An insurance company sells a policy that covers a specific type of minor surgery. The probability that a policyholder will need this surgery in a given year is 0.05. The company has 1,000 policyholders. If the surgery costs the company $2,000, what is the expected total payout for the company for these policyholders for the year?
Answer: $100,000
The problem asks for the expected total payout. This can be found by first calculating the expected number of surgeries and then multiplying by the cost per surgery. The number of surgeries, X, follows a binomial distribution B(1000, 0.05). \\ The expected number of surgeries is the mean of the distribution: E[X] = μ=np=1000×0.05=50. \\ The expected total payout is the expected number of surgeries multiplied by the cost of each: Expected Payout = 50 surgeries $2,000/surgery = $100,000. Distractor A is the expected number of claims, not the payout. Distractor B is the cost of a single claim. Distractor D is based on the number of policyholders who are not expected to have a claim (950 $2,000).
Consider two sampling scenarios: \\ I. A city has 10,000 households, 40% of which have a pet. An analyst randomly selects 50 households for a survey. Let X be the number of selected households with a pet. \\ II. A committee of 20 people has 8 members from Party A. A subcommittee of 4 people is selected at random without replacement. Let Y be the number of subcommittee members from Party A. \\ Which of the following statements is correct regarding the use of a binomial distribution to model X and Y?
Answer: A binomial model is appropriate for I but not for II.
Both scenarios involve sampling without replacement, which technically violates the independence condition of the binomial distribution. However, when the sample size is small relative to the population size (typically, when the sample is no more than 10% of the population), the effect on the probabilities from trial to trial is negligible, and a binomial distribution can be used as a good approximation. In Scenario I, the sample size (50) is only 0.5% of the population (10,000), so the binomial model is appropriate. In Scenario II, the sample size (4) is 20% of the population (20). This is a significant fraction, so the probabilities will change substantially with each selection. Thus, a binomial model is not appropriate for II; the correct model is the hypergeometric distribution.
A student is completely unprepared for a 10-question true/false quiz and guesses on every question. What is the approximate probability that the student gets at most 8 questions correct?
Answer: 0.989
Let X be the number of correct answers. X follows a binomial distribution with n=10 and p=0.5 (for random guessing on true/false). We want to find P(X ≤ 8). Calculating this directly would require summing P(X=0) through P(X=8). It is much faster to use the complement: P(X ≤ 8) = 1 - P(X > 8) = 1 - [P(X=9) + P(X=10)]. \\ P(X=9) = \(\binom{10}{9}(0.5)^9(0.5)^1 = 10(0.5)^{10} = 10/1024\\) \\ P(X=10) = \(\binom{10}{10}(0.5)^{10}(0.5)^0 = 1(0.5)^{10} = 1/1024\\) \\ P(X > 8) = 10/1024 + 1/1024 = 11/1024. \\ P(X ≤ 8) = 1 - 11/1024 = 1013/1024 ≈ 0.989. Distractor A is P(X=8) only. Distractor B is P(X>8), the complement itself. Distractor C is 1 - P(X=8).
A quality control inspector examines a small batch of 50 microchips, of which 10 are defective. The inspector randomly selects 5 chips for testing without replacement. Let X be the number of defective chips selected. A student decides to model X using a binomial distribution. Which statement best explains why this is an inappropriate application of the binomial distribution?
Answer: The probability of selecting a defective chip is not constant across the 5 trials.
A key condition for a binomial distribution is that the probability of success (p) must be constant for each trial. Because the inspector is sampling without replacement from a small population (50 chips), the probability of selecting a defective chip changes with each draw. For example, P(1st is defective) = 10/50. If the first was defective, P(2nd is defective) = 9/49. This violates the constant probability condition. While this violation can sometimes be ignored if the sample is less than 10% of the population, here the sample size (5) is exactly 10% of the population size (50), making the change in probability non-negligible. The more appropriate distribution would be the hypergeometric distribution.
A student takes a 20-question multiple-choice test. Each question has four options. The student guesses randomly on every question. However, due to fatigue, the probability of the student reading a question carefully enough to eliminate one obviously wrong answer before guessing is 0.5 for the first 10 questions, but drops to 0.2 for the last 10 questions. If the student eliminates one option, they guess from the remaining three; otherwise, they guess from all four. Why is a single binomial distribution B(20,p) inappropriate for modeling the total number of correct answers?
Answer: The probability of answering a question correctly is not the same for all trials.
A condition for the binomial distribution is that the probability of success, p, must be constant for all trials. Here, the probability of answering a question correctly changes. For the first 10 questions, P(Correct) = P(Eliminate)P(Correct|Eliminate) + P(Not Eliminate)P(Correct|Not Eliminate) = (0.5)(1/3) + (0.5)(1/4) ≈ 0.292. For the last 10 questions, P(Correct) = (0.2)(1/3) + (0.8)(1/4) ≈ 0.267. Since the probability of success is not constant for all 20 trials, a single binomial distribution is inappropriate. This setup is a Poisson-binomial distribution.
In a certain plant species, the allele for purple flowers (P) is dominant over the allele for white flowers (p). When two heterozygous (Pp) plants are crossed, the probability of an offspring having purple flowers is 0.75. An experimenter grows a number of these offspring, and the expected number of purple-flowered plants is 120. How many offspring were grown in total?
Answer: 160
The expected value (mean) of a binomial distribution is given by the formula μ=np, where n is the number of trials and p is the probability of success. In this context, n is the total number of offspring, p is the probability of having purple flowers (0.75), and μ is the expected number of purple-flowered plants (120). We need to solve for n: \\ 120 = n * 0.75 \\ n = 120 / 0.75 = 160. So, 160 offspring were grown in total. Distractor A is the result of incorrectly multiplying 120 by 0.75. Distractor D is the result of incorrectly dividing 120 by the failure probability, 0.25.
In a carnival game, a player throws darts at a board. The probability of hitting the bullseye on any given throw is 0.2. A player throws 5 darts. To win a prize, the player must hit the bullseye at least 4 times. What is the probability that a player wins a prize?
Answer: 0.00672
Let X be the number of bullseyes. X follows a binomial distribution with n=5 and p=0.2. Winning requires hitting the bullseye at least 4 times, which means X=4 or X=5. We must calculate P(X=4) + P(X=5). \\ P(X=4) = \(\binom{5}{4}(0.2)^4(0.8)^1 = 5(0.0016)(0.8) = 0.0064\\) \\ P(X=5) = \(\binom{5}{5}(0.2)^5(0.8)^0 = 1(0.00032)(1) = 0.00032\\) \\ P(Win) = P(X≥4) = 0.0064 + 0.00032 = 0.00672. Distractor A is only P(X=4). Distractor C is P(X≥3). Distractor D is the result of incorrectly swapping p=0.2 and q=0.8 in the calculation for P(X=4).
A batch of 10 electronic components is tested. Each component has an independent probability of 0.1 of being defective. Given that at least one component in the batch is defective, what is the approximate probability that exactly one component is defective?
Answer: 0.595
This is a conditional probability problem: P(X=1 | X ≥ 1). The formula is P(X=1 and X ≥ 1) / P(X ≥ 1). Since the event {X=1} is a subset of {X ≥ 1}, the intersection is just {X=1}. So we need to compute P(X=1) / P(X ≥ 1). The distribution is Binomial(n=10, p=0.1). \\ Numerator: P(X=1) = \(\binom{10}{1}(0.1)^1(0.9)^9 = 10(0.1)(0.3874) \approx 0.3874\\). \\ Denominator: P(X ≥ 1) = 1 - P(X=0). \\ P(X=0) = \(\binom{10}{0}(0.1)^0(0.9)^{10} \approx 0.3487\\). \\ So, P(X ≥ 1) = 1 - 0.3487 = 0.6513. \\ Conditional Probability = 0.3874 / 0.6513 ≈ 0.5948. Distractor A is the unconditional probability P(X=1). Distractor B is P(X=0). Distractor C is the probability of the condition, P(X≥1).
Three different researchers are independently studying a rare disease. For each researcher, the probability of making a key discovery within a year is 0.3. What is the probability that at least one researcher makes the key discovery within a year?
Answer: 0.657
When you encounter probability questions asking about "at least one" success among independent events, the most efficient approach is using the complement rule. Instead of calculating all the ways at least one researcher succeeds, calculate the probability that none succeed, then subtract from 1.
Since each researcher has a 0.3 probability of making the discovery, each has a 0.7 probability of not making it. Because the researchers work independently, you multiply these probabilities: the chance that all three fail is 0.7×0.7×0.7=0.343.
Therefore, the probability that at least one succeeds is 1−0.343=0.657. This confirms answer A is correct.
Looking at the wrong answers: B (0.027) represents the probability that all three researchers make the discovery, calculated as 0.33=0.027. This is a common confusion between "at least one" and "all succeed." C (0.900) appears to come from incorrectly adding the individual probabilities (0.3×3=0.9), but this ignores the possibility of overlap when events aren't mutually exclusive. D (0.343) is the probability that none succeed – the complement of what we want.
Remember this strategy: for "at least one" problems with independent events, always consider using the complement rule. Calculate the probability of complete failure, then subtract from 1. This approach is usually much simpler than trying to account for all the different ways success can occur.
A machine produces light bulbs with a 10% defect rate. In a random sample of 8 bulbs, what is the approximate probability that the number of defective bulbs is either 2 or 3?
Answer: 0.182
Let X be the number of defective bulbs. X follows a binomial distribution with n=8 and p=0.1. We need to find P(X=2 or X=3), which is P(X=2) + P(X=3). \\ P(X=2) = \(\binom{8}{2}(0.1)^2(0.9)^6 = 28(0.01)(0.531441) \approx 0.1488\\) \\ P(X=3) = \(\binom{8}{3}(0.1)^3(0.9)^5 = 56(0.001)(0.59049) \approx 0.0331\\) \\ The total probability is 0.1488 + 0.0331 = 0.1819, which is approximately 0.182. Distractor A is only P(X=2). Distractor B is only P(X=3). Distractor D results from forgetting the binomial coefficients in the calculation.
Let X be a binomial random variable with parameters n=20 and p=0.8. Let Y be a binomial random variable with parameters n=20 and p=0.2. Which statement accurately compares the probability distributions of X and Y?
Answer: The distribution of X is a mirror image of the distribution of Y; that is, P(X=k)=P(Y=20−k) for 0≤k≤20.
The distribution of X (p=0.8 > 0.5) will be skewed to the left, centered at μX=20(0.8)=16. The distribution of Y (p=0.2 < 0.5) will be skewed to the right, centered at μY=20(0.2)=4. They are not identical or symmetric (since n(1−p)=4 for X and np=4 for Y, which are less than the common rule-of-thumb value of 10). Let's check the mirror-image property: \\ P(X=k)=(k20)(0.8)k(0.2)20−k \\ P(Y=20−k)=(20−k20)(0.2)20−k(0.8)20−(20−k)=(k20)(0.2)20−k(0.8)k. \\ Since (kn)=(n−kn), the expressions are identical. Thus, the distributions are mirror images of each other.
A hospital emergency room admits patients at an average rate of 3 per hour. A statistician wants to calculate the probability that the ER admits exactly 5 patients between 10:00 AM and 11:00 AM. Which statement best explains why a binomial distribution is not the most appropriate model for this scenario?
Answer: The number of trials is not defined or fixed, and events can occur at any point in a continuous interval.
The primary reason a binomial distribution is inappropriate here is that it requires a fixed number of discrete trials (n). This scenario involves counting events over a continuous interval of time. There is no natural definition of a 'trial'. Is a trial every second? Every minute? The number of potential 'trials' is effectively infinite. This type of problem, counting occurrences over a fixed interval, is correctly modeled by a Poisson distribution, which does not require a fixed number of trials.
A new drug is effective in 70% of patients. In a trial with n=100 patients, let X be the number for whom the drug is effective. In a second trial with a different drug that is effective in 30% of patients, m=200 patients are enrolled. Let Y be the number for whom this second drug is effective. Which of the following statements about the standard deviations of the number of effective cases, σX and σY, is true?
Answer: σY>σX, because the variance np(1−p) is larger for the second trial.
The standard deviation of a binomial random variable is σ=np(1−p). \\ For drug X: σX=100(0.7)(0.3)=21≈4.58. \\ For drug Y: σY=200(0.3)(0.7)=42≈6.48. \\ Since 42>21, σY>σX. The variance is maximized when p=0.5, and although drug X has a higher p, the larger sample size for drug Y results in a larger overall variance and standard deviation in this case.
A public health official is studying the spread of a contagious virus in a small, isolated community of 30 people. The official wishes to model the number of people who will contract the virus in the next month using a binomial distribution. Which of the following is the most significant reason this model is likely inappropriate?
Answer: The probability of one person contracting the virus is likely not independent of whether others contract it.
One of the fundamental conditions for a binomial distribution is that the trials must be independent. In the context of a contagious disease spreading through a small community, this assumption is violated. If one person becomes infected, they are more likely to transmit the virus to others in the community, especially those they are in close contact with. Therefore, the outcome for one person (trial) is dependent on the outcomes of others, making the binomial model inappropriate.
A binomial distribution has parameters n=50 and p=0.2. If the number of trials n is increased to 200 while p remains the same, how will the shape and standard deviation of the distribution change?
Answer: The distribution will become more symmetric, and the standard deviation will increase.
Let's analyze the shape and standard deviation for both cases. \\ Original: n=50, p=0.2. σ=50(0.2)(0.8)=8≈2.83. For shape, we check np=10 and n(1−p)=40. Since np≥10, it's reasonably symmetric. \\ New: n=200, p=0.2. σ=200(0.2)(0.8)=32≈5.66. For shape, np=40 and n(1−p)=160. Both values are large, indicating the distribution will be more symmetric and bell-shaped, resembling a normal distribution more closely. \\ Comparing the two, the standard deviation increases (from 8 to 32), and the shape becomes more symmetric. Therefore, option D is correct.
A genetic trait appears in 25% of the offspring of a certain cross. An investigator examines 12 offspring from such crosses. If X represents the number of offspring exhibiting the trait, which binomial probability calculation would determine the likelihood that the trait appears in fewer than 3 of the offspring?
Answer: ∑k=02(k12)(0.25)k(0.75)12−k
The number of offspring with the trait, X, follows a binomial distribution with n=12 trials and p=0.25 probability of success (exhibiting the trait). The phrase 'fewer than 3' means X < 3, which corresponds to X being 0, 1, or 2. Therefore, we need to sum the probabilities for k=0, k=1, and k=2. The correct formula for this sum is ∑k=02P(X=k), which is ∑k=02(k12)(0.25)k(0.75)12−k. Distractor B represents 'at most 3' (X ≤ 3). Distractor C represents 'exactly 3'. Distractor D incorrectly swaps the probabilities of success (p) and failure (1-p).
A researcher is screening potential drug compounds. Each compound tested has a 0.1 probability of being a 'hit'. The researcher decides to test compounds one by one until they have found 2 'hits'. Why is a binomial distribution not suitable for modeling the total number of compounds the researcher will have to test?
Answer: The number of trials is not fixed in advance; it is the random variable of interest.
When analyzing probability distributions, you need to carefully examine whether your situation meets all the required conditions. The binomial distribution has four key requirements: fixed number of trials, independent trials, binary outcomes, and constant probability of success.
In this scenario, the researcher is testing compounds until finding exactly 2 hits. This means the number of trials isn't predetermined—it could take 2 tests (if the first two are hits) or potentially many more tests. The researcher stops when they reach their target of 2 hits, making the total number of tests the unknown quantity we're trying to model. This describes a negative binomial distribution (also called a Pascal distribution), not a binomial distribution.
Answer A correctly identifies that the number of trials is random rather than fixed in advance—this violates the fundamental binomial requirement of a predetermined sample size.
Answer B incorrectly suggests the trials aren't independent. The problem states each compound has a 0.1 probability of being a hit, implying independence between tests.
Answer C is wrong because probability values don't affect whether binomial distribution is appropriate. Low probabilities (like 0.1) are perfectly valid for binomial models—they don't make the model inaccurate.
Answer D mischaracterizes the outcomes. The problem clearly describes binary outcomes: each compound is either a "hit" or not a "hit."
Study tip: When you see "until" or "stopping when a condition is met" in probability problems, think negative binomial or geometric distributions, not binomial. The binomial distribution requires knowing your sample size upfront.
The probability that a customer service call is resolved on the first attempt is 0.8. A manager reviews a random sample of 6 recent calls. The probability that exactly 4 of these calls were resolved on the first attempt is given by the expression (46)(0.8)4(0.2)2. What does the term (0.8)4 represent in this context?
Answer: The probability of observing any single, specific sequence of 4 successes and 2 failures.
The binomial probability formula consists of two parts. The first part, (kn), is the number of combinations, representing the number of ways to arrange k successes among n trials. The second part, pk(1−p)n−k, represents the probability of any one specific sequence of k successes and n-k failures occurring. In this case, (0.8)4(0.2)2 is the probability of one such sequence (e.g., SSSSF F). The term (0.8)4 by itself represents the probability of 4 specific trials being successes. Option B is the most complete description of the probability part of the term.
A basketball player makes 60% of her free throws. Assuming all free throw attempts are independent, what is the approximate probability that she makes at least two of her next 5 attempts?
Answer: 0.913
Let X be the number of successful free throws. X follows a binomial distribution with n=5 trials and p=0.6 probability of success. The question asks for P(X ≥ 2). It is easier to calculate the complement, P(X < 2), and subtract from 1. \\ P(X < 2) = P(X=0) + P(X=1). \\ P(X=0) = \(\binom{5}{0}(0.6)^0(0.4)^5 = 1(1)(0.01024) = 0.01024\\) \\ P(X=1) = \(\binom{5}{1}(0.6)^1(0.4)^4 = 5(0.6)(0.0256) = 0.0768\\) \\ P(X < 2) = 0.01024 + 0.0768 = 0.08704. \\ Therefore, P(X ≥ 2) = 1 - P(X < 2) = 1 - 0.08704 = 0.91296, which is approximately 0.913. Distractor A is P(X=2) only. Distractor B is P(X ≤ 2). Distractor C is 1 - P(X ≤ 2), or P(X > 2).
