Business Statistics Quiz: Binomial And Geometric Models
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Binomial And Geometric ModelsQuestion 1 of 20

An inspector samples 5 items from a large batch with a 30% defect rate. It is known that the sample contains at least one defective item. Given this information, what is the probability that the sample contains exactly two defective items?

0.309
0.371
0.832
0.857
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Business Statistics Quiz

Business Statistics Quiz: Binomial And Geometric Models

Practice Binomial And Geometric Models in Business Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Binomial And Geometric Models, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

An inspector samples 5 items from a large batch with a 30% defect rate. It is known that the sample contains at least one defective item. Given this information, what is the probability that the sample contains exactly two defective items?

  1. 0.309
  2. 0.371 (correct answer)
  3. 0.832
  4. 0.857
Explanation: This is a conditional probability problem using the binomial distribution (n=5,p=0.3n=5, p=0.3). Let A be the event X=2X=2 (exactly two defects) and B be the event X≥1X \ge 1 (at least one defect). We want P(A∣B)=P(A∩B)/P(B)P(A|B) = P(A \cap B) / P(B). Since A is a subset of B, A∩B=AA \cap B = A. So we need to calculate P(X=2)/P(X≥1)P(X=2) / P(X \ge 1). Numerator: P(X=2)=(52)(0.3)2(0.7)3=10(0.09)(0.343)=0.3087P(X=2) = \binom{5}{2}(0.3)^2(0.7)^3 = 10(0.09)(0.343) = 0.3087. Denominator: P(X≥1)=1−P(X=0)=1−(50)(0.3)0(0.7)5=1−0.16807=0.83193P(X \ge 1) = 1 - P(X=0) = 1 - \binom{5}{0}(0.3)^0(0.7)^5 = 1 - 0.16807 = 0.83193. Conditional probability: 0.3087/0.83193≈0.37110.3087 / 0.83193 \approx 0.3711.

Question 2

A factory produces 10,000 plastic bottles per day, and the probability of a bottle having a minor defect is 0.08. What is the approximate probability that between 770 and 820 bottles (inclusive) are defective on a given day, using a normal approximation to the binomial distribution?

  1. 0.55
  2. 0.65 (correct answer)
  3. 0.75
  4. 0.87
Explanation: First, verify the normal approximation is appropriate: np=10000×0.08=800>10np = 10000 \times 0.08 = 800 > 10 and n(1−p)=9200>10n(1-p) = 9200 > 10. The mean is μ=np=800\mu = np = 800. The standard deviation is σ=np(1−p)=10000×0.08×0.92=736≈27.13\sigma = \sqrt{np(1-p)} = \sqrt{10000 \times 0.08 \times 0.92} = \sqrt{736} \approx 27.13. We want P(770≤X≤820)P(770 \le X \le 820). Applying the continuity correction, we find the probability for the range [769.5, 820.5]. Convert to z-scores: z1=(769.5−800)/27.13≈−1.12z_1 = (769.5 - 800) / 27.13 \approx -1.12 and z2=(820.5−800)/27.13≈0.76z_2 = (820.5 - 800) / 27.13 \approx 0.76. We need P(−1.12≤Z≤0.76)P(-1.12 \le Z \le 0.76), which is Φ(0.76)−Φ(−1.12)=0.7764−0.1314=0.645\Phi(0.76) - \Phi(-1.12) = 0.7764 - 0.1314 = 0.645.

Question 3

A financial advisor finds that the probability of converting a prospect into a client is 0.10 on any given meeting. What is the probability that the first conversion occurs after the 3rd meeting but on or before the 6th meeting?

  1. 0.198 (correct answer)
  2. 0.279
  3. 0.531
  4. 0.729
Explanation: This scenario follows a geometric distribution with success probability p=0.10p=0.10. We want the probability that the first success (conversion) occurs on trial X=4,5,or 6X=4, 5, \text{or } 6. We can calculate these probabilities and sum them. The formula is P(X=k)=(1−p)k−1pP(X=k) = (1-p)^{k-1}p. P(X=4)=(0.9)3(0.1)=0.0729P(X=4) = (0.9)^3(0.1) = 0.0729 P(X=5)=(0.9)4(0.1)=0.06561P(X=5) = (0.9)^4(0.1) = 0.06561 P(X=6)=(0.9)5(0.1)=0.059049P(X=6) = (0.9)^5(0.1) = 0.059049 The total probability is 0.0729+0.06561+0.059049≈0.19760.0729 + 0.06561 + 0.059049 \approx 0.1976. Alternatively, this is P(X>3)−P(X>6)=(0.9)3−(0.9)6=0.729−0.531441≈0.198P(X > 3) - P(X > 6) = (0.9)^3 - (0.9)^6 = 0.729 - 0.531441 \approx 0.198.

Question 4

An insurance agent contacts 8 potential clients. The probability of selling a policy to any one client is 0.2. The agent earns a $500 commission for each policy sold, but incurs a fixed cost of $600 for the batch of 8 contacts. What is the probability that the agent incurs a net loss from this batch of contacts?

  1. 0.168
  2. 0.336
  3. 0.503 (correct answer)
  4. 0.797
Explanation: A net loss occurs if total commission is less than the fixed cost. Commission per sale is $500. Cost is $600. Let XX be the number of sales. If X=0X=0, revenue = $0, loss = $600. If X=1X=1, revenue = $500, loss = $100. If X=2X=2, revenue = $1000, profit = $400. A net loss occurs if X=0X=0 or X=1X=1. This is a binomial problem with n=8,p=0.2n=8, p=0.2. We need P(X≤1)=P(X=0)+P(X=1)P(X \le 1) = P(X=0) + P(X=1). P(X=0)=(80)(0.2)0(0.8)8≈0.1678P(X=0) = \binom{8}{0}(0.2)^0(0.8)^8 \approx 0.1678 P(X=1)=(81)(0.2)1(0.8)7≈0.3355P(X=1) = \binom{8}{1}(0.2)^1(0.8)^7 \approx 0.3355 The probability of a net loss is 0.1678+0.3355=0.50330.1678 + 0.3355 = 0.5033.

Question 5

A market researcher is surveying 100 customers selected from a small, local store with only 250 loyalty members. Customers are not replaced in the sampling pool after being surveyed. The researcher is modeling the number of "highly satisfied" customers using a binomial distribution. Which of the following assumptions of the binomial model is most likely violated in this scenario?

  1. The number of trials is not fixed.
  2. There are more than two possible outcomes for each trial.
  3. The probability of success, p, is unknown.
  4. The trials are not independent of each other. (correct answer)
Explanation: When you encounter questions about binomial distributions, you need to verify that all four key assumptions are met: fixed number of trials, exactly two outcomes per trial, constant probability of success, and independence between trials. In this scenario, the researcher is sampling 100 customers from a small pool of only 250 loyalty members without replacement. This creates a dependency problem. When the first customer is surveyed and removed from the pool, it changes the composition of the remaining population. If that customer was "highly satisfied," there's now one fewer satisfied customer available for subsequent selections, which slightly alters the probability for all remaining trials. Each selection affects the probability of success for the next selection. Let's examine why the other options don't represent violations: Option A is incorrect because the number of trials is clearly fixed at 100 customers. Option B is wrong since each customer can only be classified as either "highly satisfied" or "not highly satisfied" – exactly two outcomes. Option C misses the point entirely; while the true probability might be unknown to the researcher, this doesn't violate the binomial assumption that the probability remains constant throughout the experiment. Option D correctly identifies the independence violation. Without replacement sampling from a small population creates dependencies between trials, making the binomial model inappropriate. Study tip: Remember the "10% rule" – binomial approximation is generally acceptable when sampling without replacement only if your sample size is less than 10% of the population. Here, 100/250 = 40%, far exceeding this threshold.

Question 6

A software development team estimates a 60% chance of successfully completing a feature in a one-week sprint. Sprints are independent. What is the probability that the team's third successful feature completion occurs in the fifth sprint?

  1. 0.078
  2. 0.207 (correct answer)
  3. 0.346
  4. 0.600
Explanation: This is a negative binomial problem that can be solved with binomial reasoning. For the 3rd success to occur in the 5th sprint, two conditions must be met: 1) The 5th sprint must be a success (probability = 0.6). 2) The first four sprints must contain exactly two successes. We can calculate the probability of the second condition using the binomial formula with n=4n=4, k=2k=2, and p=0.6p=0.6: P(2 successes in 4 sprints)=(42)(0.6)2(0.4)2=6×0.36×0.16=0.3456P(\text{2 successes in 4 sprints}) = \binom{4}{2}(0.6)^2(0.4)^2 = 6 \times 0.36 \times 0.16 = 0.3456. Since the sprints are independent, we multiply the probabilities of the two conditions: 0.3456×0.6=0.207360.3456 \times 0.6 = 0.20736.

Question 7

A factory produces electronic chips in batches of 500. The long-run probability of a chip being defective is 0.02. A batch is flagged for having unusually high variability if the number of defects is more than two standard deviations above the mean. What is the minimum integer number of defects required for a batch to be flagged?

  1. 16
  2. 17 (correct answer)
  3. 20
  4. 30
Explanation: This is a binomial distribution with n=500n=500 and p=0.02p=0.02. The mean is μ=np=500×0.02=10\mu = np = 500 \times 0.02 = 10. The variance is σ2=np(1−p)=500×0.02×0.98=9.8\sigma^2 = np(1-p) = 500 \times 0.02 \times 0.98 = 9.8. The standard deviation is σ=9.8≈3.13\sigma = \sqrt{9.8} \approx 3.13. The threshold for being flagged is μ+2σ=10+2(3.13)=10+6.26=16.26\mu + 2\sigma = 10 + 2(3.13) = 10 + 6.26 = 16.26. Since the number of defects must be an integer and the condition is "more than" 16.26, the minimum integer number of defects is 17.

Question 8

A manufacturing process produces defective items with probability 0.06. Quality inspectors sample 50 items. Using normal approximation to the binomial, what is the probability that the number of defective items differs from the expected value by more than 2 standard deviations?

  1. Approximately 0.05, since this represents the typical Type I error rate in hypothesis testing
  2. Approximately 0.32, calculated as 2×P(Z>2)=2×0.0228=0.04562 \times P(Z > 2) = 2 \times 0.0228 = 0.0456
  3. Approximately 0.05, calculated as 2×P(Z>2)=2×0.0228=0.04562 \times P(Z > 2) = 2 \times 0.0228 = 0.0456 (correct answer)
  4. Approximately 0.95, since this represents the area within 2 standard deviations of the mean
Explanation: With n=50, p=0.06, we have μ=3 and σ=√(50×0.06×0.94)≈1.68. We want P(|X-μ|>2σ) = P(Z<-2 or Z>2) = 2×P(Z>2) ≈ 2×0.0228 ≈ 0.0456. Choice A gives the right answer but wrong reasoning (not about Type I error). Choice B has wrong final calculation. Choice D gives the complement probability.

Question 9

A marketing firm observes that 15% of visitors to a client's landing page make a purchase. The firm launches a campaign that drives 20 new, independent visitors to the page. The campaign is considered a failure if two or fewer visitors make a purchase. What is the approximate probability that the campaign is considered a failure?

  1. 0.176
  2. 0.229
  3. 0.405 (correct answer)
  4. 0.595
Explanation: This scenario follows a binomial distribution with n=20n = 20 trials (visitors) and a success probability of p=0.15p = 0.15 (purchase). The campaign is a failure if the number of successes, XX, is X≤2X \le 2. We need to calculate P(X≤2)=P(X=0)+P(X=1)+P(X=2)P(X \le 2) = P(X=0) + P(X=1) + P(X=2). P(X=0)=(200)(0.15)0(0.85)20≈0.0388P(X=0) = \binom{20}{0}(0.15)^0(0.85)^{20} \approx 0.0388 P(X=1)=(201)(0.15)1(0.85)19≈0.1368P(X=1) = \binom{20}{1}(0.15)^1(0.85)^{19} \approx 0.1368 P(X=2)=(202)(0.15)2(0.85)18≈0.2293P(X=2) = \binom{20}{2}(0.15)^2(0.85)^{18} \approx 0.2293 Summing these probabilities gives 0.0388+0.1368+0.2293=0.4049≈0.4050.0388 + 0.1368 + 0.2293 = 0.4049 \approx 0.405.

Question 10

A manufacturer produces widgets with a 4% defect rate. A quality control inspector checks widgets one by one until a defective one is found. The cost of inspecting each widget is $2.00. A new process is proposed that would reduce the defect rate to 3% but increase the inspection cost to $2.50 per widget. What is the expected change in the total inspection cost to find the first defect if the new process is adopted?

  1. An increase of $8.33
  2. A decrease of $16.67
  3. An increase of $33.33 (correct answer)
  4. A decrease of $4.17
Explanation: This scenario is modeled by the geometric distribution, where the expected number of trials until the first success is E[X]=1/pE[X] = 1/p. For the current process, p1=0.04p_1 = 0.04, so the expected number of inspections is 1/0.04=251/0.04 = 25. The expected cost is 25 \times \2.00 = $50.00.Forthenewprocess,. For the new process, p_2 = 0.03,sotheexpectednumberofinspectionsis, so the expected number of inspections is 1/0.03 \approx 33.33.Theexpectedcostis. The expected cost is 33.33 \times $2.50 = $83.33.Thechangeinexpectedcostisanincreaseof. The change in expected cost is an increase of $83.33 - $50.00 = $33.33$.

Question 11

An e-commerce site has a 8% conversion rate for visitors who view a product page. A data scientist is modeling the number of page views until the 3rd conversion occurs. If X represents this random variable, what is the coefficient of variation for X?

  1. CV=3(1−0.08)3/0.08=2.7637.5≈0.044\text{CV} = \frac{\sqrt{3(1-0.08)}}{3/0.08} = \frac{\sqrt{2.76}}{37.5} \approx 0.044
  2. CV=3(1−0.08)/0.0823/0.08=431.2537.5≈0.555\text{CV} = \frac{\sqrt{3(1-0.08)/0.08^2}}{3/0.08} = \frac{\sqrt{431.25}}{37.5} \approx 0.555 (correct answer)
  3. CV=3×0.083/0.08=0.2437.5≈0.013\text{CV} = \frac{\sqrt{3 \times 0.08}}{3/0.08} = \frac{\sqrt{0.24}}{37.5} \approx 0.013
  4. CV=37.537.5=137.5≈0.163\text{CV} = \frac{\sqrt{37.5}}{37.5} = \frac{1}{\sqrt{37.5}} \approx 0.163
Explanation: For negative binomial distribution with r=3 successes and p=0.08, E[X]=r/p=3/0.08=37.5 and Var(X)=r(1-p)/p²=3(0.92)/(0.08)²=431.25. Therefore CV=σ/μ=√431.25/37.5≈0.555. Choice A uses wrong variance formula. Choice C uses p instead of (1-p). Choice D oversimplifies the calculation incorrectly.

Question 12

A streaming service models customer churn using a geometric distribution. There is a 15% probability that a customer will cancel their subscription in any given month, independent of previous months. A promotion is offered to customers who stay subscribed for at least one full year (12 months). What is the probability that a new customer will qualify for this promotion?

  1. 0.021
  2. 0.142 (correct answer)
  3. 0.833
  4. 0.858
Explanation: To qualify for the promotion, a customer must not cancel their subscription for each of the first 12 months. The probability of cancellation in a given month is p=0.15p=0.15. The probability of not cancelling (i.e., renewing) in a given month is 1−p=0.851-p = 0.85. Since the months are independent, the probability of not cancelling for 12 consecutive months is (0.85)12(0.85)^{12}. Calculating this value gives (0.85)12≈0.1422(0.85)^{12} \approx 0.1422. This is the probability that the first cancellation occurs after month 12, P(X>12)P(X > 12).

Question 13

An e-commerce site runs two conversion experiments. Experiment 1 consists of showing an ad to 100 users, with a conversion probability p=0.1p=0.1. Experiment 2 shows a different ad to 200 users, also with a conversion probability p=0.1p=0.1. Which of the following statements correctly compares the resulting binomial distributions of the number of conversions?

  1. The distribution for Experiment 2 will have a larger mean and be less spread out.
  2. The distributions for both experiments will have the same degree of skewness.
  3. The distributions for both experiments will have the same variance since p is the same.
  4. The distribution for Experiment 2 will have a larger mean and be more spread out. (correct answer)
Explanation: When comparing binomial distributions, you need to examine how sample size (n) and probability (p) affect the three key characteristics: mean, variance, and shape. For any binomial distribution, the mean equals npnp and the variance equals np(1−p)np(1-p). In Experiment 1, with n=100 and p=0.1, the mean is 10 and variance is 9. In Experiment 2, with n=200 and p=0.1, the mean is 20 and variance is 18. Since Experiment 2 has both a larger mean and larger variance, it will be more spread out (higher standard deviation of 18≈4.24\sqrt{18} \approx 4.24 versus 9=3\sqrt{9} = 3). Answer D correctly identifies that Experiment 2 has both a larger mean and greater spread. Answer A incorrectly claims Experiment 2 will be "less spread out" – this confuses the concept that while the distribution becomes more normal-looking with larger n, the actual spread (measured by standard deviation) still increases when variance increases. Answer B is wrong because skewness depends on both n and p; while both experiments have the same p=0.1, the different sample sizes create different degrees of skewness. Answer C makes a fundamental error by suggesting equal variances – variance depends on both n and p, not just p alone. Study tip: Remember the binomial formulas: mean = np, variance = np(1-p). When sample size increases while p stays constant, both mean and variance increase proportionally, making the distribution more spread out despite becoming more symmetric.

Question 14

A sales representative has a 20% success rate for closing a deal on any given call, with calls being independent events. The representative's performance on the next set of calls is being analyzed. Which of the following business questions is best answered using a geometric model?

  1. What is the probability that the representative's second sale occurs on the eighth call?
  2. What is the average number of sales the representative can expect from the next 15 calls?
  3. What is the probability that the representative will need to make exactly 5 calls to get their first sale? (correct answer)
  4. What is the probability of the representative securing at least 3 sales in the next 15 calls?
Explanation: The geometric distribution models the number of trials required to achieve the first success. Option C directly asks this question. Option A involves the second success, which is modeled by the negative binomial distribution. Options B and D involve a fixed number of trials (15 calls) and are concerned with the number of successes within that fixed set, which are classic applications of the binomial distribution (expected value and cumulative probability, respectively).

Question 15

A web designer is A/B testing two page layouts. The established conversion rate for the original layout (Layout A) is 20%. The designer tests a new layout (Layout B) on a sample of 10 visitors. The probability of observing zero conversions among these 10 visitors with Layout B was found to be 0.0282. Assuming the number of conversions follows a binomial distribution, what is the approximate improvement in the conversion rate (as a percentage point difference) offered by Layout B over Layout A?

  1. 3.0%
  2. 7.0%
  3. 10.0% (correct answer)
  4. 30.0%
Explanation: Let pBp_B be the conversion rate for Layout B. The probability of zero conversions in n=10n=10 trials is given by P(X=0)=(100)pB0(1−pB)10=(1−pB)10P(X=0) = \binom{10}{0} p_B^0 (1-p_B)^{10} = (1-p_B)^{10}. We are given that this probability is 0.0282. So, (1−pB)10=0.0282(1-p_B)^{10} = 0.0282. To find 1−pB1-p_B, we take the 10th root of 0.0282: 1−pB=(0.0282)1/10≈0.701-p_B = (0.0282)^{1/10} \approx 0.70. This implies pB=1−0.70=0.30p_B = 1 - 0.70 = 0.30, or 30%. The conversion rate for Layout A is 20%. The improvement is the difference: 30%−20%=10%30\% - 20\% = 10\%.

Question 16

A company sends out promotional mailers. 2% of recipients make a purchase, and 5% request more information. These are mutually exclusive events. What is the expected number of mailers the company must send to get its first response of any kind (either a purchase or a request for information)?

  1. 7.0
  2. 14.3 (correct answer)
  3. 35.0
  4. 70.0
Explanation: This is a geometric distribution problem. A "success" is defined as getting any kind of response. Since the events (purchase, information request) are mutually exclusive, the total probability of success on any one trial is the sum of their individual probabilities: p=0.02+0.05=0.07p = 0.02 + 0.05 = 0.07. The expected number of trials until the first success in a geometric distribution is given by E[X]=1/pE[X] = 1/p. Therefore, the expected number of mailers is 1/0.07≈14.2861 / 0.07 \approx 14.286.

Question 17

A customer service center finds that 25% of incoming calls are for technical support. Calls are independent events. An operator has just handled four consecutive non-technical calls. What is the probability that the very next call will be for technical support?

  1. 0.079
  2. 0.250 (correct answer)
  3. 0.316
  4. 0.750
Explanation: This question tests the memoryless property of Bernoulli trials, which are the foundation of the geometric and binomial distributions. Because the trials (calls) are independent, the outcome of previous calls has no impact on the outcome of the next call. The probability that any given call is for technical support remains constant at 0.25.

Question 18

A quality control process involves testing items from a production line where 10% of items are defective. An inspector tests 10 items. If more than one item is defective, the entire batch is subjected to a full inspection at a cost of $1000. If one or zero items are defective, the batch passes at no extra cost. What is the expected cost of full inspection for any given batch of 10 items?

  1. $263.90 (correct answer)
  2. $387.40
  3. $612.60
  4. $736.10
Explanation: The expected cost is the cost of the event multiplied by the probability of the event. The cost of $1000 is incurred if the number of defects XX is greater than 1 (i.e., X≥2X \ge 2). We need to calculate P(X≥2)P(X \ge 2) using the binomial distribution with n=10n=10 and p=0.1p=0.1. It's easiest to calculate the complement, P(X≤1)=P(X=0)+P(X=1)P(X \le 1) = P(X=0) + P(X=1). P(X=0)=(100)(0.1)0(0.9)10≈0.3487P(X=0) = \binom{10}{0}(0.1)^0(0.9)^{10} \approx 0.3487 P(X=1)=(101)(0.1)1(0.9)9≈0.3874P(X=1) = \binom{10}{1}(0.1)^1(0.9)^9 \approx 0.3874 So, P(X≤1)=0.3487+0.3874=0.7361P(X \le 1) = 0.3487 + 0.3874 = 0.7361. The probability of needing inspection is P(X≥2)=1−0.7361=0.2639P(X \ge 2) = 1 - 0.7361 = 0.2639. Expected cost = 1000 \times P(X \ge 2) = 1000 \times 0.2639 = \263.90$.

Question 19

In a clinical trial for a new drug, the probability of a patient responding positively is 0.3. Patients are treated sequentially and independently. What is the probability that the first positive response is observed after the fourth patient?

  1. 0.072
  2. 0.103
  3. 0.240 (correct answer)
  4. 0.760
Explanation: This scenario is modeled by the geometric distribution, with a success probability of p=0.3p=0.3. The question asks for the probability that the first success occurs after the 4th trial, i.e., P(X>4)P(X > 4). This is equivalent to the probability of having 4 consecutive failures. The probability of a single failure (a non-response) is 1−p=1−0.3=0.71-p = 1-0.3 = 0.7. The probability of four independent failures in a row is (0.7)4=0.2401(0.7)^4 = 0.2401.

Question 20

A company sources a component from two suppliers. Supplier A has a 5% defect rate and sends components in crates of 20. Supplier B has a 3% defect rate and sends components in crates of 30. A crate is rejected if it contains two or more defective components. Which supplier is more likely to have a crate rejected, and by approximately what margin?

  1. Supplier A, by a margin of 0.037 (correct answer)
  2. Supplier B, by a margin of 0.037
  3. Supplier A, by a margin of 0.024
  4. Supplier B, by a margin of 0.024
Explanation: We need to calculate the probability of rejection, P(X≥2)P(X \ge 2), for each supplier using the binomial model. This is equal to 1−[P(X=0)+P(X=1)]1 - [P(X=0) + P(X=1)]. For Supplier A (n=20,p=0.05n=20, p=0.05): P(XA≥2)=1−[(200)(0.05)0(0.95)20+(201)(0.05)1(0.95)19]=1−[0.3585+0.3774]=1−0.7359=0.2641P(X_A \ge 2) = 1 - [\binom{20}{0}(0.05)^0(0.95)^{20} + \binom{20}{1}(0.05)^1(0.95)^{19}] = 1 - [0.3585 + 0.3774] = 1 - 0.7359 = 0.2641. For Supplier B (n=30,p=0.03n=30, p=0.03): P(XB≥2)=1−[(300)(0.03)0(0.97)30+(301)(0.03)1(0.97)29]=1−[0.4010+0.3721]=1−0.7731=0.2269P(X_B \ge 2) = 1 - [\binom{30}{0}(0.03)^0(0.97)^{30} + \binom{30}{1}(0.03)^1(0.97)^{29}] = 1 - [0.4010 + 0.3721] = 1 - 0.7731 = 0.2269. Supplier A has a higher rejection probability (0.2641 vs 0.2269). The margin is 0.2641−0.2269=0.03720.2641 - 0.2269 = 0.0372.