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Binomial probability distribution

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52334712
A basketball player makes \(80\%\) of free throws. The player takes a series of \(12\) shots. Explain the assumptions needed to model this situation as a binomial experiment with \(n=12\).

Hints

- Identify the possible outcomes of one shot. - Consider whether the player's success probability is assumed to change during the series. - Decide whether one shot affects another. - Recall all conditions required for a binomial experiment.

Solution

1. Each shot must have exactly two relevant outcomes: make or miss. 2. The probability of a make must remain constant at \(p=0.80\) for all \(12\) shots. 3. The shots must be independent, so the result of one shot does not change the probability of making another shot. 4. Because the same type of trial is repeated \(12\) times, the number of trials is \(n=12\).

Answer

The model is binomial if each shot has two outcomes, the shots are independent, and the probability of a make stays constant at \(p=0.80\). The number of trials is \(n=12\).
52357512
Determine whether each situation can be modeled as a binomial experiment. If it can, define a success and state \(n\) and \(p\). If it cannot, briefly explain why. a) A fair six-sided die is rolled \(12\) times. The number of prime-number results is counted. b) A box contains \(10\) red pens and \(10\) black pens. Four pens are selected without replacement. c) A person shoots a basketball until making \(3\) baskets. d) A spinner with three equal blue, yellow, and red sections is spun \(20\) times. The number of yellow results is counted.

Hints

- Check whether each trial has two categories, success and failure. - Check whether the success probability remains constant. - Check whether the trials are independent. - Check whether the number of trials is fixed before the experiment begins.

Solution

1. In part a, define success as rolling a prime number, \(2\), \(3\), or \(5\). The rolls are independent, \(n=12\), and \(p=\frac{3}{6}=0.5\), so the experiment is binomial. 2. Part b is not binomial. Without replacement, the selections are dependent and the probability of selecting a red pen changes. 3. Part c is not binomial because the number of trials is not fixed in advance; it depends on when the third basket is made. 4. In part d, define success as landing on yellow. The spins are independent, \(n=20\), and \(p=\frac{1}{3}\), so the experiment is binomial.

Answer

a) Yes; success: rolling a prime number; \(n=12\); \(p=0.5\) b) No; without replacement, the trials are dependent and \(p\) changes. c) No; the number of trials is not fixed in advance. d) Yes; success: landing on yellow; \(n=20\); \(p=\frac{1}{3}\)
52872312
A standard six-sided die is rolled once. For each description, decide whether the trial is a Bernoulli trial. a) The six results \(1\) through \(6\) are treated as distinct outcomes. b) The result is classified as prime \((2, 3, 5)\) or not prime \((1, 4, 6)\). c) The result is classified as greater than \(4\) or not greater than \(4\).

Hints

- Count the outcome categories used in each description. - Several die results may be grouped into one event. - A Bernoulli trial separates outcomes into an event and its complement.

Solution

1. A Bernoulli trial has exactly two relevant outcome categories, often called success and failure. 2. Part a has six distinct outcome categories, so it is not a Bernoulli trial. 3. Part b has two complementary categories, prime and not prime, so it is a Bernoulli trial. 4. Part c also has two complementary categories, greater than \(4\) and not greater than \(4\), so it is a Bernoulli trial.

Answer

a) No b) Yes c) Yes
52333112
A gardener plants three seeds of a special flower variety. Each seed has a \(70\%\) chance of sprouting, independently of the other seeds. Find the probability of each event: a) All three seeds sprout. b) None of the seeds sprout. c) At least two seeds sprout.

Hints

- Identify the success probability and the number of independent trials. - For a specific sequence of outcomes, multiply the probabilities along the sequence. - For “at least two,” determine which possible values of the number of successes meet the condition. - A complement may be useful when an event includes many outcomes.

Solution

Let \(X\) be the number of seeds that sprout. This is a binomial setting with \(n = 3\), \(p = 0.7\), and \(1-p = 0.3\). 1. \(P(X=3) = 0.7^3 = 0.343\). 2. \(P(X=0) = 0.3^3 = 0.027\). 3. \(P(X\ge 2) = P(X=2)+P(X=3)\). Also, \(P(X=2)=\binom{3}{2}\cdot(0.7)^2\cdot0.3=0.441\), so \(P(X\ge 2)=0.441+0.343=0.784\).

Answer

a) \(0.343\), or \(34.3\%\) b) \(0.027\), or \(2.7\%\) c) \(0.784\), or \(78.4\%\)
52333212
A factory finds that \(5\%\) of its electronic components are defective. An inspector independently tests four components, with a constant probability of \(0.05\) that any component is defective. Find the probability that: a) None of the four components are defective. b) Exactly one component is defective. c) Exactly two components are defective.

Hints

- First find the probability that a component is not defective. - Use the binomial probability formula for an exact number of defective components. - The binomial coefficient counts the possible positions of the defective components. - Keep full precision during the calculation and round only at the end.

Solution

Let \(X\) be the number of defective components. Then \(X\) has a binomial distribution with \(n = 4\), \(p = 0.05\), and \(1-p = 0.95\). 1. \(P(X=0)=(0.95)^4=0.81450625\). 2. \(P(X=1)=\binom{4}{1}\cdot0.05\cdot(0.95)^3=0.171475\). 3. \(P(X=2)=\binom{4}{2}\cdot(0.05)^2\cdot(0.95)^2=0.0135375\).

Answer

a) \(0.81450625\), or about \(81.45\%\) b) \(0.171475\), or about \(17.15\%\) c) \(0.0135375\), or about \(1.35\%\)
52333512
A basketball player makes a free throw with probability \(0.60\). The player takes four independent free throws, and the probability of making each shot remains constant. a) Explain why this situation can be modeled by a binomial experiment. State \(n\) and \(p\). b) Find the probabilities of the following events: \(A\): All four shots are made. \(B\): The first two shots are missed and the last two are made. \(C\): At least one shot is made. \(D\): Exactly two shots are made.

Hints

- Recall the conditions required for a binomial experiment. - For a particular ordered sequence, multiply the four trial probabilities. - For “at least one,” consider the complementary event. - For exactly two makes, account for every possible placement of the two successful shots.

Solution

1. Each trial has two outcomes, make or miss. The trials are independent, and the success probability is constant. Therefore, \(n=4\) and \(p=0.60\). 2. \(P(A)=(0.60)^4=0.1296\). 3. \(P(B)=(0.40)^2\cdot(0.60)^2=0.0576\). 4. Use the complement of making no shots: \(P(C)=1-(0.40)^4=0.9744\). 5. \(P(D)=\binom{4}{2}\cdot(0.60)^2\cdot(0.40)^2=0.3456\).

Answer

a) There are two outcomes per trial, the trials are independent, and the success probability is constant; \(n=4\) and \(p=0.60\). b) \(P(A)=0.1296\); \(P(B)=0.0576\); \(P(C)=0.9744\); \(P(D)=0.3456\).
52333612
A glass manufacturer finds that \(10\%\) of its drinking glasses are defective. For quality control, five glasses are selected independently from a very large production run. a) State \(n\) and \(p\) for a binomial model in which a defective glass is a success. b) Find the probabilities of the following events: \(E\): No glass is defective. \(F\): The first three glasses are not defective and the last two are defective. \(G\): At least one glass is defective. \(H\): Exactly one glass is defective.

Hints

- Identify what counts as a success in this setting. - The probabilities of “defective” and “not defective” must add to \(1\). - For event \(F\), decide whether the order is fixed. - For event \(G\), use the complement of no defective glasses. - For exactly one defective glass, count the possible positions of that glass.

Solution

1. The number of trials is \(n=5\), and the probability of a defective glass is \(p=0.10\). 2. \(P(E)=(0.90)^5=0.59049\). 3. \(P(F)=(0.90)^3\cdot(0.10)^2=0.00729\). 4. \(P(G)=1-P(E)=1-0.59049=0.40951\). 5. \(P(H)=\binom{5}{1}\cdot0.10\cdot(0.90)^4=0.32805\).

Answer

a) \(n=5\); \(p=0.10\) b) \(P(E)=0.59049\); \(P(F)=0.00729\); \(P(G)=0.40951\); \(P(H)=0.32805\).
52333712
An archer hits the center of a target with probability \(0.40\). The archer takes three independent shots. Record a hit as \(H\) and a miss as \(M\). a) List all eight possible outcomes, using notation such as \(HMM\). b) Find \(P(HMM)\) and \(P(HHM)\). c) Create a probability distribution table for \(X\), the number of hits.

Hints

- Each shot has two possible outcomes. Use the fundamental counting principle to determine the sample-space size. - Multiply probabilities along a specific ordered outcome. - Group outcomes that have the same number of hits. - The probabilities in the distribution must add to \(1\).

Solution

1. The sample space is \(\{HHH, HHM, HMH, MHH, HMM, MHM, MMH, MMM\}\). 2. \(P(HMM)=0.40\cdot(0.60)^2=0.144\), and \(P(HHM)=(0.40)^2\cdot0.60=0.096\). 3. For \(X\), use the binomial probability formula: \(P(X=0)=(0.60)^3=0.216\). \(P(X=1)=\binom{3}{1}\cdot0.40\cdot(0.60)^2=0.432\). \(P(X=2)=\binom{3}{2}\cdot(0.40)^2\cdot0.60=0.288\). \(P(X=3)=(0.40)^3=0.064\).

Answer

a) \(\{HHH, HHM, HMH, MHH, HMM, MHM, MMH, MMM\}\) b) \(P(HMM)=0.144\); \(P(HHM)=0.096\) c) <table> <tr><td>Number of hits \(k\)</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>
52333812
At a factory, \(10\%\) of electronic components are defective \((D)\), and the remaining components are acceptable \((A)\). Three components are tested independently, with a constant defective rate. a) Explain why the outcomes \(ADA\) and \(AAD\) have the same probability, and state that probability. b) Find the probability that at least two of the three components are defective. c) Give the probability distribution for \(X\), the number of defective components.

Hints

- Compare the factors in the probabilities of the two ordered outcomes. - “At least two” includes exactly two and exactly three. - For the distribution, match each possible value of \(X\) with its probability.

Solution

1. Each outcome contains two acceptable components and one defective component. Therefore, \(P(ADA)=0.90\cdot0.10\cdot0.90=0.081\) and \(P(AAD)=(0.90)^2\cdot0.10=0.081\). 2. \(P(X\ge 2)=P(X=2)+P(X=3)\). Thus, \(P(X\ge 2)=\binom{3}{2}\cdot(0.10)^2\cdot0.90+(0.10)^3=0.027+0.001=0.028\). 3. The distribution is \(P(X=0)=(0.90)^3=0.729\), \(P(X=1)=\binom{3}{1}\cdot0.10\cdot(0.90)^2=0.243\), \(P(X=2)=0.027\), and \(P(X=3)=0.001\).

Answer

a) Both outcomes contain the same factors, so \(P(ADA)=P(AAD)=0.081\). b) \(P(X\ge 2)=0.028\). c) <table> <tr><td>Number of defective components \(k\)</td><td>0</td><td>1</td><td>2</td><td>3</td></tr> <tr><td>\(P(X=k)\)</td><td>\(0.729\)</td><td>\(0.243\)</td><td>\(0.027\)</td><td>\(0.001\)</td></tr> </table>
52335512
For a certain flower variety, each seed independently has a \(15\%\) chance of not sprouting. A nursery plants \(12\) seeds. Find the probability that: a) Exactly two seeds do not sprout. b) Only the first two seeds do not sprout.

Hints

- Decide whether the order of the outcomes matters in each part. - Use a binomial probability when any two of the seeds may fail to sprout. - Multiply along one specific ordered sequence when the first two positions are fixed. - Compare “any two” with “only the first two.”

Solution

1. Let \(X\) be the number of seeds that do not sprout. Then \(X\) has a binomial distribution with \(n=12\) and \(p=0.15\). 2. \(P(X=2)=\binom{12}{2}\cdot(0.15)^2\cdot(0.85)^{10}\approx 0.2924\). 3. In part b, the order is fixed: the first two seeds do not sprout and the other ten do. Therefore, \(P=(0.15)^2\cdot(0.85)^{10}\approx 0.0044\).

Answer

a) About \(0.2924\), or \(29.24\%\) b) About \(0.0044\), or \(0.44\%\)
52335612
A basketball player makes a free throw with probability \(0.80\). The player takes \(10\) independent free throws. Find the probability that: a) Exactly \(8\) shots are made. b) Only the last two shots are missed.

Hints

- Identify the success probability and the number of trials. - Decide when a binomial coefficient is needed. - Compare any arrangement of eight makes with one specified arrangement. - In part b, determine what “only the last two” requires for the other shots.

Solution

1. Let \(X\) be the number of made shots. Then \(X\) has a binomial distribution with \(n=10\) and \(p=0.80\). 2. \(P(X=8)=\binom{10}{8}\cdot(0.80)^8\cdot(0.20)^2\approx 0.3020\). 3. In part b, the order is fixed: eight makes followed by two misses. Thus, \(P=(0.80)^8\cdot(0.20)^2\approx 0.0067\).

Answer

a) About \(0.3020\), or \(30.20\%\) b) About \(0.0067\), or \(0.67\%\)
52348112
A basketball player makes a free throw with probability \(0.40\). The player takes independent free throws. a) Find the probability of making at least one of three shots. Is this event more likely than missing all three shots? Support your answer with calculations. b) What is the minimum number of shots the player must take so that the probability of making at least one shot is greater than \(90\%\)?

Hints

- Describe the complement of “at least one make.” - Write the probability of a sequence of misses. - In part b, form an inequality with the number of shots in the exponent. - You may solve the inequality with logarithms or test consecutive whole-number values.

Solution

1. The probability of missing all three shots is \((0.60)^3=0.216\). 2. Therefore, the probability of at least one make is \(1-0.216=0.784\). Since \(0.784>0.216\), at least one make is more likely. 3. For \(n\) shots, require \(1-(0.60)^n>0.90\), so \((0.60)^n<0.10\). 4. Solving gives \(n>\frac{\ln(0.10)}{\ln(0.60)}\approx 4.5076\). 5. The least whole number that satisfies the inequality is \(n=5\).

Answer

a) The probability of at least one make is \(0.784\), or \(78.4\%\). This is greater than the probability of missing all three, \(0.216\), or \(21.6\%\). b) The player must take at least \(5\) shots.
52348212
A spinner has \(10\) equal sections, and only one section is a winning section. The spinner is spun independently several times. a) For five spins, is at least one win more likely than no wins? b) Explain why no finite number of spins can guarantee a win with probability exactly \(100\%\). c) Find the least number of spins for which the probability of at least one win is greater than \(80\%\).

Hints

- Find the probability of not winning on one spin. - Use the complement to calculate “at least one win.” - For part b, consider whether the probability of losing every spin can equal zero for a finite number of spins. - Use logarithms or test consecutive whole-number values in part c.

Solution

1. The probability of a win on one spin is \(0.10\), and the probability of no win is \(0.90\). 2. For five spins, \(P(\text{no wins})=(0.90)^5=0.59049\). 3. Thus, \(P(\text{at least one win})=1-0.59049=0.40951\). No wins is more likely. 4. For every finite \(n\), the probability \((0.90)^n\) of losing every spin is greater than \(0\). Therefore, \(1-(0.90)^n\) is always less than \(1\). 5. Require \(1-(0.90)^n>0.80\), so \((0.90)^n<0.20\). 6. Solving gives \(n>\frac{\ln(0.20)}{\ln(0.90)}\approx 15.2755\), so the least possible value is \(n=16\).

Answer

a) No wins is more likely: \(P(\text{no wins})\approx 59.05\%\), while \(P(\text{at least one win})\approx 40.95\%\). b) For any finite number of spins, losing every spin still has a positive probability, so the probability of at least one win remains below \(100\%\). c) At least \(16\) spins are required.
52872612
A box contains \(100\) tickets labeled as grand prize, small prize, or no prize. a) Explain why drawing one ticket is not a Bernoulli trial when all three ticket types are treated as separate outcomes. b) Redefine the outcomes in two different ways so that the draw becomes a Bernoulli trial. c) A student says, “The number of ticket types does not matter. For any event of interest, the outcomes can be grouped into that event and its complement.” Evaluate the statement.

Hints

- Recall how many outcome categories a Bernoulli trial uses. - Combine the three ticket types into two complementary groups. - Think of an event and its complement.

Solution

1. When grand prize, small prize, and no prize are treated separately, there are three outcome categories rather than two. 2. One possible definition is success = any prize and failure = no prize. Another is success = grand prize and failure = not a grand prize. 3. The statement is correct. Once a specific event \(E\) is chosen, every outcome either belongs to \(E\) or to its complement \(E^c\). This creates two complementary categories for that question.

Answer

a) There are three outcome categories instead of two. b) For example: any prize versus no prize; grand prize versus no grand prize. c) The statement is correct for a specified event \(E\), because all outcomes can be separated into \(E\) and \(E^c\).
53746312
Three components are tested one at a time. The component outcomes are independent. Each component works with probability \(p\) and is defective with probability \(q\), where \(p+q=1\). a) Use the probability tree to identify every path with exactly two working components. Write each path probability in terms of \(p\) and \(q\). b) Add the path probabilities. What coefficient multiplies \(p^2q\)? c) Where does this coefficient appear in Pascal's triangle? Explain how it is related to the number of paths.
Figure for problem 537463

Hints

- Multiply probabilities along one path. - List the different orders of two working results and one defective result. - Pascal's triangle gives the coefficients in \((p+q)^n\). - Relate the coefficient to \(\binom{3}{2}\).

Solution

1. The paths with exactly two working components are \((W,W,D)\), \((W,D,W)\), and \((D,W,W)\). 2. Each path has probability \(p^2q\), because multiplication along the path gives two factors of \(p\) and one factor of \(q\). 3. Adding the three path probabilities gives \(p^2q+p^2q+p^2q=3p^2q\), so the coefficient is \(3\). 4. The coefficient is \(\binom{3}{2}=3\). It is the entry for \(n=3\) and \(k=2\) in Pascal's triangle, which is the third entry in the fourth row when the top row is counted as row \(0\). It counts the three ways to choose which two of the three positions are working.

Answer

a) \((W,W,D)\), \((W,D,W)\), and \((D,W,W)\); each has probability \(p^2q\). b) \(3p^2q\); the coefficient is \(3\). c) The coefficient is \(\binom{3}{2}=3\), the entry at \(n=3\), \(k=2\) in Pascal's triangle. It counts the three qualifying paths.
53755612
Three independent sensors each produce a false alarm with probability \(0.1\). Find the probability that at least two sensors produce a false alarm at the same time.
Figure for problem 537556

Hints

- Split “at least two” into exactly two and exactly three. - Count the possible positions of the one sensor without a false alarm in the first case. - Add the two disjoint probabilities.

Solution

1. Exactly two false alarms can occur in three ways. Its probability is \(3\cdot(0.1)^2\cdot 0.9=0.027\). 2. The probability of three false alarms is \((0.1)^3=0.001\). 3. Therefore, \(P(\text{at least two false alarms})=0.027+0.001=0.028\).

Answer

\(0.028=2.8\%\)
52708912
A running-shoe company manufactures shoes at two factories. Factory 1 produces \(60\%\) of the shoes, and Factory 2 produces the rest. The defect rate is \(2\%\) at Factory 1 and \(4\%\) at Factory 2. a) A shoe is selected at random from all production. Find the probability that it is defective. b) A selected shoe is defective. Find the probability that it came from Factory 1. c) A random sample of \(150\) shoes from Factory 2 is inspected. Let \(X\) be the number of defective shoes. Use a binomial model to find: \(E\): exactly \(5\) shoes are defective; \(F\): at least \(4\) but fewer than \(10\) shoes are defective.

Hints

- For part a, identify the two paths that end with a defective shoe. - For part b, divide the probability of “Factory 1 and defective” by the total probability of a defect. - In part c, identify \(n\) and \(p\) for the binomial model. - Translate “at least \(4\) but fewer than \(10\)” into a range of integer values for \(X\).

Solution

1. Let \(D\) mean defective. By the law of total probability, \(P(D)=0.60\cdot 0.02+0.40\cdot 0.04=0.028\). 2. The joint probability that a shoe is from Factory 1 and defective is \(0.60\cdot 0.02=0.012\). Thus, \(P(\text{Factory 1}\mid D)=\frac{0.012}{0.028}=\frac{3}{7}\approx 0.4286\). 3. For Factory 2, \(X\sim\operatorname{Binomial}(150,0.04)\). Then \(P(E)=P(X=5)=\binom{150}{5}(0.04)^5(0.96)^{145}\approx 0.1628\). 4. For event \(F\), \(P(F)=P(4\le X<10)=\sum_{k=4}^{9}\binom{150}{k}(0.04)^k(0.96)^{150-k}\approx 0.7745\).

Answer

a) \(P(D)=0.028\), or \(2.8\%\). b) \(P(\text{Factory 1}\mid D)=\frac{3}{7}\approx 0.4286\), or about \(42.86\%\). c) \(P(E)\approx 0.1628\); \(P(F)\approx 0.7745\).
52877412
At an electronics factory, \(40\%\) of components receive a special coating. A random sample of \(15\) components is selected independently. 1. Find the probability that exactly \(6\) sampled components have the coating. 2. Find every value of \(k\) for which \(P(X=k)<1\%\), where \(X\) is the number of coated components in the sample.

Hints

- Identify \(n\) and \(p\) for the binomial distribution. - The most likely values are near the mean, so begin checking at both tails. - Compare each individual probability with \(0.01\). - Use the binomial probability formula.

Solution

1. The random variable \(X\) has a binomial distribution with \(n=15\) and \(p=0.40\). Therefore, \(P(X=6)=\binom{15}{6}\cdot(0.40)^6\cdot(0.60)^9\approx 0.2066\). 2. Evaluate \(P(X=k)=\binom{15}{k}\cdot(0.40)^k\cdot(0.60)^{15-k}\) at the tails of the distribution. The probabilities for \(k=0\) and \(k=1\) are less than \(0.01\), while \(P(X=2)\approx 0.0219\). At the upper tail, \(P(X=10)\approx 0.0245\), while the probabilities for \(k=11,12,13,14,15\) are each less than \(0.01\). 3. Thus, the required values are \(k\in\{0,1,11,12,13,14,15\}\).

Answer

1. \(P(X=6)\approx 0.2066\), or \(20.66\%\) 2. \(k\in\{0,1,11,12,13,14,15\}\)
53755512
A biased coin lands heads with probability \(q\). When the coin is tossed twice, the probability of getting exactly one head is \(0.42\). Find all possible values of \(q\).
Figure for problem 537555

Hints

- Write the two orders that produce exactly one head. - Express their total probability in terms of \(q\). - Solve the resulting quadratic equation and check that both solutions are valid probabilities.

Solution

1. Exactly one head occurs as heads-tails or tails-heads. Therefore, \(q(1-q)+(1-q)q=0.42\). 2. Simplify: \(2q(1-q)=0.42\) \(q^2-q+0.21=0\). 3. Solve the quadratic equation: \(q=\frac{1\pm\sqrt{1-4(0.21)}}{2}=\frac{1\pm0.4}{2}\). Thus, \(q=0.3\) or \(q=0.7\).

Answer

\(q=0.3\) or \(q=0.7\)

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