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Binomial probability distribution

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55143712
A random variable is written as \(X\sim\operatorname{Bin}(8,0.25)\). How many trials are in the binomial experiment?

Hints

- Recall what the two parameters in \(\operatorname{Bin}(n,p)\) represent. - Identify the parameter that gives the trial count.

Solution

1. In \(\operatorname{Bin}(n,p)\), the first parameter \(n\) is the number of trials. 2. Here, \(n=8\).

Answer

\(8\) trials
55143812
The spinner shown is spun independently 6 times. Landing on S is called a success. What is the probability of success on one spin?
Figure for problem 551438

Hints

- Compare the number of equal sectors labeled S with the total number of sectors. - The question asks for the probability on one spin, not the probability of a six-spin outcome.

Solution

1. The spinner has four equal sectors, and one sector is labeled S. 2. Therefore, the success probability is \(p=\frac{1}{4}\).

Answer

\(p=\frac{1}{4}\)
55584212
A bag contains \(4\) red balls and \(6\) blue balls. Four balls are drawn without replacement, and \(X\) is the number of red balls drawn. Is \(X\) a binomial random variable? Answer yes or no and give the key reason.

Hints

- Check whether the success probability stays the same from draw to draw. - Ask what happens to the bag composition after the first draw. - Compare that behavior with the independence requirement for a binomial model.

Solution

1. A binomial model requires independent trials with a constant success probability. 2. Because the balls are not replaced, the bag composition changes after each draw, so the red probability can change. 3. Therefore, the draws are dependent and \(X\) is not binomial.

Answer

No. Without replacement, the success probability changes and the draws are dependent.
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

a) Let \(X\) be the number of seeds that sprout. With \(n=3\) and \(p=0.7\), \(P(X=3)=(0.7)^3=0.343\). b) The probability that a seed does not sprout is \(0.3\), so \(P(X=0)=(0.3)^3=0.027\). c) At least two seeds sprout when \(X=2\) or \(X=3\). Since \(P(X=2)=\binom{3}{2}(0.7)^2(0.3)=0.441\), \(P(X\ge 2)=0.441+0.343=0.784\).

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

a) Let \(X\) be the number of defective components. With \(n=4\) and \(p=0.05\), \(P(X=0)=(0.95)^4=0.81450625\). b) \(P(X=1)=\binom{4}{1}(0.05)(0.95)^3=0.171475\). c) \(P(X=2)=\binom{4}{2}(0.05)^2(0.95)^2=0.0135375\).

Answer

a) \(0.81450625\), or about \(81.45\%\) b) \(0.171475\), or about \(17.15\%\) c) \(0.0135375\), or about \(1.35\%\)
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

a) Yes. Define success as rolling a prime number, \(2\), \(3\), or \(5\). The rolls are independent, \(n=12\), and \(p=\frac{3}{6}=0.5\). b) No. Without replacement, the selections are dependent and the probability of selecting a red pen changes. c) No. The number of trials is not fixed in advance; it depends on when the third basket is made. d) Yes. Define success as landing on yellow. The spins are independent, \(n=20\), and \(p=\frac{1}{3}\).

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

a) No. There are six distinct outcome categories, rather than two relevant categories. b) Yes. Prime and not prime are two complementary outcome categories. c) Yes. Greater than \(4\) and not greater than \(4\) are two complementary outcome categories.

Answer

a) No b) Yes c) Yes
53746512
A factory produces electronic components. Historically, \(10\%\) are defective \((D)\), and the rest are good \((G)\). Three components are selected independently from a very large production run and tested. a) Explain why this is a binomial experiment. Give \(n\) and the success probability \(p\) for “defective.” b) Use the tree diagram and path rules to find the probability that at most one selected component is defective.
Figure for problem 537465

Hints

- What conditions define a binomial experiment? - Which defect counts are included in “at most one”? - Multiply probabilities along each path and add the probabilities of the qualifying paths.

Solution

1. Each selection has two relevant outcomes, defective or good. The selections are independent, and the defect probability remains \(0.10\). Therefore, \(n=3\) and \(p=0.10\). 2. “At most one defective” includes \(GGG\), \(DGG\), \(GDG\), and \(GGD\). 3. The all-good path has probability \((0.90)^3=0.729\). 4. Each path with exactly one defective component has probability \(0.10\cdot(0.90)^2=0.081\). There are three such paths, so their total probability is \(3\cdot0.081=0.243\). 5. Add the disjoint path probabilities: \(0.729+0.243=0.972\).

Answer

a) The trials have two outcomes, constant probability, and independence. \(n=3\), \(p=0.10\). b) \(P(X\le1)=0.972\)
55584312
The two bar charts show binomial probability distributions with the same number of trials, \(n=4\). Which distribution has the larger success probability \(p\)? Explain using the location of the probability mass in the graphs.
Figure for problem 555843

Hints

- Compare where the tallest bars occur in the two graphs. - Think about how a larger chance of success on each trial changes the typical number of successes. - The number of trials is the same in both graphs, so focus on the horizontal shift in the probability mass.

Solution

1. In graph a), most of the probability mass is concentrated at smaller values of \(X\), especially \(0\), \(1\), and \(2\). 2. In graph b), the distribution is shifted toward larger values, especially \(2\), \(3\), and \(4\). 3. With the same number of trials, a larger success probability makes larger success counts more typical, so graph b) has the larger \(p\).

Answer

Graph b) has the larger success probability \(p\) because its probability mass is shifted toward larger numbers of successes.
55585012
The bar chart shows the probability distribution of a binomial random variable \(X\), the number of successes in repeated trials. Based on the shape of the displayed distribution, decide whether the success probability satisfies \(p<0.5\), \(p=0.5\), or \(p>0.5\). Explain using the graph.
Figure for problem 555850

Hints

- Compare bars that are equally far from the center of the graph. - Decide whether the probability mass is shifted toward smaller values, shifted toward larger values, or balanced symmetrically. - For a binomial distribution, think about what value of \(p\) treats success and failure symmetrically.

Solution

1. The bars occur at \(X=0,1,\ldots,6\), and the distribution is symmetric about \(X=3\). 2. In a binomial distribution, replacing \(p\) by \(1-p\) reflects the distribution across the midpoint. A distribution that is its own reflection has \(p=1-p\). 3. Therefore, \(p=0.5\).

Answer

\(p=0.5\). The displayed binomial distribution is symmetric about \(X=3\).
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

a) Each trial has two outcomes, make or miss. The trials are independent, and the success probability is constant. Therefore, \(n=4\) and \(p=0.60\). b) For the requested events: \(P(A)=(0.60)^4=0.1296\). \(P(B)=(0.40)^2(0.60)^2=0.0576\). \(P(C)=1-(0.40)^4=0.9744\). \(P(D)=\binom{4}{2}(0.60)^2(0.40)^2=0.3456\).

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) \(P(A)=0.1296\); \(P(B)=0.0576\); \(P(C)=0.9744\); \(P(D)=0.3456\).
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

a) The sample space is \(\{HHH, HHM, HMH, MHH, HMM, MHM, MMH, MMM\}\). b) \(P(HMM)=0.40(0.60)^2=0.144\), and \(P(HHM)=(0.40)^2(0.60)=0.096\). c) Group outcomes by the number of hits. The probabilities are \(P(X=0)=(0.60)^3=0.216\), \(P(X=1)=\binom{3}{1}(0.40)(0.60)^2=0.432\), \(P(X=2)=\binom{3}{2}(0.40)^2(0.60)=0.288\), and \(P(X=3)=(0.40)^3=0.064\).

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

a) Each outcome contains two acceptable components and one defective component, so \(P(ADA)=0.90(0.10)(0.90)=0.081\) and \(P(AAD)=(0.90)^2(0.10)=0.081\). b) \(P(X\ge 2)=P(X=2)+P(X=3)=\binom{3}{2}(0.10)^2(0.90)+(0.10)^3=0.028\). c) \(P(X=0)=0.729\), \(P(X=1)=0.243\), \(P(X=2)=0.027\), and \(P(X=3)=0.001\).

Answer

a) Both outcomes contain the same probability 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

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

Answer

a) About \(0.2924\), or \(29.24\%\) b) About \(0.0044\), or \(0.44\%\)
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

a) The probability of no win on one spin is \(0.90\). For five spins, \(P(\text{no wins})=(0.90)^5=0.59049\), so \(P(\text{at least one win})=1-0.59049=0.40951\). No wins is more likely. b) For every finite \(n\), the probability \((0.90)^n\) of losing every spin is greater than \(0\). Therefore, \(1-(0.90)^n<1\). c) Require \(1-(0.90)^n>0.80\), so \((0.90)^n<0.20\). Then \(n>\frac{\ln(0.20)}{\ln(0.90)}\approx 15.2755\), so the least whole number is \(16\).

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 positive probability, so the probability of at least one win remains below \(100\%\). c) At least \(16\) spins are required.
52708912
A running-shoe factory has a defect rate of \(4\%\). Assume the defect status of each selected shoe is independent of the others. A random sample of \(150\) shoes is inspected, and \(X\) is the number of defective shoes. a) State the binomial model for \(X\), including \(n\) and \(p\). b) Find the probability that exactly \(5\) shoes are defective. c) Find the probability that at least \(4\) but fewer than \(10\) shoes are defective.

Hints

- Check the four binomial conditions before choosing a formula. - For exactly \(5\) defective shoes, account for both the defective and nondefective shoes in the sample. - Translate “at least \(4\) but fewer than \(10\)” into the integer values that \(X\) can take.

Solution

1. The number of trials is fixed at \(150\), each shoe is defective or not defective, the defect probability is constant at \(0.04\), and the selections are independent. Thus \(X\sim\operatorname{Binomial}(150,0.04)\). 2. For exactly \(5\) defective shoes, \(P(X=5)=\binom{150}{5}(0.04)^5(0.96)^{145}\approx0.1628\). 3. “At least \(4\) but fewer than \(10\)” means \(4\le X\le9\), so \(P(4\le X\le9)=\sum_{k=4}^{9}\binom{150}{k}(0.04)^k(0.96)^{150-k}\approx0.7745\).

Answer

a) \(X\sim\operatorname{Binomial}(150,0.04)\), so \(n=150\) and \(p=0.04\). b) \(P(X=5)\approx0.1628\) c) \(P(4\le X\le9)\approx0.7745\)
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) Selin 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. Selin’s statement is correct. Once a specific event \(E\) is chosen, every outcome either belongs to \(E\) or to its complement \(E^c\), creating 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
The probability tree is the only description of a repeated experiment. Each edge label is the probability of the outcome written at the end of that edge. a) How many stages does the experiment have? Then list every complete path containing exactly two \(W\) outcomes and write the probability of each path. b) Add and simplify the probabilities from part a). c) Identify the entry in Pascal's triangle that matches the numerical coefficient in your simplified expression. Explain why the tree produces that coefficient.
Figure for problem 537463

Hints

- Count how many branch decisions occur from the root to any terminal node. - Trace complete paths and keep only those containing exactly two \(W\) labels. - Multiply the edge labels along each retained path before combining the probabilities. - Compare the number of retained paths with an entry in the corresponding row of Pascal's triangle.

Solution

1. Reading from root to terminal nodes shows three stages. The complete paths with exactly two \(W\) outcomes are \((W,W,D)\), \((W,D,W)\), and \((D,W,W)\). 2. The edge labels on each of these paths multiply to \(p^2q\). Therefore, their total probability is \(p^2q+p^2q+p^2q=3p^2q\). 3. The coefficient \(3\) is \(\binom{3}{2}\), the entry at \(n=3\), \(k=2\) in Pascal's triangle. It counts the three tree paths obtained by choosing which two of the three stages have outcome \(W\).

Answer

a) Three stages; \((W,W,D)\), \((W,D,W)\), and \((D,W,W)\), each with probability \(p^2q\). b) \(3p^2q\) c) \(\binom{3}{2}=3\), the \(n=3\), \(k=2\) entry; 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.

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 \(\binom{3}{2}=3\) position patterns. Its probability is \(3\cdot(0.1)^2\cdot0.9=0.027\). 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

\(0.028=2.8\%\)
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. \(X\) has a binomial distribution with \(n=15\) and \(p=0.40\). Therefore, \(P(X=6)=\binom{15}{6}(0.40)^6(0.60)^9\approx 0.2066\). 2. Using \(P(X=k)=\binom{15}{k}(0.40)^k(0.60)^{15-k}\), the probabilities for \(k=0\) and \(k=1\) are below \(0.01\), 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 below \(0.01\). Thus \(k\in\{0,1,11,12,13,14,15\}\).

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 \(q\).

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: \(2q(1-q)=0.42\), so \(q^2-q+0.21=0\). 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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