IB Mathematics: Applications and Interpretation Quiz: Binomial Distribution
20 questions · exam conditions
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Binomial DistributionQuestion 1 of 20

An insurance company sells a policy that pays out if a specific event occurs. The company estimates the probability of the event occurring for any policyholder in a year is 0.02. The company sells 100 such policies.

What is the probability, to three significant figures, that the company has to pay out on 4 or 5 of these policies in a year?

0.0353
0.0902
0.126
0.985
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Binomial Distribution

Practice Binomial Distribution in IB Mathematics: Applications and Interpretation 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 Distribution, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

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.

All questions

Question 1

An insurance company sells a policy that pays out if a specific event occurs. The company estimates the probability of the event occurring for any policyholder in a year is 0.02. The company sells 100 such policies.

What is the probability, to three significant figures, that the company has to pay out on 4 or 5 of these policies in a year?

  1. 0.0353
  2. 0.0902
  3. 0.126 (correct answer)
  4. 0.985
Explanation: Let X be the number of payouts. X follows a binomial distribution, X ~ B(100, 0.02). We need to find the probability of 4 or 5 payouts, which is P(X=4 or X=5). Since these are mutually exclusive events, we can add their probabilities: P(X=4) + P(X=5). Using a calculator's binompdf function: P(X=4) = binompdf(100, 0.02, 4) ≈ 0.09021. P(X=5) = binompdf(100, 0.02, 5) ≈ 0.03534. The sum is 0.09021 + 0.03534 ≈ 0.12555. To three significant figures, this is 0.126.

Question 2

A coffee shop owner knows that 40% of customers order a specialty coffee. On a morning, she predicts she will have 50 customers.

Find the smallest number of specialty coffees, k, she should have ingredients for, to be at least 95% certain she can meet the demand.

  1. 20
  2. 25
  3. 26 (correct answer)
  4. 27
Explanation: Let X be the number of customers who order a specialty coffee. X ~ B(50, 0.4). We need to find the smallest integer k such that P(X ≤ k) ≥ 0.95. This is a reverse lookup on the cumulative distribution. Using a calculator's inverse binomial function (e.g., invBinom(0.95, 50, 0.4)) or by testing values: P(X ≤ 25) = binomcdf(50, 0.4, 25) ≈ 0.9427 (which is less than 0.95). P(X ≤ 26) = binomcdf(50, 0.4, 26) ≈ 0.9712 (which is greater than 0.95). Therefore, the smallest number of coffees is 26.

Question 3

A survey of 200 randomly selected voters in a town found that 88 intended to vote for Candidate A. Later, a new random sample of 12 voters is taken.

Based on the initial survey, what is the probability that in the new sample of 12 voters, at least half of them intend to vote for Candidate A?

  1. 0.225
  2. 0.440
  3. 0.457 (correct answer)
  4. 0.543
Explanation: First, estimate the probability of success (intending to vote for Candidate A) from the large survey: p = 88/200 = 0.44. Now, model the new sample with a binomial distribution. Let X be the number of voters for A in the new sample of 12. X ~ B(12, 0.44). We need to find the probability that at least half (6 or more) vote for A, which is P(X ≥ 6). This is calculated as 1 - P(X ≤ 5). Using a calculator: P(X ≤ 5) = binomcdf(12, 0.44, 5) ≈ 0.5428. Therefore, P(X ≥ 6) = 1 - 0.5428 = 0.4572.

Question 4

A student takes a 15-question multiple-choice quiz. Each question has four options, only one of which is correct. The student guesses randomly on every question.

What is the probability, to three significant figures, that the student gets at most 3 questions correct?

  1. 0.236
  2. 0.461 (correct answer)
  3. 0.539
  4. 0.999
Explanation: Let X be the number of correct answers. X follows a binomial distribution, X ~ B(n, p), with n = 15 trials and probability of success p = 1/4 = 0.25. The question asks for the probability of getting at most 3 correct answers, which is P(X ≤ 3). Using a calculator's binomial cumulative distribution function (binomcdf), P(X ≤ 3) = binomcdf(n=15, p=0.25, x=3) ≈ 0.461.

Question 5

An archer hits the target with a probability of 0.8. She takes 10 shots. Let X be the number of hits.

Given that she hits the target at least 9 times, what is the probability, to three significant figures, that she hits it exactly 10 times?

  1. 0.107
  2. 0.286 (correct answer)
  3. 0.376
  4. 0.401
Explanation: This is a conditional probability problem: P(X=10 | X≥9). The formula is P(X=10) / P(X≥9). We have X ~ B(10, 0.8). First, calculate the individual probabilities: P(X=10) = binompdf(10, 0.8, 10) ≈ 0.10737. P(X=9) = binompdf(10, 0.8, 9) ≈ 0.26844. Then, P(X≥9) = P(X=9) + P(X=10) ≈ 0.26844 + 0.10737 ≈ 0.37581. Finally, the conditional probability is P(X=10) / P(X≥9) ≈ 0.10737 / 0.37581 ≈ 0.2857. To three significant figures, this is 0.286.

Question 6

The probability that a randomly chosen person is left-handed is 0.12. An experiment consists of selecting a random group of 8 people. This experiment is repeated 100 times.

What is the expected number of times that a sample of 8 people would contain exactly 2 left-handed people?

  1. 1
  2. 2
  3. 19 (correct answer)
  4. 24
Explanation: This is a two-stage problem. First, find the probability of the specific event: a sample of 8 people containing exactly 2 left-handers. Let Y ~ B(8, 0.12). The probability is P(Y=2) = binompdf(8, 0.12, 2) ≈ 0.1896. Now, this event becomes the 'success' in a new binomial experiment where n=100 (repeated 100 times) and p = 0.1896. The expected number of successes in this new experiment is E(X) = np = 100 * 0.1896 = 18.96. The closest integer is 19.

Question 7

A factory produces computer chips, and on average 4% are defective. A quality control inspector randomly selects 20 chips for testing.

What is the probability, to three significant figures, that the inspector finds at least two defective chips?

  1. 0.141
  2. 0.190 (correct answer)
  3. 0.810
  4. 0.950
Explanation: Let X be the number of defective chips. X follows a binomial distribution, X ~ B(n, p), with n = 20 and p = 0.04. The probability of finding at least two defective chips is P(X ≥ 2). This is calculated as 1 - P(X ≤ 1). Using a calculator: P(X ≤ 1) = binomcdf(n=20, p=0.04, x=1) ≈ 0.8103. Therefore, P(X ≥ 2) = 1 - 0.8103 ≈ 0.1897. To three significant figures, this is 0.190.

Question 8

In a binomial experiment with 12 trials, the expected number of successes is 4.2.

What is the probability of obtaining exactly 5 successes in this experiment, correct to three significant figures?

  1. 0.190 (correct answer)
  2. 0.222
  3. 0.350
  4. 0.420
Explanation: First, find the probability of success, p. The expected value E(X) = np. We are given n=12 and E(X)=4.2. So, 12p = 4.2, which gives p = 4.2 / 12 = 0.35. Now we need to find the probability of exactly 5 successes, P(X=5), for X ~ B(12, 0.35). Using the binomial probability formula or a calculator's binompdf function: P(X=5) = binompdf(n=12, p=0.35, k=5) ≈ 0.190.

Question 9

An archer hits the target with a probability of 0.8. She takes 10 shots. Let X be the number of hits.

Given that she hits the target at least 9 times, what is the probability, to three significant figures, that she hits it exactly 10 times?

  1. 0.107
  2. 0.286 (correct answer)
  3. 0.376
  4. 0.401
Explanation: This is a conditional probability problem: P(X=10 | X≥9). The formula is P(X=10) / P(X≥9). We have X ~ B(10, 0.8). First, calculate the individual probabilities: P(X=10) = binompdf(10, 0.8, 10) ≈ 0.10737. P(X=9) = binompdf(10, 0.8, 9) ≈ 0.26844. Then, P(X≥9) = P(X=9) + P(X=10) ≈ 0.26844 + 0.10737 ≈ 0.37581. Finally, the conditional probability is P(X=10) / P(X≥9) ≈ 0.10737 / 0.37581 ≈ 0.2857. To three significant figures, this is 0.286.

Question 10

A seed company claims that 92% of its tomato seeds germinate. A gardener plants 150 seeds.

Assuming the company's claim is true, what are the mean and standard deviation, respectively, for the number of seeds that germinate?

  1. 138 and 3.32 (correct answer)
  2. 138 and 11.04
  3. 12 and 3.32
  4. 12 and 11.04
Explanation: Let X be the number of seeds that germinate. X ~ B(150, 0.92). The mean (expected value) is E(X) = np = 150 * 0.92 = 138. The variance is Var(X) = np(1-p) = 150 * 0.92 * (1 - 0.92) = 150 * 0.92 * 0.08 = 11.04. The standard deviation is the square root of the variance, σ = sqrt(11.04) ≈ 3.32. Thus, the mean is 138 and the standard deviation is 3.32.

Question 11

A call centre agent makes 30 calls per hour. The probability of making a sale on any given call is 0.05.

What is the probability that the agent makes more than their expected number of sales in a randomly chosen hour?

  1. 0.187
  2. 0.255
  3. 0.447 (correct answer)
  4. 0.553
Explanation: Let X be the number of sales. X ~ B(30, 0.05). First, find the expected number of sales: E(X) = np = 30 * 0.05 = 1.5. The question asks for the probability of making more than 1.5 sales. Since the number of sales must be an integer, this means P(X ≥ 2). This is calculated as 1 - P(X ≤ 1). Using a calculator: P(X ≤ 1) = binomcdf(30, 0.05, 1) ≈ 0.5535. So, P(X ≥ 2) = 1 - 0.5535 = 0.4465, which is approximately 0.447.

Question 12

A biology test consists of 25 true/false questions. To pass, a student must answer at least 15 questions correctly. A student with no knowledge of the subject decides to guess every answer.

What is the probability, to three significant figures, that the student passes the test?

  1. 0.115
  2. 0.212 (correct answer)
  3. 0.322
  4. 0.788
Explanation: Let X be the number of correct answers. Since the student is guessing on true/false questions, the probability of success on each question is p = 0.5. The number of trials is n = 25. So, X ~ B(25, 0.5). To pass, the student must answer at least 15 questions correctly, so we need to find P(X ≥ 15). This is calculated as 1 - P(X ≤ 14). Using a calculator: P(X ≤ 14) = binomcdf(25, 0.5, 14) ≈ 0.7878. Therefore, P(X ≥ 15) = 1 - 0.7878 = 0.2122.

Question 13

A machine produces items, and historical data shows that 10% of them are defective. A new batch of 50 items is produced and 10 of them are found to be defective. Let X be the number of defective items in a batch of 50, modelled by X ~ B(50, 0.1).

Which statement is the most accurate interpretation of finding 10 defective items in the new batch?

  1. This outcome is impossible under the proposed model.
  2. This outcome is close to the expected value, so the process is stable.
  3. This outcome is expected, as the defect rate is 10% and 10/50 is 20%.
  4. This is a statistically unlikely outcome, which may suggest the defect rate has increased. (correct answer)
Explanation: When you encounter a binomial distribution problem asking you to interpret an observed outcome, you need to compare what actually happened against what the model predicts should typically happen. For X ~ B(50, 0.1), the expected value is E(X)=np=50×0.1=5E(X) = np = 50 \times 0.1 = 5 defective items. The standard deviation is np(1−p)=50×0.1×0.9=4.5≈2.12\sqrt{np(1-p)} = \sqrt{50 \times 0.1 \times 0.9} = \sqrt{4.5} \approx 2.12. Finding 10 defective items means the observed value is about 2.4 standard deviations above the mean—this falls well outside the typical range of outcomes and would occur less than 5% of the time under normal conditions. Answer D correctly identifies this as statistically unlikely, suggesting the defect rate may have increased from the historical 10%. Answer A is wrong because while unlikely, this outcome isn't impossible—it's just improbable under the current model. Answer B incorrectly claims 10 is "close to" the expected value of 5, when it's actually quite far from it. Answer C contains a logical error: it correctly states that 10/50 = 20%, but fails to recognize this is double the expected 10% rate, making it concerning rather than expected. Study tip: When interpreting binomial outcomes, always calculate how many standard deviations the observed result is from the expected value. If it's more than 2 standard deviations away, treat it as statistically significant and consider whether your model assumptions might need revision.

Question 14

A company uses two independent machines, Machine A and Machine B, to produce widgets. Machine A has a 3% defect rate and produces 15 widgets per hour. Machine B has a 5% defect rate and produces 10 widgets per hour.

What is the probability, to three significant figures, that in one hour of operation, there are no defective widgets from Machine A and exactly one defective widget from Machine B?

  1. 0.200 (correct answer)
  2. 0.315
  3. 0.633
  4. 0.948
Explanation: Let X_A be the number of defects from Machine A, and X_B be the number of defects from Machine B. We have X_A ~ B(15, 0.03) and X_B ~ B(10, 0.05). We want to find P(X_A=0 AND X_B=1). Since the machines are independent, this is P(X_A=0) × P(X_B=1). Calculate each probability: P(X_A=0) = binompdf(15, 0.03, 0) ≈ 0.63326. P(X_B=1) = binompdf(10, 0.05, 1) ≈ 0.31512. The combined probability is 0.63326 × 0.31512 ≈ 0.19955. To three significant figures, this is 0.200.

Question 15

A seed company claims that 92% of its tomato seeds germinate. A gardener plants 150 seeds.

Assuming the company's claim is true, what are the mean and standard deviation, respectively, for the number of seeds that germinate?

  1. 138 and 3.32 (correct answer)
  2. 138 and 11.04
  3. 12 and 3.32
  4. 12 and 11.04
Explanation: Let X be the number of seeds that germinate. X ~ B(150, 0.92). The mean (expected value) is E(X) = np = 150 * 0.92 = 138. The variance is Var(X) = np(1-p) = 150 * 0.92 * (1 - 0.92) = 150 * 0.92 * 0.08 = 11.04. The standard deviation is the square root of the variance, σ = sqrt(11.04) ≈ 3.32. Thus, the mean is 138 and the standard deviation is 3.32.

Question 16

A factory produces computer chips, and on average 4% are defective. A quality control inspector randomly selects 20 chips for testing.

What is the probability, to three significant figures, that the inspector finds at least two defective chips?

  1. 0.141
  2. 0.190 (correct answer)
  3. 0.810
  4. 0.950
Explanation: Let X be the number of defective chips. X follows a binomial distribution, X ~ B(n, p), with n = 20 and p = 0.04. The probability of finding at least two defective chips is P(X ≥ 2). This is calculated as 1 - P(X ≤ 1). Using a calculator: P(X ≤ 1) = binomcdf(n=20, p=0.04, x=1) ≈ 0.8103. Therefore, P(X ≥ 2) = 1 - 0.8103 ≈ 0.1897. To three significant figures, this is 0.190.

Question 17

A new drug is effective in 60% of cases. A pharmaceutical company wants to conduct a trial with n patients.

What is the minimum number of patients, n, required for the probability of having at least one successful treatment to be greater than 99.9%?

  1. 4
  2. 7
  3. 8 (correct answer)
  4. 11
Explanation: Let X be the number of successful treatments. X ~ B(n, 0.6). The probability of failure is 1 - 0.6 = 0.4. We want P(X ≥ 1) > 0.999. The complement is P(X = 0), which is the probability of no successes. So, 1 - P(X = 0) > 0.999, which means P(X = 0) < 0.001. The probability of zero successes is (0.4)^n. We need to solve (0.4)^n < 0.001. Taking logarithms of both sides: n * log(0.4) < log(0.001). Since log(0.4) is negative, we must reverse the inequality sign when dividing: n > log(0.001) / log(0.4) ≈ 7.53. Since n must be an integer, the minimum number of patients is 8.

Question 18

In a binomial experiment with 12 trials, the expected number of successes is 4.2.

What is the probability of obtaining exactly 5 successes in this experiment, correct to three significant figures?

  1. 0.190 (correct answer)
  2. 0.222
  3. 0.350
  4. 0.420
Explanation: First, find the probability of success, p. The expected value E(X) = np. We are given n=12 and E(X)=4.2. So, 12p = 4.2, which gives p = 4.2 / 12 = 0.35. Now we need to find the probability of exactly 5 successes, P(X=5), for X ~ B(12, 0.35). Using the binomial probability formula or a calculator's binompdf function: P(X=5) = binompdf(n=12, p=0.35, k=5) ≈ 0.190.

Question 19

A call centre agent makes 30 calls per hour. The probability of making a sale on any given call is 0.05.

What is the probability that the agent makes more than their expected number of sales in a randomly chosen hour?

  1. 0.187
  2. 0.255
  3. 0.447 (correct answer)
  4. 0.553
Explanation: Let X be the number of sales. X ~ B(30, 0.05). First, find the expected number of sales: E(X) = np = 30 * 0.05 = 1.5. The question asks for the probability of making more than 1.5 sales. Since the number of sales must be an integer, this means P(X ≥ 2). This is calculated as 1 - P(X ≤ 1). Using a calculator: P(X ≤ 1) = binomcdf(30, 0.05, 1) ≈ 0.5535. So, P(X ≥ 2) = 1 - 0.5535 = 0.4465, which is approximately 0.447.

Question 20

A basketball player has a free throw success rate of 78%. In an upcoming tournament, she will attempt 18 free throws.

What is the most likely number of successful free throws she will make?

  1. 13
  2. 14 (correct answer)
  3. 14.04
  4. 15
Explanation: Let X be the number of successful free throws, so X ~ B(18, 0.78). The most likely number of successes is the mode of the distribution. The mode is the integer value k that maximizes P(X=k). The mode is always close to the mean, E(X) = np = 18 * 0.78 = 14.04. Since the number of successes must be an integer, we test the integers around 14.04. P(X=14) = binompdf(18, 0.78, 14) ≈ 0.245. P(X=15) = binompdf(18, 0.78, 15) ≈ 0.215. Since P(X=14) is higher, 14 is the most likely number of successes. The mode can also be found by calculating floor((n+1)p) = floor(19 * 0.78) = floor(14.82) = 14.