-P(X=5)}{2}=\frac{193}{512}\). 5. Since \(\frac{193}{512}>\frac{63}{256}\), more than \(5\) correct answers is more likely.

a) \(\frac{63}{256}\). b) \(\frac{7}{128}\). c) More than \(5\) correct answers is more likely because \(P(X>5)=\frac{193}{512}>\frac{63}{256}=P(X=5)\).

- 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 \((0.5)^n\).

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, \(P(X\le 3)=\frac{\binom{8}{0}+\binom{8}{1}+\binom{8}{2}+\binom{8}{3}}{2^8}=\frac{93}{256}\approx 0.363281\).

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 \(36.33\%\).

- 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.

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 \(1\).

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 \(1\).

- 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.

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. \(P(X\ge2)=P(X=2)+P(X=3)+P(X=4)=0.1808\). 4. The coefficient is \(\binom{4}{2}=6\).

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) \(\binom{4}{2}=6\)

- First find the success probability for one five-shot series. - Use that series-level probability in a second binomial model. - Preserve an exact expression for the very small zero-success probability.

1. For one series, \(p_S=\binom54(0.8)^4(0.2)+(0.8)^5=0.73728.\) 2. Let \(Y\sim\operatorname{Bin}(12,0.73728)\). Then \(P(Y=8)=\binom{12}{8}(0.73728)^8(0.26272)^4\approx0.2059.\) 3. \(P(Y=0)=(0.26272)^{12}\approx1.08\times10^{-7}.\)

1. \(0.73728\). 2. \(0.2059\). 3. \((0.26272)^{12}\approx1.08\times10^{-7}\).

- First find the probability that a ten-chip package contains at least one defect. - Treat each package as one Bernoulli trial in the second-stage model. - Keep the package-level probability unrounded in later calculations.

1. A package is accepted only if all ten chips are nondefective, so \(p_R=1-(0.95)^{10}\approx0.4013\). 2. Let \(Y\) be the number of rejected packages. Then \(Y\sim\operatorname{Bin}(20,p_R)\). 3. Using the unrounded value of \(p_R\), \(P(Y=5)=\binom{20}{5}p_R^5(1-p_R)^{15}\approx0.0735\). 4. \(P(Y\le1)=(1-p_R)^{20}+20p_R(1-p_R)^{19}\approx0.000505\).

a) \(p_R\approx0.4013\). b) \(0.0735\). c) \(0.000505\).

- Translate each condition on owners into a condition on nonowners. - Remember that \(X+Y=50\). - Track the inequality direction carefully when converting between \(X\) and \(Y\).

1. Since \(Y=50-X\), \(P(X=42)=P(Y=8)\approx0.1169.\) 2. \(P(X\ge38)=P(Y\le12)\approx0.8139.\) 3. \(P(X\le44)=P(Y\ge6)=1-P(Y\le5)\approx0.9520.\)

(1) \(0.1169\). (2) \(0.8139\). (3) \(0.9520\).

- Translate strict inequalities to inclusive integer endpoints. - Use a difference of cumulative probabilities for the interval. - Use a complement for the upper tail.

1. \(X\sim\operatorname{Bin}(100,0.75)\). 2. \(P(X<70)=P(X\le69)\approx0.1038.\) 3. \(P(72\le X\le80)=P(X\le80)-P(X\le71)\approx0.6929.\) 4. \(P(X>80)=1-P(X\le80)\approx0.0995.\)

(1) \(0.1038\). (2) \(0.6929\). (3) \(0.0995\).

- Use cumulative probabilities for the tail and interval events. - Translate the number with landlines into the complementary count without landlines. - Check whether part b) describes a new event or the same event in different words.

1. \(X\sim\operatorname{Bin}(100,0.12)\). 2. \(P(X\le10)\approx0.3337.\) 3. \(P(X>15)=1-P(X\le15)\approx0.1414.\) 4. \(P(8\le X\le14)=P(X\le14)-P(X\le7)\approx0.7079.\) 5. At least \(90\) households with a landline is equivalent to at most \(10\) without a landline, so the probability is again \(0.3337\).

a) (1) \(0.3337\); (2) \(0.1414\); (3) \(0.7079\). b) \(0.3337\); it is the same event as at most \(10\) households without a landline.

- 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.

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, \(\frac{P(E_1)}{P(E_3)}=\frac{27}{20}=1.35\).

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 \(P(E_1)\ne2P(E_3)\).

- Use the binomial probability formula for the exact-count events. - For “at least one,” use the complement of no wins. - An expected value is an average, not a guaranteed outcome.

1. The success probability is \(p=\frac14\). 2. \(P(A)=\binom41(\frac14)(\frac34)^3=\frac{27}{64}=0.421875\approx0.4219\). 3. \(P(B)=\binom82(\frac14)^2(\frac34)^6=\frac{5103}{16384}\approx0.3115\). The probabilities are not equal. 4. Expected value is a long-run average, not a guarantee for every group of four spins. 5. \(P(C)=1-(\frac34)^4=\frac{175}{256}=0.68359375\approx0.6836<1\), so \(C\) is not certain.

a) \(P(A)=\frac{27}{64}\approx0.4219\) and \(P(B)=\frac{5103}{16384}\approx0.3115\); they are not equal. b) The claim is false. \(P(C)=\frac{175}{256}\approx0.6836\), not \(1\).

- A complement contains exactly the outcomes not in the original event. - Track whether each boundary value is included. - Use cumulative probabilities after writing the complement correctly.

1. \(X\sim\operatorname{Bin}(25,0.45)\). 2. The complement of \(E_1\) is “at most \(10\) made shots,” so \(P(\overline{E_1})=P(X\le10)\approx0.3843.\) 3. The complement of \(E_2\) is “at least \(15\) made shots,” so \(P(\overline{E_2})=P(X\ge15)=1-P(X\le14)\approx0.0960.\)

(1) At most \(10\) made shots; \(0.3843\). (2) At least \(15\) made shots; \(0.0960\).

- Write each original event as an inequality before finding its complement. - The complement of a bounded interval consists of two separated ranges. - Event and complement probabilities add to \(1\).

1. \(X\sim\operatorname{Bin}(50,0.20)\). 2. The complement of \(E_1\) is \(X\le11\), meaning at most eleven people support the statement. Thus, \(P(E_1^c)\approx0.7107\). 3. Event \(E_2\) is \(6\le X\le12\). Its complement is \(X\le5\) or \(X\ge13\). 4. \(P(E_2)=P(X\le12)-P(X\le5)\), so \(P(E_2^c)=1-P(E_2)\approx0.2341\).

a) At most \(11\) people support the statement; \(P(E_1^c)\approx0.7107\). b) At most \(5\) or at least \(13\) people support the statement; \(P(E_2^c)\approx0.2341\).

- 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.

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.

The network must support at least \(38\) households simultaneously.

- Use arrival probability \(0.90\), not the no-show probability, as \(p\). - In part b), use the complement relationship from part a). - Use a point probability for exactly eighty arrivals and a CDF difference for the interval.

1. Each guest arrives with probability \(0.90\), so \(X\sim\operatorname{Bin}(85,0.90)\). 2. \(P(X\le80)\approx0.9357\). 3. The event \(X>80\) is the complement of \(X\le80\), so \(P(X>80)=1-P(X\le80)\approx0.0643\). 4. \(P(X=80)=\binom{85}{80}(0.90)^{80}(0.10)^5\approx0.0717\). 5. \(P(70\le X\le75)=P(X\le75)-P(X\le69)\approx0.3338\).

a) \(0.9357\). b) \(0.0643\); the two events are complements. c) \(0.0717\). d) \(0.3338\).

- Use cumulative probabilities for the tail and interval events. - Find \(np\) before translating the event in part d). - Use the binomial point formula once the mean count is identified.

1. \(X\sim\operatorname{Bin}(200,0.04)\). 2. \(P(X\le5)\approx0.1856\). 3. \(P(X>10)=1-P(X\le10)\approx0.1800\). 4. \(P(6\le X\le12)=P(X\le12)-P(X\le5)\approx0.7545\). 5. The mean is \(\mu=np=8\), so the final event is \(X=8\). Thus, \(P(X=8)=\binom{200}{8}(0.04)^8(0.96)^{192}\approx0.1425\).

a) \(0.1856\). b) \(0.1800\). c) \(0.7545\). d) \(0.1425\).

- State all four conditions for a binomial setting, including the fixed number of trials. - Define success in this context. - Use the binomial formula for an exact number of successes.

1. A binomial model requires a fixed number of trials, two relevant outcomes on each trial, independent trials, and a constant success probability. 2. Here, the fixed number is \(20\) packs. Each pack either contains a limited-edition card or does not. Because the production run is very large, the success probability is approximately constant at \(1/7\), and the selections can be treated as approximately independent. 3. Thus, \(n=20\) and \(p=\frac17\). 4. \(P(X=3)=\binom{20}{3}\left(\frac17\right)^3\left(\frac67\right)^{17}\approx0.2418.\)

a) Fixed \(n=20\), two relevant outcomes, approximately independent selections, and approximately constant success probability \(p=\frac17\). b) \(n=20\) and \(p=\frac17\). c) \(\binom{20}{3}(1/7)^3(6/7)^{17}\approx0.2418\).

- Use the binomial probability formula for each value of \(k\). - Organize the values and probabilities in a table. - The expected value is the weighted average of all possible values. - Multiply each value \(k\) by its probability and add.

1. Using \(P(X=k)=\binom{5}{k}(0.4)^k(0.6)^{5-k}\), the probabilities are: \(P(X=0)=0.07776\), \(P(X=1)=0.2592\), \(P(X=2)=0.3456\), \(P(X=3)=0.2304\), \(P(X=4)=0.0768\), and \(P(X=5)=0.01024\). 2. The expected value from the distribution is \(E(X)=0(0.07776)+1(0.2592)+2(0.3456)+3(0.2304)+4(0.0768)+5(0.01024)=2\).

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>\(P(X=k)\)</td><td>\(0.07776\)</td><td>\(0.2592\)</td><td>\(0.3456\)</td><td>\(0.2304\)</td><td>\(0.0768\)</td><td>\(0.01024\)</td></tr> </table> b) \(E(X)=2\)

- Use the expected-value formula for a binomial random variable. - Apply the point-probability formula at the mean and its neighboring values. - A mode must have probability at least as large as the probabilities at neighboring values.

1. The success probability is \(p=0.25\), so \(\mu=np=40(0.25)=10.\) 2. \(P(X=10)=\binom{40}{10}(0.25)^{10}(0.75)^{30}\approx0.1444.\) 3. \(P(X=9)=\binom{40}{9}(0.25)^9(0.75)^{31}\approx0.1397,\) and \(P(X=11)=\binom{40}{11}(0.25)^{11}(0.75)^{29}\approx0.1312.\) 4. Since \(0.1444\) exceeds both neighboring probabilities, \(10\) is the most likely value.

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 \(P(X=10)\).

- Find \(\mu\) and \(q\) before forming the stated interval. - Identify the integer contained in that interval. - Compare the candidate mode with its immediate neighbors using binomial point probabilities.

1. The expected value is \(\mu=np=15(0.70)=10.5,\) and \(q=1-0.70=0.30\). 2. The stated interval is \([\mu-q,\mu+p]=[10.2,11.2].\) Its only integer is \(k=11\). 3. \(P(X=10)\approx0.2061,\qquad P(X=11)\approx0.2186,\qquad P(X=12)\approx0.1700.\) 4. Since \(P(X=11)\) is greater than both neighboring probabilities, \(11\) is the mode.

a) \(\mu=10.5\) and \(q=0.30\). b) \([10.2,11.2]\); \(k=11\). c) \(P(X=10)\approx0.2061\), \(P(X=11)\approx0.2186\), and \(P(X=12)\approx0.1700\). Thus, \(11\) is the most likely value.

- Use the same binomial model for all one-day probabilities. - “More than \(3\)” begins at \(4\). - For the expected number of days, multiply the one-day probability by \(365\); an expected count need not be an integer.

1. \(X\sim\operatorname{Bin}(300,0.008)\). 2. \(P(X=0)=(0.992)^{300}\approx0.0898,\) \(P(X=1)=300(0.008)(0.992)^{299}\approx0.2174,\) and \(P(X=2)=\binom{300}{2}(0.008)^2(0.992)^{298}\approx0.2621.\) 3. \(P(X>3)=1-P(X\le3)\approx0.2208.\) 4. Let \(I_d\) indicate that day \(d\)'s sample has no defective chips. Then \(E(I_d)=P(X=0)\). 5. By linearity of expectation, the expected number of such days is \(365(0.992)^{300}\approx32.79.\)

a) \(P(X=0)\approx0.0898\), \(P(X=1)\approx0.2174\), and \(P(X=2)\approx0.2621\). b) \(P(X>3)\approx0.2208\). c) Approximately \(32.79\) days.

- Identify \(n\) and \(p\). - Use the binomial formula for zero and one success. - For at least two successes, subtract the zero- and one-success cases from \(1\).

1. Let \(X\sim\operatorname{Bin}(200,\frac{1}{200})\). 2. \(P(X=0)=\left(\frac{199}{200}\right)^{200}\approx0.3670.\) This is about \(0.0009\) less than \(1/e\approx0.3679\). 3. \(P(X=1)=200\left(\frac1{200}\right)\left(\frac{199}{200}\right)^{199} =\left(\frac{199}{200}\right)^{199} \approx0.3688.\) 4. \(P(X\ge2)=1-P(X=0)-P(X=1)\approx0.2642.\)

a) \(P(X=0)\approx0.3670\), about \(0.0009\) less than \(\frac1e\). b) \(P(X=1)\approx0.3688\). c) \(P(X\ge2)\approx0.2642\).

- First model the number of bubbles in one specified section. - For the expected number of empty sections, introduce one indicator for each section. - Add the indicator expectations rather than trying to model all empty-section events jointly.

1. For one specified section, each bubble lands there with probability \(1/80\), so its bubble count \(X\sim\operatorname{Bin}(60,1/80)\). 2. \(P(X>2)=1-[P(X=0)+P(X=1)+P(X=2)]\approx0.0395\). 3. A specified section is empty with probability \(P(X=0)=(79/80)^{60}\). 4. Let \(I_j\) indicate that section \(j\) is empty, and let \(T=I_1+\cdots+I_{80}\). Then \(E(T)=\sum_{j=1}^{80}E(I_j)=80(79/80)^{60}\approx37.61\). 5. Linearity of expectation holds whether or not the indicators \(I_j\) are independent, so independence between empty-section events is not needed.

a) \(0.0395\). b) \(80(79/80)^{60}\approx37.61\) empty sections. c) Linearity of expectation does not require independence.

- Convert one member's booth-use time into a probability at a randomly selected time. - Use a complement for “more than \(6\).” - For the capacity decision, compare consecutive cumulative probabilities around the target.

1. Each member is using a booth with probability \(p=15/60=0.25\), and the simultaneous-use indicators are assumed independent. Hence, \(X\sim\operatorname{Bin}(15,0.25)\). 2. \(P(X\le4)\approx0.6865\). 3. \(P(X>6)=1-P(X\le6)\approx0.0566\). 4. For the capacity requirement, seek the smallest integer \(k\) such that \(P(X\le k)\ge0.95\). 5. Since \(P(X\le6)\approx0.9434<0.95\) and \(P(X\le7)\approx0.9827\ge0.95\), the minimum is \(7\) booths.

a) \(0.6865\). b) \(0.0566\). c) \(7\) booths.

- Convert the six-minute usage interval into a probability at a randomly selected time. - Distinguish an exact-count event from an upper-tail event. - Use a complement for “more than \(4\).” - Verify the minimum capacity by comparing consecutive cumulative probabilities around \(0.99\).

1. Each employee is using a pallet jack with probability \(p=6/60=0.10\), and the simultaneous-use indicators are assumed independent. Thus, \(X\sim\operatorname{Bin}(20,0.10)\). 2. For part a), \(P(X=2)=\binom{20}{2}(0.10)^2(0.90)^{18}\approx0.2852\). 3. For part b), \(P(X>4)=1-P(X\le4)\approx0.0432\). 4. For part c), seek the smallest integer \(k\) such that \(P(X\le k)\ge0.99\). 5. Since \(P(X\le5)\approx0.9887<0.99\) and \(P(X\le6)\approx0.9976\ge0.99\), the minimum is \(6\).

a) \(0.2852\). b) \(0.0432\). c) \(6\) pallet jacks.

- Convert the exact 15-minute usage duration into a random-instant occupancy probability. - Translate a shortage into an inequality involving simultaneous users and licenses. - Compare consecutive cumulative probabilities to prove the minimum.

1. Each employee is using the software with probability \(p=15/60=0.25\), and the simultaneous-use indicators are assumed independent. Thus, \(X\sim\operatorname{Bin}(50,0.25)\). 2. If \(k\) licenses are available, a shortage occurs when \(X>k\). The requirement is \(P(X>k)\le0.02\), equivalently \(P(X\le k)\ge0.98\). 3. \(P(X\le18)\approx0.9713<0.98\), while \(P(X\le19)\approx0.9861\ge0.98\). 4. Therefore, the minimum number of licenses is \(19\).

\(19\) licenses.

- Convert the ten minutes of use into a random-instant occupancy probability. - Use a complement for the current ten-machine shortage probability. - For the minimum capacity, compare consecutive cumulative probabilities around the target.

1. Each participant wants a rowing machine with probability \(p=10/60=1/6\), and the simultaneous-demand indicators are assumed independent. Thus, \(X\sim\operatorname{Bin}(40,1/6)\). 2. \(P(X>10)=1-P(X\le10)\approx0.0584\). 3. For part b), seek the smallest integer \(k\) such that \(P(X>k)\le0.05\), equivalently \(P(X\le k)\ge0.95\). 4. Since \(P(X\le10)\approx0.9416<0.95\) and \(P(X\le11)\approx0.9739\ge0.95\), the minimum is \(11\) rowing machines.

a) \(0.0584\). b) \(11\) rowing machines.

- Use the supplied probability that one patient is present at the selected time. - Overcrowding means more than ten simultaneous patients. - Compare the calculated probability with the stated one-percent criterion.

1. Each patient is in the waiting room with probability \(p=15/300=0.05\), and the simultaneous waiting-room indicators are assumed independent. Therefore, \(X\sim\operatorname{Bin}(80,0.05)\). 2. Overcrowding occurs when \(X>10\). 3. \(P(X>10)=1-P(X\le10)\approx0.0021\). 4. Since \(0.0021=0.21\%<1\%\), ten seats meet the stated requirement.

The overcrowding probability is approximately \(0.0021\), or \(0.21\%\). Therefore, \(10\) seats are sufficient under the stated criterion.

- Find the binomial mean and standard deviation first. - Check the large-count condition before using the normal approximation. - Multiply the standard deviation by the central-\(95\%\) critical value.

1. Let \(X\sim\operatorname{Bin}(600,0.15).\) The mean is \(\mu=np=90.\) 2. The standard deviation is \(\sigma=\sqrt{np(1-p)} =\sqrt{600(0.15)(0.85)} =\sqrt{76.5} \approx8.746.\) The large-count condition is satisfied. 3. A central \(95\%\) normal interval uses \(z^*\approx1.96.\) Therefore, \(r=z^*\sigma\approx1.96(8.746)\approx17.14.\)

\(r\approx17.14\), giving the approximate continuous interval \([90-17.14,90+17.14]=[72.86{,}107.14].\)

- Use the binomial mean and standard-deviation formulas. - Remember that the binomial random variable takes integer values. - Check the large-count condition before invoking a normal approximation. - Match a two-standard-deviation interval with the Empirical Rule.

1. Since \(X\sim\operatorname{Bin}(400,0.3),\) \(\mu=np=120\) and \(\sigma=\sqrt{np(1-p)}=\sqrt{84}\approx9.17.\) 2. The radius is \(2\sigma=2\sqrt{84}\approx18.33,\) so the continuous interval is approximately \([101.67{,}138.33].\) The attainable integer values are \(102{,}103,\ldots,138\). 3. The large-count condition is satisfied, and the Empirical Rule gives approximately \(95.4\%\) within two standard deviations for the normal approximation.

1. \(\mu=120\), \(\sigma=\sqrt{84}\approx9.17\). 2. Approximately \([101.67{,}138.33]\); integer values \(102\) through \(138\). 3. Approximately \(95.4\%\).

- Find the continuous one-standard-deviation interval and identify its integer values. - Extend the integer interval by \(0.5\) at each end for the continuity correction. - Compare the final percentage with both proposed benchmarks.

1. The mean is \(\mu=np=40,\) and the standard deviation is \(\sigma=\sqrt{100(0.4)(0.6)}=\sqrt{24}\approx4.899.\) 2. The one-standard-deviation interval is approximately \([35.101,44.899],\) so its integer values are \(36\) through \(44\). 3. With continuity correction, approximate \(P(36\le X\le44)\) by the normal probability over \(35.5\le Y\le44.5,\) where \(Y\) has mean \(40\) and standard deviation \(\sqrt{24}\). 4. The standard scores are approximately \(-0.919\) and \(0.919\), giving \(P(36\le X\le44)\approx0.6417=64.2\%.\) 5. This is \(3.8\) percentage points from \(68\%\) and \(5.8\) percentage points from \(70\%\), so it is closer to \(68\%\).

The integer values are \(36\) through \(44\). The normal approximation gives about \(64.2\%\), which is closer to \(68\%\).

- Use the binomial formulas for mean and standard deviation. - Check both expected successes and expected failures. - Identify the integer values inside the continuous \(1.96\sigma\) interval. - Compare the exact binomial probability with the normal benchmark.

1. Since \(X\sim\operatorname{Bin}(400,0.15),\) \(\mu=60,\qquad \sigma=\sqrt{51}\approx7.14.\) The large-count condition is satisfied because \(np=60\) and \(n(1-p)=340\). 2. The radius is \(1.96\sigma\approx13.997,\) so the integer values inside the interval are \(47\) through \(73\). 3. Using the binomial distribution, \(P(47\le X\le73)\approx0.9417.\) 4. The exact binomial probability, about \(94.17\%\), is close to the central normal value of \(95\%\), but not identical because the binomial distribution is discrete.

1. \(\mu=60\), \(\sigma=\sqrt{51}\approx7.14\); the normal approximation is appropriate. 2. \(P(47\le X\le73)\approx0.9417\). 3. \(94.17\%\) is close to \(95\%\).

- Find the binomial mean and standard deviation. - Check the large-count condition before using a normal approximation. - Use the appropriate central-normal critical value. - Distinguish the continuous approximation interval from the attainable integer counts.

1. In part a, \(\mu=np=120\) and \(\sigma=\sqrt{np(1-p)}=\sqrt{96}\approx9.798\). The large-count condition is satisfied. 2. For a central \(90\%\) interval, \(z^*\approx1.645\). Thus, \(120\pm1.645(9.798)\approx[103.88,136.12]\), containing integer outcomes \(104\) through \(136\). 3. In part b, \(\mu=280\) and \(\sigma=\sqrt{84}\approx9.165\). The large-count condition is satisfied. 4. For a central \(95\%\) interval, \(z^*\approx1.96\). Thus, \(280\pm1.96(9.165)\approx[262.04,297.96]\), containing integer outcomes \(263\) through \(297\).

a) Continuous approximation \([103.88,136.12]\); integer outcomes \(104\) through \(136\). b) Continuous approximation \([262.04,297.96]\); integer outcomes \(263\) through \(297\).

- Find the binomial mean and standard deviation. - Check the large-count condition before using a normal approximation. - Use the critical value for a central \(95\%\) normal interval. - Relate greater captured probability to a larger critical value.

1. The binomial mean is \(\mu=np=200\), and the standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{168}\approx12.961\). The large-count condition is satisfied. 2. For a central \(95\%\) interval, \(z^*\approx1.96\). 3. Therefore, \(200\pm1.96(12.961)\approx[174.60,225.40]\), containing integer outcomes \(175\) through \(225\). 4. A central \(99\%\) interval must include more probability, so it uses a larger critical value and is wider.

a) Continuous approximation \([174.60,225.40]\); integer outcomes \(175\) through \(225\). b) The \(99\%\) interval is wider.

- Find the binomial mean and standard deviation. - Check the large-count condition before using the normal approximation. - Use the critical value for a central \(95\%\) normal interval. - Distinguish the continuous approximation from the attainable integer counts.

1. Let \(X\sim\operatorname{Bin}(1000,0.04)\). The mean is \(\mu=np=40\). 2. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{38.4}\approx6.197\). Since \(np=40\) and \(n(1-p)=960\), the normal approximation is appropriate. 3. A central \(95\%\) interval uses \(z^*\approx1.96\). 4. The radius is \(1.96(6.197\ldots)\approx12.15\), so the continuous interval is approximately \([27.85,52.15]\). 5. The integer counts inside this interval are \(28\) through \(52\).

Continuous approximation: \([27.85,52.15]\). Integer counts: \(28\) through \(52\).

- Use the complement of finding no defective bulbs. - In part b), the unknown sample size appears in an exponent. - Solve the exponential inequality with logarithms and verify the neighboring integers.

1. The probability that one bulb does not have the defect is \(0.96\). 2. \(P(\text{at least one defect in 25})=1-(0.96)^{25}\approx0.6396\). 3. For part b), require \(1-(0.96)^n>0.90\), equivalently \((0.96)^n<0.10\). 4. Taking logarithms gives \(n>\frac{\ln(0.10)}{\ln(0.96)}\approx56.406\). 5. Therefore, the minimum integer is \(57\). Indeed, \(n=56\) gives a probability below \(0.90\), while \(n=57\) gives a probability above \(0.90\).

a) \(0.6396\). b) \(57\) bulbs.

- Ask how much a sample of ten changes a very large population. - “At most one” includes zero and one. - Treat the binomial model here as an approximation, not an exact without-replacement model.

1. Removing only \(10\) chips from a very large shipment changes the remaining defect proportion negligibly. The selections can therefore be treated as approximately independent with approximately constant defect probability \(p=0.04\). 2. Use \(X\approx\operatorname{Bin}(10,0.04)\). 3. \(P(X=1)=\binom{10}{1}(0.04)(0.96)^9\approx0.2770.\) 4. \(P(X\le1)=(0.96)^{10}+10(0.04)(0.96)^9\approx0.9418.\)

a) The sample is negligible compared with the shipment, so the defect probability is approximately constant and the selections are approximately independent. b) \(0.2770\). c) \(0.9418\).

- 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.

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.

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 \((p+(1-p))^n=1\).

- Compare each summand with the binomial probability formula. - Use \((a+b)^n\) with \(a=0.25\) and \(b=0.75\). - If the zero-success term is omitted, subtract exactly that term from the complete probability sum.

1. Each summand has the form \(\binom{n}{k}p^k(1-p)^{n-k},\) so \(n=12\) and \(p=0.25\). 2. Because the sum includes every possible number of successes, \(T\) is the total probability of the complete binomial distribution. 3. By the binomial theorem, \(T=(0.25+0.75)^{12}=1.\) 4. Starting at \(k=1\) omits only the \(k=0\) term, \((0.75)^{12}\). 5. Therefore, the new value is \(1-(0.75)^{12}.\)

a) \(n=12\), \(p=0.25\), and \(T\) is the sum of all probabilities in the binomial distribution. b) \(T=1\). c) \(1-(0.75)^{12}\).

- Count failures instead of successes. - Relate the two counts by their fixed total of \(20\). - Convert the resulting upper-tail event to a complement if useful.

1. Let \(Y\) be the number of seeds that do not germinate. Then \(Y\sim\operatorname{Bin}(20,0.20).\) 2. The event \(X\le14\) is equivalent to \(Y\ge6\). 3. Therefore, \(P(X\le14)=P(Y\ge6)=1-P(Y\le5)\approx0.1958.\)

\(P(X\le14)\approx0.1958\).

- Translate the interval for correct answers into one for incorrect answers. - Remember that increasing \(X\) decreases \(Y=25-X\). - Use two cumulative probabilities after finding the transformed integer interval.

1. A cumulative distribution lets an inclusive interval probability be found by subtracting two cumulative probabilities instead of adding many individual point probabilities. 2. Since \(Y=25-X\), the condition \(15\le X\le20\) is equivalent to \(5\le Y\le10.\) 3. Therefore, \(P(15\le X\le20)=P(5\le Y\le10) =P(Y\le10)-P(Y\le4) \approx0.7566.\)

a) An inclusive interval can be found as \(P(X\le b)-P(X\le a-1)\), avoiding a long sum of point probabilities. b) \(P(15\le X\le20)\approx0.7566\).

- Interpret the binomial mean as an expected count. - Use the binomial standard-deviation formula. - Build an interval symmetric about the mean. - Check the large-count condition before applying the Empirical Rule to the normal approximation.

1. Let \(X\sim\operatorname{Bin}(2400,0.05)\). The mean is \(\mu=np=120\). 2. The standard deviation is \(\sigma=\sqrt{2400(0.05)(0.95)}=\sqrt{114}\approx10.68\). 3. The radius is \(3\sigma\approx32.03\), so the continuous interval is approximately \([87.97,152.03]\), containing integer counts \(88\) through \(152\). 4. The large-count condition is satisfied, so a normal approximation is appropriate. 5. By the Empirical Rule, a three-standard-deviation interval contains approximately \(99.7\%\) of the normal approximation.

a) \(120\) packages. b) Continuous interval approximately \([87.97,152.03]\); integer counts \(88\) through \(152\); approximate coverage \(99.7\%\).

- Replacement makes the one-draw probability stay at \(1/25\). - Use the binomial point formula for zero, one, and two draws of the specified ball. - Use the complement of at most two for part b), and \(np\) for part c).

1. Let \(X\) be the number of times the specified ball is drawn. Replacement makes the draws independent with constant probability \(p=1/25=0.04\), so \(X\sim\operatorname{Bin}(25,0.04)\). 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. \(P(X>2)=1-P(X\le2)\approx0.0765\). 6. \(E(X)=np=25(0.04)=1\).

a) \(0.3604,\ 0.3754,\ 0.1877\), respectively. b) \(0.0765\). c) \(1\).

- Identify the binomial parameters. - Use the formulas for the binomial mean and standard deviation. - Check the large-count condition before using a normal approximation. - Match central \(95\%\) and \(99\%\) intervals with their normal critical values.

1. Let \(X\sim\operatorname{Bin}(6000,0.015)\). The mean is \(\mu=np=90\). 2. The standard deviation is \(\sigma=\sqrt{6000(0.015)(0.985)}=\sqrt{88.65}\approx9.42\). The large-count condition is satisfied. 3. For a central \(95\%\) interval, use \(z^*\approx1.96\): \(90\pm1.96(9.415\ldots)\approx[71.55,108.45]\), containing integer counts \(72\) through \(108\). 4. For a central \(99\%\) interval, use \(z^*\approx2.576\): \(90\pm2.576(9.415\ldots)\approx[65.75,114.25]\), containing integer counts \(66\) through \(114\).

a) \(\mu=90\), \(\sigma\approx9.42\). b) \(95\%\): continuous interval approximately \([71.55,108.45]\), integer counts \(72\) through \(108\). \(99\%\): continuous interval approximately \([65.75,114.25]\), integer counts \(66\) through \(114\).

- Identify the binomial model. - Find its mean and standard deviation. - Check the large-count condition. - Use the critical value for a central \(95\%\) normal interval.

1. Let \(X\sim\operatorname{Bin}(2500,0.04)\). The mean is \(\mu=np=100\). 2. The standard deviation is \(\sigma=\sqrt{2500(0.04)(0.96)}=\sqrt{96}\approx9.798\). The large-count condition is satisfied. 3. A central \(95\%\) interval uses \(z^*\approx1.96\). Thus, \(100\pm1.96(9.798\ldots)\approx[80.80,119.20]\). 4. The integer counts inside the interval are \(81\) through \(119\).

Continuous approximation: \([80.80,119.20]\). Integer counts: \(81\) through \(119\).

- Find the binomial mean and standard deviation. - Check the large-count condition. - Use the critical values for central \(90\%\) and \(99\%\) normal intervals. - Distinguish continuous endpoints from attainable integer counts.

1. Let \(X\sim\operatorname{Bin}(600,0.60)\). The mean is \(\mu=360\), and the standard deviation is \(\sigma=\sqrt{600(0.60)(0.40)}=12\). The large-count condition is satisfied. 2. For a central \(90\%\) interval, use \(z^*\approx1.645\): \(360\pm1.645(12)\approx[340.26,379.74]\), containing integer counts \(341\) through \(379\). 3. For a central \(99\%\) interval, use \(z^*\approx2.576\): \(360\pm2.576(12)\approx[329.09,390.91]\), containing integer counts \(330\) through \(390\).

Central \(90\%\): continuous interval approximately \([340.26,379.74]\), integer counts \(341\) through \(379\). Central \(99\%\): continuous interval approximately \([329.09,390.91]\), integer counts \(330\) through \(390\).

- Identify the binomial parameters first. - Check the large-count condition before using a normal approximation. - Distinguish a central \(95\%\) critical-value interval from a two-standard-deviation Empirical-Rule interval. - A deviation from the mean can occur in either direction.

1. Let \(X\sim\operatorname{Bin}(1600,0.20)\). Then \(\mu=np=320\) and \(\sigma=\sqrt{1600(0.20)(0.80)}=16\). The large-count condition is satisfied. 2. A central \(95\%\) interval uses \(z^*\approx1.96\), giving \(320\pm1.96(16)=[288.64,351.36]\). The integer counts inside are \(289\) through \(351\). 3. A deviation of \(32\) is \(2\sigma\). The Empirical Rule places approximately \(95.4\%\) within two standard deviations, so the probability of a larger deviation is approximately \(4.6\%\).

1. \(\mu=320\), \(\sigma=16\). 2. Continuous interval \([288.64,351.36]\); integer counts \(289\) through \(351\). 3. Approximately \(4.6\%\).

- Find the binomial mean and standard deviation first. - Match approximately \(32\%\) outside with the complement of the central one-standard-deviation region. - For a central \(95\%\) approximation, use \(1.96\) standard deviations. - Compare the given deviation in part c with \(\sigma\).

1. \(X\sim\operatorname{Bin}(900,0.10)\), so \(\mu=np=90\) and \(\sigma=\sqrt{900(0.10)(0.90)}=9\). The large-count condition is satisfied. 2. The probability outside the central one-standard-deviation interval is about \(31.7\%\), so the deviation in part a is \(9\). 3. A central \(95\%\) interval uses \(z^*\approx1.96\): \(90\pm1.96(9)=[72.36,107.64]\). The integer counts inside are \(73\) through \(107\). 4. Since \(27=3\sigma\), the probability of a deviation greater than \(27\) is approximately the probability outside three standard deviations, or \(0.3\%\).

a) \(9\). b) \(73\) through \(107\). c) Approximately \(0.3\%\).

- Find the binomial mean and standard deviation. - Compute each continuous interval first. - To list integer values that lie inside a continuous interval, use the first integer at or above the lower endpoint and the last integer at or below the upper endpoint. - Do not enlarge the interval by rounding outward.

1. \(X\sim\operatorname{Bin}(1800,1/6)\), so \(\mu=300\) and \(\sigma=\sqrt{250}\approx15.811\). The large-count condition is satisfied. 2. The one-standard-deviation interval is approximately \([284.189,315.811]\). The integer values inside it are \(285\) through \(315\), with approximate normal probability \(68.3\%\). 3. The two-standard-deviation interval is approximately \([268.377,331.623]\). The integer values inside it are \(269\) through \(331\), with approximate normal probability \(95.4\%\). 4. The three-standard-deviation interval is approximately \([252.566,347.434]\). The integer values inside it are \(253\) through \(347\), with approximate normal probability \(99.7\%\).

1. \(285\) through \(315\); approximately \(68.3\%\). 2. \(269\) through \(331\); approximately \(95.4\%\). 3. \(253\) through \(347\); approximately \(99.7\%\).

- Check the expected numbers of successes and failures. - Match a two-standard-deviation interval with the Empirical Rule. - List only integer counts that lie inside the continuous interval.

1. \(X\sim\operatorname{Bin}(2000,0.03)\), so \(\mu=np=60\) and \(\sigma=\sqrt{2000(0.03)(0.97)}=\sqrt{58.2}\approx7.629\). 2. The large-count condition is satisfied because \(np=60\) and \(n(1-p)=1940\), so a normal approximation is appropriate. 3. The continuous two-standard-deviation interval is approximately \(60\pm2(7.629)=[44.742,75.258]\). 4. The integer counts inside are \(45\) through \(75\). The Empirical Rule gives approximate normal coverage \(95.4\%\).

a) Yes; the large-count condition is satisfied. b) \(45\) through \(75\) defective chips; approximately \(95.4\%\).

- Find the binomial mean and standard deviation. - Check the large-count condition. - Build the interval with radius \(3\sigma\). - Apply the criterion stated in the question rather than treating “consistent” as an undefined judgment.

1. Let \(X\sim\operatorname{Bin}(4000,0.025)\). The mean is \(\mu=100\), and the standard deviation is \(\sigma=\sqrt{4000(0.025)(0.975)}=\sqrt{97.5}\approx9.874\). The large-count condition is satisfied. 2. The continuous three-standard-deviation interval is \(100\pm3(9.874\ldots)\approx[70.38,129.62]\). The integer counts inside are \(71\) through \(129\). 3. The observed count \(160\) is above this interval. 4. Under the stated criterion, an observation outside three standard deviations is treated as inconsistent with the reported rate; such outcomes have total normal-approximation probability about \(0.3\%\).

a) Continuous interval approximately \([70.38,129.62]\); integer counts \(71\) through \(129\). b) No. The count \(160\) lies outside the three-standard-deviation interval.

- Read the number of green sections from the displayed spinner. - State all four conditions for a binomial setting. - Use the binomial point formula for exactly two green results. - “At most \(1\)” includes zero and one.

1. The experiment has a fixed \(6\) spins, each spin has two relevant outcomes (green or not green), the spins are independent, and the probability of green is constant. 2. Reading the spinner gives one green section out of five, so \(p=\frac15=0.20\). Thus, \(X\sim\operatorname{Bin}(6,0.20).\) 3. \(P(X=2)=\binom62(0.20)^2(0.80)^4=0.24576.\) 4. \(P(X\le1)=(0.80)^6+6(0.20)(0.80)^5=0.65536.\)

a) Fixed \(n=6\), two relevant outcomes, independence, and constant \(p=\frac15=0.20\); hence \(X\sim\operatorname{Bin}(6,0.20)\). b) \(0.24576\). c) \(0.65536\).

- Read the green and red section counts from the displayed spinner. - Treat the named color as success in a separate binomial model for each part. - Use a complement for “at least \(2\).”

1. The spinner shows \(3\) green sections out of \(6\), so \(p_G=1/2\). If \(X\sim\operatorname{Bin}(8,1/2)\), then \(P(X\le3)=\frac{\binom80+\binom81+\binom82+\binom83}{2^8}=\frac{93}{256}\approx36.33\%\). 2. The spinner shows \(1\) red section out of \(6\), so \(p_R=1/6\). If \(Y\sim\operatorname{Bin}(8,1/6)\), then \(P(Y\ge2)=1-P(Y\le1)=\frac{663991}{1679616}\approx39.53\%\).

a) \(\frac{93}{256}\approx36.33\%\). b) \(\frac{663991}{1679616}\approx39.53\%\).

- 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 \(P(a\le X\le b)=P(X\le b)-P(X\le a-1)\).

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, \(P(12\le B\le 18)=P(B\le 18)-P(B\le 11)\approx 0.7995=79.95\%\).

a) About \(43.15\%\) b) About \(79.95\%\)

- Read the red and blue section counts from the displayed spinner. - State all four binomial conditions in part a). - For part c), list every four-spin arrangement with exactly two consecutive blue results. - Do not use a binomial coefficient blindly when adjacency matters.

1. The experiment has a fixed \(10\) spins, two relevant outcomes for \(X\) (red or not red), independent spins, and a constant red probability. 2. Reading the spinner gives \(2\) red sections out of \(8\), so \(p=\frac14\). Thus, \(X\sim\operatorname{Bin}(10,0.25).\) 3. \(P(X\le2)=\sum_{k=0}^{2}\binom{10}{k}(0.25)^k(0.75)^{10-k}\approx0.5256.\) 4. For part c), blue also has probability \(2/8=1/4\). The valid four-spin patterns are BBNN, NBBN, and NNBB, where B means blue and N means not blue. 5. Each valid pattern has probability \(\left(\frac14\right)^2\left(\frac34\right)^2.\) Therefore, the total probability is \(3\left(\frac14\right)^2\left(\frac34\right)^2=\frac{27}{256}.\)

a) Fixed \(n=10\), two relevant outcomes, independence, and constant \(p=\frac14\); hence \(X\sim\operatorname{Bin}(10,0.25)\). b) \(P(X\le2)\approx0.5256\). c) \(\frac{27}{256}\).

- 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.

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.

Expressions (3) and (4) are correct.

- Match \(0.30\) with purple and \(0.70\) with orange. - Compare the first expression with a binomial point probability. - A leading \(1-\) indicates a complement. - Read the summation limits as the included values of \(X\).

1. The probability of purple is \(0.30\), and the probability of orange is \(0.70\). 2. Expression 1 is \(P(X=2)\): purple occurs exactly twice. 3. Expression 2 is \(P(X=0)\): every result is orange. 4. Expression 3 is \(1-P(X=0)=P(X\ge1)\): purple occurs at least once. 5. Expression 4 is \(P(X\le2)\): purple occurs at most twice.

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.

- Read each needed color probability from the displayed spinner. - Distinguish an exact-count event from a cumulative event. - Match the \(0.125\) in part b) to the appropriate color on the spinner.

1. Reading the spinner gives \(P(\text{green})=\frac38\), \(P(\text{red})=\frac14\), and \(P(\text{yellow})=\frac18=0.125\). 2. For event \(A\), \(P(A)=\binom{50}{20}\left(\frac38\right)^{20}\left(\frac58\right)^{30}\approx0.107230,\) or \(10.72\%\). 3. For event \(B\), \(P(B)=\sum_{k=0}^{10}\binom{50}{k}\left(\frac14\right)^k\left(\frac34\right)^{50-k}\approx0.262202,\) or \(26.22\%\). 4. In part b), the success probability \(0.125\) corresponds to yellow, and the summation from \(0\) through \(8\) represents at most eight yellow results in \(60\) spins.

a) \(P(A)\approx10.72\%\) and \(P(B)\approx26.22\%\). b) In \(60\) spins, the spinner lands on yellow at most \(8\) times.

- Use the same exact-count binomial formula under each proposed value of \(p\). - Compare the two probabilities for the observed count. - Distinguish “more plausible under these models” from “proved correct.”

1. Under Model A, \(P(X=17)=\binom{20}{17}(0.95)^{17}(0.05)^3\approx0.05958.\) 2. Under Model B, \(P(X=17)=\binom{20}{17}(0.80)^{17}(0.20)^3\approx0.20536.\) 3. The observed count has the larger point probability under Model B, so Model B makes this particular observation more plausible. 4. This comparison does not prove that Model B is true; it compares only how likely this observed count is under the two specified models.

a) Model A: \(0.05958\); Model B: \(0.20536\). b) Model B makes exactly \(17\) germinations more plausible, but the comparison does not prove that Model B is correct.

- Identify the success and failure probabilities for one guessed question. - Use the binomial formula for each possible value of \(X\). - Compare the binomial coefficients with the number of arrangements having the same number of correct answers.

1. Here, \(X\sim\operatorname{Bin}(3,0.25)\). 2. \(P(X=0)=0.75^3=0.421875\), \(P(X=1)=3(0.25)(0.75)^2=0.421875\), \(P(X=2)=3(0.25)^2(0.75)=0.140625\), and \(P(X=3)=0.25^3=0.015625\). 3. In \((p+q)^3=p^3+3p^2q+3pq^2+q^3\), the four terms give the probabilities of \(3,2,1,0\) correct answers, respectively. 4. The binomial coefficients count the possible orders of correct and incorrect answers with each success count.

a) <table> <tr><th>\(k\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th></tr> <tr><th>\(P(X=k)\)</th><td>\(0.421875\)</td><td>\(0.421875\)</td><td>\(0.140625\)</td><td>\(0.015625\)</td></tr> </table> b) The terms \(0.25^3\), \(3(0.25)^2(0.75)\), \(3(0.25)(0.75)^2\), and \(0.75^3\) are \(P(X=3)\), \(P(X=2)\), \(P(X=1)\), and \(P(X=0)\), respectively.

- Treat each independently generated interval as a capture or miss. - Use the confidence level as the capture probability. - Standardize the observed count using the binomial mean and standard deviation.

1. Each interval captures the parameter with probability \(0.92\), so \(Y\sim\operatorname{Bin}(1000,0.92).\) 2. The mean is \(E(Y)=1000(0.92)=920.\) 3. The standard deviation is \(\sigma_Y=\sqrt{1000(0.92)(0.08)}=\sqrt{73.6}\approx8.58.\) 4. For \(900\) captures, \(z=\frac{900-920}{\sqrt{73.6}}\approx-2.33.\) 5. A count more than two standard deviations below the mean is reasonably described as unusual, although it remains possible.

a) \(920\). b) \(\sqrt{73.6}\approx8.58\). c) \(2.33\) standard deviations below the mean; yes, reasonably unusual.

- Translate the bounds on the sample proportion into allowable whole-number success counts. - For the approximation, extend the count interval by half a unit at each endpoint. - Use the count mean and standard deviation when standardizing the corrected boundaries. - Compare the two final probabilities using their difference.

1. The event \(0.45\le\hat p\le0.55\) is equivalent to \(18\le X\le22\), where \(X\sim\operatorname{Bin}(40,0.50)\). 2. The exact binomial probability is \(\sum_{k=18}^{22}\binom{40}{k}(0.50)^{40}\approx0.57041\). 3. For the continuity correction, use \(17.5\le Y\le22.5\), where \(Y\) is normal with mean \(20\) and standard deviation \(\sqrt{10}\). 4. The corresponding standardized bounds are approximately \(-0.7906\) and \(0.7906\), giving a normal approximation of \(0.57080\). 5. The approximation exceeds the exact probability by \(0.5708047\ldots-0.5704095\ldots=0.0003952\ldots\approx0.00040\) using the requested rounded values.

a) \(\sum_{k=18}^{22}\binom{40}{k}(0.50)^{40}\approx0.57041\). b) \(0.57080\). c) The approximation is higher by approximately \(0.00040\).

- Translate “fewer than half” into the largest allowable whole-number success count. - Use a boundary halfway between adjacent counts when moving to a continuous model. - Standardize the corrected boundary using the count mean and standard deviation.

1. Fewer than half of \(200\) means at most \(99\) successes. 2. The success count has mean \(np=200(0.58)=116\) and standard deviation \(\sqrt{200(0.58)(0.42)}\approx6.9800\). 3. With a continuity correction, use the boundary \(99.5\). 4. The standardized boundary is \(z=\frac{99.5-116}{6.979971\ldots}\approx-2.364\). 5. The lower-tail probability is approximately \(0.00904\).

Approximately \(0.00904\), or \(0.904\%\).

- 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.”

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.

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.

- 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.

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 \(n=31\).

a) \(n=19\) b) \(n=31\)

- 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.

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 \(p\approx 0.25\).

\(p\approx 0.25\)

- 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.

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, \(p\approx 0.54\).

\(p\approx 0.54\)

- 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.

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.

The success probability must be approximately \(0.30\), or \(30\%\).

- 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 \(p\).

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 \(p\approx 0.884\).

The success probability must be at least approximately \(0.884\).

- 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.

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 \(p\le 0.0269141\).

The boundary value is \(p\approx 0.027\). More precisely, the acceptance probability is at least \(90\%\) when \(p\le 0.0269141\).

- 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.

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 \(20\).

The gardener must plant at least \(20\) seeds.

- 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.

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 \(0.85\).

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 \(0.85\).

- 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.

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 \(0.91\).

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 \(0.91\).

- 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 \(0.15\).

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, \(p\approx 0.092\).

The probability that a bottle contains a small air bubble is approximately \(0.092\), or \(9.2\%\).

- Use a binomial model for the fixed sample. - “At most \(2\)” includes \(0,1,2\). - For “at least \(1\),” use the complement of zero defects. - Use logarithms when the unknown sample size appears in an exponent.

1. For parts a) and b), \(X\sim\operatorname{Bin}(20,0.04)\). 2. \(P(X=1)=\binom{20}{1}(0.04)(0.96)^{19}\approx0.3683.\) 3. \(P(X\le2)=\sum_{k=0}^{2}\binom{20}{k}(0.04)^k(0.96)^{20-k}\approx0.9561.\) 4. For part c), \(1-(0.96)^n\ge0.99\), so \(n\ge\frac{\ln(0.01)}{\ln(0.96)}\approx112.811.\) The minimum integer is \(113\).

a) \(\binom{20}{1}(0.04)(0.96)^{19}\approx0.3683\). b) \(\sum_{k=0}^{2}\binom{20}{k}(0.04)^k(0.96)^{20-k}\approx0.9561\). c) \(113\) chips.

- Use complements for the “at least” events. - Use logarithms for part b). - For part c), test consecutive integer sample sizes around the threshold.

1. \(P(X\ge1)=1-(0.96)^{25}\approx0.6396.\) 2. For part b), \(1-(0.96)^n\ge0.90\), giving \(n\ge56.406\ldots\), so the minimum is \(57\). 3. For part c), require \(P(X\le1)=(0.96)^n+n(0.04)(0.96)^{n-1}\le0.10.\) 4. \(n=95\) fails and \(n=96\) succeeds, so the minimum is \(96\).

a) \(1-(0.96)^{25}\approx0.6396\). b) \(57\) bulbs. c) \(96\) bulbs.

- Use a complement for at least two defects. - For the threshold, test consecutive integer values of \(k\). - The first tail below \(1\%\) determines the answer.

1. \(X\sim\operatorname{Bin}(20,0.05)\). 2. \(P(X\ge2)=1-P(X=0)-P(X=1)\approx0.2642.\) 3. For the cutoff, compare consecutive upper tails. 4. \(P(X\ge4)\approx0.0159>0.01\), while \(P(X\ge5)\approx0.0026<0.01\). 5. Therefore, the smallest qualifying value is \(k=5\).

a) \(P(X\ge2)\approx0.2642\). b) \(k=5\).

- Standardize the \(495\,\text{g}\) cutoff. - Use the one-package underweight probability as the binomial success probability. - Use the complement of no underweight packages for part 3.

1. The standard score is \(-1.25\), so \(p=P(X<495)=\Phi(-1.25)\approx0.1056.\) 2. Let \(Y\sim\operatorname{Bin}(100,p)\). Then \(P(Y=10)=\binom{100}{10}p^{10}(1-p)^{90}\approx0.1296.\) 3. For \(n\) packages, \(P(Y\ge1)=1-(1-p)^n.\) Requiring this to exceed \(0.99\) gives \(n>\frac{\ln(0.01)}{\ln(1-p)}\approx41.24.\) The minimum integer is \(42\).

1. \(\Phi(-1.25)\approx0.1056\). 2. \(P(Y=10)\approx0.1296\). 3. \(42\) packages.

- At each spin, count how many outcomes would make the new partial sum a multiple of \(5\). - Use the resulting one-game probability as the success probability for the repeated games. - Apply the binomial formula for exactly five successful games.

1. At each stage exactly one of the five possible next values would make the current partial sum divisible by \(5\). 2. Therefore, \(p=P(\text{no partial sum divisible by }5)=\left(\frac45\right)^3=\frac{64}{125}=0.512.\) 3. Let \(K\sim\operatorname{Bin}(10,64/125)\). Then \(P(K=5)=\binom{10}{5}\left(\frac{64}{125}\right)^5\left(\frac{61}{125}\right)^5\approx0.2454.\)

a) \(\frac{64}{125}=0.512\). b) \(\binom{10}{5}(64/125)^5(61/125)^5\approx0.2454\).

- 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.

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 \(k=6\).

\(P(X=k)\) is greatest for \(k=5\) and \(k=6\).

- Find the probability of one correct guess first. - Translate conditions on incorrect answers into conditions on the number correct when useful. - In part c), separate the first three fixed successes from the remaining seventeen questions. - In part d), treat the first five and last five questions as independent blocks.

1. The probability of a correct answer is \(p=0.25\), and the probability of an incorrect answer is \(0.75\). 2. Part a) specifies one fixed sequence, so \(P=(0.25)^5(0.75)^{15}=1.305025489\ldots\times10^{-5}\approx1.305\times10^{-5}.\) 3. More than \(15\) incorrect answers means fewer than \(5\) correct answers. If \(X\) is the total number correct, \(P(X\le4)\approx0.4148.\) 4. In part c), the first \(3\) answers are already correct, so exactly \(3\) of the remaining \(17\) must be correct: \(P=\binom{17}{3}(0.25)^6(0.75)^{14}\approx0.0030.\) 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. \(P(Y_1\ge4)=\binom54(0.25)^4(0.75)+(0.25)^5=0.015625\). 7. \(P(Y_2\le1)=(0.75)^5+5(0.25)(0.75)^4=0.6328125\). 8. The two blocks are independent, so \(P(Y_1\ge4,\ Y_2\le1)=0.015625(0.6328125)\approx0.0099.\)

a) \(P\approx1.305\times10^{-5}\). b) \(P\approx0.4148\). c) \(P\approx0.0030\). d) \(P\approx0.0099\).

- 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.

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 \(n=14\).

The study needs at least \(14\) patients.

- For “at most \(10\),” a complement may shorten the calculation. - Exactly two failures means exactly ten germinations. - In part c), first find the one-seed probability of both germination and pest effect. - Use a complement and logarithms to solve the minimum-\(n\) condition.

1. For part a), let \(X\sim\operatorname{Bin}(12,0.80)\). 2. \(P(C)=P(X\le10)\approx0.7251.\) 3. Event \(D\) means exactly \(10\) seeds germinate, so \(P(D)=\binom{12}{10}(0.80)^{10}(0.20)^2\approx0.2835.\) 4. \((0.8)^{12}\) is the probability that all \(12\) seeds germinate, so \(1-(0.8)^{12}\) is the probability that at least one seed does not germinate. 5. For one planted seed, the probability of germinating and then producing an affected plant is \((0.80)(0.02)=0.016\). 6. For \(n\) seeds, require \(1-(0.984)^n>0.95.\) 7. Thus, \(n>\frac{\ln(0.05)}{\ln(0.984)}\approx185.731,\) so the minimum integer is \(186\).

a) \(P(C)\approx0.7251\) and \(P(D)\approx0.2835\). b) It is the probability that at least one of the twelve seeds does not germinate. c) \(186\) seeds.

- Start with the probability that all glasses in one box are nondefective. - For part b), use the complement of finding no acceptable boxes. - For part c), express the all-nondefective probability in terms of \(1-p\) and solve the inequality.

1. A glass is nondefective with probability \(0.92\), so an acceptable three-glass box has probability \(p_B=(0.92)^3=0.778688\approx0.7787.\) 2. For \(n\) independent boxes, the probability of finding at least one acceptable box is \(1-(1-p_B)^n.\) Require \(1-(1-p_B)^n\ge0.95.\) This gives \(n\ge\frac{\ln(0.05)}{\ln(1-0.778688)}\approx1.986,\) so the minimum integer is \(2\). 3. For part c), require \((1-p)^{20}\ge0.90.\) Therefore, \(p\le1-(0.90)^{1/20}\approx0.005254.\) As a percentage, the maximum is approximately \(0.525\%\).

a) \((0.92)^3=0.778688\approx0.7787\). b) \(2\) boxes. c) \(p\le1-(0.90)^{1/20}\), so the maximum defect probability is approximately \(0.525\%\).

- In part a), separate the fixed first-three outcome from the binomial count among the remaining seven. - For part b), use the complement event in which no seeds germinate. - For part c), write the all-ten-germinate probability as \(p^{10}\).

1. For part a), the probability that the first \(3\) seeds germinate is \((0.75)^3\). 2. If \(Y\sim\operatorname{Bin}(7,0.75)\), then \(P(Y\ge4)\approx0.9294.\) 3. By independence, the required probability is \((0.75)^3P(Y\ge4)\approx0.3921.\) 4. For part b), require \(1-(0.25)^n\ge0.99.\) Thus, \(n\ge\frac{\ln(0.01)}{\ln(0.25)}\approx3.322,\) so the minimum integer is \(4\). 5. For part c), require \(p^{10}\ge0.95.\) Therefore, \(p\ge(0.95)^{1/10}\approx0.994884.\) As a percentage, the minimum is approximately \(99.488\%\).

a) \(0.3921\). b) \(4\) seeds. c) Approximately \(99.488\%\).

- 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.

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 \(77\).

The minimum sample size is \(77\) components.

- 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.

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 \(74\).

At least \(74\) juvenile animals must be released.

- 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.

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 \([0,4]\).

a) \(\mu=2\) b) \([0,4]\), with probability approximately \(0.9568\), or \(95.68\%\)

- 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.

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, \(P(X=2)\approx\frac{1}{2e}\).

a) \(\frac{P(X=1)}{P(X=0)}=\frac{n}{n-1}\), so \(P(X=1)=\frac{n}{n-1}P(X=0)\). b) \(P(X=2)=\frac{n-1}{2n}\left(1-\frac1n\right)^{n-2}\approx\frac12\cdot\frac1e=\frac{1}{2e}\).

- 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 \(5\%\).

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 \(143\).

At most \(143\) employees can participate.

- Compare the variance formulas after interchanging \(p\) and \(1-p\). - Relate the red and blue counts using their fixed total. - Transform the interval for \(X\) into the corresponding interval for \(Y\).

1. For \(X\), \(\sigma_X=\sqrt{192(0.25)(0.75)}=6.\) For \(Y\), \(\sigma_Y=\sqrt{192(0.75)(0.25)}=6.\) 2. The mean of \(X\) is \(\mu_X=192(0.25)=48\), so the two-standard-deviation interval is \([36,60]\). 3. \(P(36\le X\le60)=P(X\le60)-P(X\le35)\approx0.9632.\) 4. Since \(Y=192-X\), the event \(36\le X\le60\) is equivalent to \(132\le Y\le156\). 5. Because \(\mu_Y=144\) and \(\sigma_Y=6\), \([132{,}156]\) is exactly the two-standard-deviation interval for \(Y\). Thus, the two probabilities are equal.

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.

- Convert the sample-proportion event into a whole-number count event. - For the approximation, adjust the count boundary by half a unit before standardizing. - Compare the exact and approximate probabilities numerically. - Relate the discrepancy to how close the expected rare-outcome count is to the usual large-count threshold.

1. The event \(\hat p\le0.010\) is equivalent to \(X\le5\). 2. Since \(X\sim\operatorname{Bin}(500,0.02)\), the exact binomial probability is \(\sum_{k=0}^{5}\binom{500}{k}(0.02)^k(0.98)^{500-k}\approx0.06519\). 3. The count distribution has mean \(\mu_X=500(0.02)=10\) and standard deviation \(\sigma_X=\sqrt{500(0.02)(0.98)}\approx3.1305\). 4. With continuity correction, \(P(X\le5)\) is approximated by \(P(Y\le5.5)\) for a normal variable \(Y\) with that mean and standard deviation. 5. The standardized boundary is \(z=\frac{5.5-10}{3.1305\ldots}\approx-1.437\), giving \(P(X\le5)\approx0.07529\). 6. The expected success count is exactly \(10\), so the usual large-count condition is only just met; the noticeable discrepancy between the exact and approximate probabilities is therefore unsurprising.

a) \(\sum_{k=0}^{5}\binom{500}{k}(0.02)^k(0.98)^{500-k}\approx0.06519\). b) \(0.07529\). c) The normal approximation is reasonably close but not highly accurate; the expected success count is exactly at the usual minimum large-count threshold.

- Determine which whole-number success counts round to the displayed percentage. - Work with the success-count distribution so that the continuity correction can be applied naturally. - Convert the corrected count boundaries to standardized values before finding the middle area.

1. Reporting \(40\%\) requires the success count to be \(99\), \(100\), or \(101\), because these correspond to sample percentages \(39.6\%\), \(40.0\%\), and \(40.4\%\). 2. For \(X\sim\operatorname{Bin}(250,0.37)\), the mean is \(92.5\) and the standard deviation is \(\sqrt{250(0.37)(0.63)}\approx7.6338\). 3. With a continuity correction, approximate \(P(99\le X\le101)\) by \(P(98.5<Y<101.5)\) for a normal variable \(Y\) with that mean and standard deviation. 4. The standardized bounds are approximately \(0.786\) and \(1.179\). 5. The normal approximation is approximately \(0.0967\).

Approximately \(0.0967\).

- Treat each repeated study as a success-or-failure trial based on the original extremeness threshold. - Identify the one-study probability from the p-value. - Use a count model for a fixed number of independent repetitions. - Connect the Bernoulli success definition back to the interpretation of a p-value.

1. Under the true null hypothesis, each repetition has probability \(0.08\) of producing a result at least as extreme as the original result. 2. If \(Y\) is the number of such repetitions, then \(Y\sim\operatorname{Bin}(50,0.08)\). 3. The expected count is \(E(Y)=50(0.08)=4\). 4. The probability of exactly \(4\) is \(P(Y=4)=\binom{50}{4}(0.08)^4(0.92)^{46}\approx0.2037\). 5. The repeated-study model treats the p-value as the long-run probability, under the null hypothesis, of obtaining a statistic at least as extreme as the observed statistic.

a) \(4\) repetitions. b) \(0.2037\). c) Under the null hypothesis, the p-value \(0.08\) is the probability that one repetition produces a statistic at least as extreme as the observed statistic.