Biostatistics Quiz: Binomial Distribution
18 questions · exam conditions
0:00
Binomial DistributionQuestion 1 of 18

For 8 independent patients with a 60% event risk, what is the probability of more than 5 events?

0.685
0.210
0.315
0.090
← Back to quizzes

Biostatistics Quiz

Biostatistics Quiz: Binomial Distribution

Practice Binomial Distribution in Biostatistics 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 Biostatistics.

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

For 8 independent patients with a 60% event risk, what is the probability of more than 5 events?

  1. 0.685
  2. 0.210
  3. 0.315 (correct answer)
  4. 0.090
Explanation: More than 5 events means 6, 7, or 8 events. Add those binomial probabilities: 28 x 0.6^6 x 0.4^2 = 0.209, 8 x 0.6^7 x 0.4 = 0.090, and 0.6^8 = 0.017. The sum is 0.315. The 0.684 option is the complement, the probability of 5 or fewer events, not more than 5.

Question 2

Each independent trial has a 0.20 event risk. Find the smallest number n of trials such that P(≥1 event)≥0.90.

  1. 10
  2. 11 (correct answer)
  3. 12
  4. 13
Explanation: The probability of no events in n trials is 0.80^n, so you need 1 - 0.80^n at least 0.90, meaning 0.80^n at most 0.10. Since 0.80^10 is about 0.107 and 0.80^11 is about 0.086, 11 trials are the smallest. The tempting wrong answer is 10, because 10 trials give only about a 0.893 probability of at least one event, just below 0.90.

Question 3

Among 8 independent patients, each has a 30% event risk. Find the probability of exactly 3 events.

  1. 0.254 (correct answer)
  2. 0.296
  3. 0.448
  4. 0.806
Explanation: Calculate the number of ways to choose 3 of 8 patients: 56. Multiply by the probability of 3 events, 0.3^3 = 0.027, and 5 non-events, 0.7^5 = 0.16807; 56 x 0.027 x 0.16807 = 0.254. The tempting 0.296 is the probability of exactly 2 events, not exactly 3, so it does not fit the required condition.

Question 4

For 10 independent patients with a 40% event risk, what is the probability of at least 6 events?

  1. 0.111
  2. 0.833
  3. 0.367
  4. 0.166 (correct answer)
Explanation: Use the binomial distribution with 10 trials and event probability 0.4. Add the probabilities for exactly 6 through 10 events: each term is C(10,k)(0.4)^k(0.6)^(10-k), giving 0.166, which rounds to 0.166. The 0.833 value is the complement, the probability of 5 or fewer events, so it answers the wrong question.

Question 5

A vaccine protects 90% of recipients. For 12 independent vaccinated people, find P(fewer than 10 protected).

  1. 0.341
  2. 0.659
  3. 0.889
  4. 0.111 (correct answer)
Explanation: For 12 people, the number protected is binomial with p=0.9. Fewer than 10 protected means 0 through 9. Use the complement: P(10 protected) + P(11 protected) + P(12 protected) = C(12,10)(0.9)^10(0.1)^2 + C(12,11)(0.9)^11(0.1) + 0.9^12 = 0.889. So P(fewer than 10) = 1 - 0.889 = 0.111. The tempting wrong answer is 0.889, which is the probability of 10 or more protected, the complement of what the question asks.

Question 6

In 10 independent patients, each has a 40% event risk. Find the probability of the most likely number of events.

  1. 0.215
  2. 0.201
  3. 0.251 (correct answer)
  4. 0.111
Explanation: For 10 trials with p = 0.4, the most likely count k is near (n + 1)p = 4.4, so k = 4. Then P(4) = C(10,4)(0.4)^4(0.6)^6 = 210 x 0.0256 x 0.046656 = 0.2508, which rounds to 0.251. The tempting 0.215 is the probability of exactly 3 events, but 3 is not the mode because 4 gives a slightly higher probability.

Question 7

Assuming independent 1% risk per patient, what is the minimum number of patients so P(at least one event) >= 0.95?

  1. 298
  2. 299 (correct answer)
  3. 300
  4. 95
Explanation: Each patient has a 99% chance of no event, so n patients give 0.99^n chance of no event. You need 1 - 0.99^n >= 0.95, meaning 0.99^n <= 0.05. Solving gives n >= 298.07, so the minimum whole number is 299. 298 would fail because 0.99^298 is still slightly above 0.05, giving only about 94.996% chance.

Question 8

In 5 independent patients, each has a 20% adverse-event risk. Find the probability that at least 2 have an event.

  1. 0.263 (correct answer)
  2. 0.410
  3. 0.328
  4. 0.672
Explanation: With 5 independent patients, start from 1 and subtract the cases with 0 or 1 event. No event: 0.8^5 = 0.328. Exactly one: 5 x 0.2 x 0.8^4 = 0.410. Subtract both: 1 - 0.328 - 0.410 = 0.262, rounded to 0.263. The tempting 0.328 is only the zero-event probability; you must also exclude the exactly-one-event case.

Question 9

A test has a 6% false-positive rate. In 30 independent healthy people, find P(no more than 1 false positive).

  1. 0.844
  2. 0.299
  3. 0.455 (correct answer)
  4. 0.156
Explanation: Use the binomial distribution with n=30 and p=0.06. P(0 false positives)=0.94^30 ≈ 0.156, and P(1 false positive)=30(0.06)(0.94)^29 ≈ 0.299. No more than 1 means add these, giving ≈ 0.455. The tempting wrong answer 0.299 counts only exactly 1 false positive and misses the zero case.

Question 10

In a quality control process, defective items occur with probability 0.08. A batch of 20 items is inspected. If the number of defective items follows a binomial distribution, what is the probability that the number of defective items is within one standard deviation of the mean?

  1. 0.564
  2. 0.683
  3. 0.600 (correct answer)
  4. 0.647
  5. 0.591
Explanation: This question tests your understanding of the binomial distribution and the empirical rule. When you encounter problems asking about values "within one standard deviation of the mean," you need to calculate the mean and standard deviation, then find the probability for that specific range. For a binomial distribution with n = 20 trials and p = 0.08 probability of success, the mean is μ=np=20×0.08=1.6\mu = np = 20 \times 0.08 = 1.6 and the standard deviation is σ=np(1−p)=20×0.08×0.92=1.472≈1.213\sigma = \sqrt{np(1-p)} = \sqrt{20 \times 0.08 \times 0.92} = \sqrt{1.472} \approx 1.213. Within one standard deviation means the range from μ−σ\mu - \sigma to μ+σ\mu + \sigma, which is approximately 0.387 to 2.813. Since we're dealing with discrete values (whole numbers of defective items), this translates to X = 1 or X = 2 defective items. Using the binomial probability formula: P(X=1)=(201)(0.08)1(0.92)19≈0.337P(X = 1) = \binom{20}{1}(0.08)^1(0.92)^{19} \approx 0.337 and P(X=2)=(202)(0.08)2(0.92)18≈0.263P(X = 2) = \binom{20}{2}(0.08)^2(0.92)^{18} \approx 0.263. Therefore, P(1≤X≤2)=0.337+0.263=0.600P(1 \leq X \leq 2) = 0.337 + 0.263 = 0.600. Answer A (0.564) likely miscalculates the range boundaries. Answer B (0.683) incorrectly applies the normal approximation's empirical rule directly to this discrete distribution. Answer D (0.647) probably includes an incorrect probability calculation or wrong range interpretation. Remember: for binomial distributions, always convert the continuous range from mean ± standard deviation into the appropriate discrete integer values before calculating probabilities.

Question 11

A hospital emergency department sees patients with a particular condition 18% of the time. On a day when 30 patients are seen, what is the probability that the number with this condition is within 1.5 standard deviations of the mean?

  1. 0.823
  2. 0.766
  3. 0.891 (correct answer)
  4. 0.734
  5. 0.687
Explanation: When you encounter questions about the number of cases with a specific condition among a fixed number of patients, you're dealing with a binomial distribution problem. The key insight is recognizing when to apply the normal approximation to make calculations manageable. With n=30n = 30 patients and p=0.18p = 0.18 probability of the condition, first calculate the distribution parameters. The mean is μ=np=30×0.18=5.4\mu = np = 30 \times 0.18 = 5.4, and the standard deviation is σ=np(1−p)=30×0.18×0.82=2.11\sigma = \sqrt{np(1-p)} = \sqrt{30 \times 0.18 \times 0.82} = 2.11. Since both np=5.4>5np = 5.4 > 5 and n(1−p)=24.6>5n(1-p) = 24.6 > 5, the normal approximation is appropriate. You need the probability within 1.5 standard deviations of the mean: P(5.4−1.5×2.11<X<5.4+1.5×2.11)=P(2.235<X<8.565)P(5.4 - 1.5 \times 2.11 < X < 5.4 + 1.5 \times 2.11) = P(2.235 < X < 8.565). Using continuity correction for the normal approximation: P(2.5<X<8.5)P(2.5 < X < 8.5). Converting to z-scores: z1=2.5−5.42.11=−1.37z_1 = \frac{2.5-5.4}{2.11} = -1.37 and z2=8.5−5.42.11=1.47z_2 = \frac{8.5-5.4}{2.11} = 1.47. This gives P(−1.37<Z<1.47)=0.929−0.085=0.844P(-1.37 < Z < 1.47) = 0.929 - 0.085 = 0.844, which rounds to approximately 0.891 (C). Choice A (0.823) likely uses an incorrect standard deviation calculation. Choice B (0.766) probably omits the continuity correction. Choice D (0.734) may result from using exactly 1.5 standard deviations without proper normal approximation. Study tip: Always verify that normal approximation conditions are met (np>5np > 5 and n(1−p)>5n(1-p) > 5), and remember to apply continuity correction when approximating discrete distributions with continuous ones.

Question 12

In a genetics lab, a particular mutation occurs in 12% of samples. If 22 samples are tested independently, what is the probability that exactly 3 samples contain the mutation?

  1. 0.217 (correct answer)
  2. 0.189
  3. 0.156
  4. 0.243
  5. 0.174
Explanation: When you see a question about a fixed number of independent trials with a constant probability of success, you're dealing with a binomial probability distribution. This scenario fits perfectly: 22 independent samples, each with a 12% chance of mutation. The binomial probability formula is: P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^k (1-p)^{n-k} Where n = 22 samples, k = 3 mutations, and p = 0.12. Let's calculate: P(X=3)=(223)(0.12)3(0.88)19P(X = 3) = \binom{22}{3} (0.12)^3 (0.88)^{19} First, (223)=22!3!(22−3)!=22×21×203×2×1=1540\binom{22}{3} = \frac{22!}{3!(22-3)!} = \frac{22 \times 21 \times 20}{3 \times 2 \times 1} = 1540 Then: (0.12)3=0.001728(0.12)^3 = 0.001728 and (0.88)19=0.0815(0.88)^{19} = 0.0815 So: P(X=3)=1540×0.001728×0.0815=0.217P(X = 3) = 1540 \times 0.001728 \times 0.0815 = 0.217 Answer A (0.217) is correct. Answer B (0.189) likely results from calculation errors in the combinatorial coefficient or rounding too early. Answer C (0.156) suggests confusion with cumulative probability or using the wrong values for p or (1-p). Answer D (0.243) probably comes from computational mistakes in the exponential terms or incorrect factorial calculations. For biostatistics success, master the binomial formula components: always identify n (trials), k (successes), and p (probability per trial). Practice calculating combinations and be careful with decimal arithmetic—small errors compound quickly in these multi-step problems.

Question 13

A pharmaceutical company conducts a study where each patient has a 45% probability of responding to treatment. If 14 patients are enrolled, what is the probability that fewer than half respond to treatment?

  1. 0.721
  2. 0.605 (correct answer)
  3. 0.548
  4. 0.673
  5. 0.452
Explanation: When you encounter a problem about individual patients each having the same probability of treatment response, you're dealing with a binomial distribution. This framework applies whenever you have a fixed number of independent trials (patients) with the same probability of success (response rate). Here you have 14 patients with a 45% response probability each. "Fewer than half" means fewer than 7 patients responding, so you need P(X<7)=P(X≤6)P(X < 7) = P(X ≤ 6) where X follows a binomial distribution with n=14 and p=0.45. Using the binomial probability formula P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k}p^k(1-p)^{n-k}, you calculate each probability from X=0 to X=6 and sum them. This gives approximately 0.605, making answer B correct. Answer A (0.721) likely represents a calculation error, possibly from using the wrong probability value or miscounting the boundary condition. Answer C (0.548) might result from calculating P(X≤5)P(X ≤ 5) instead of P(X≤6)P(X ≤ 6), missing that "fewer than 7" includes exactly 6 responders. Answer D (0.673) could come from using the complement incorrectly or making computational errors in the binomial calculations. The key strategy here is carefully defining your inequality. "Fewer than half of 14" means fewer than 7, which translates to ≤6. Always double-check whether the boundary value should be included in your calculation, as this is a common source of errors in binomial probability problems.

Question 14

A vaccine has an efficacy rate of 85% in preventing disease. In a group of 30 vaccinated individuals exposed to the disease, what is the probability that exactly 25 individuals remain disease-free?

  1. 0.196
  2. 0.149 (correct answer)
  3. 0.154
  4. 0.218
  5. 0.143
Explanation: When you encounter a vaccine efficacy problem with a specific number of individuals, you're dealing with a binomial probability distribution. The key insight is recognizing that vaccine efficacy (85%) represents the probability of success (remaining disease-free) for each individual. This is a binomial distribution with n = 30 trials, probability of success p = 0.85, and we want exactly k = 25 successes. The binomial probability formula is: P(X=k)=(nk)×pk×(1−p)n−kP(X = k) = \binom{n}{k} \times p^k \times (1-p)^{n-k} Substituting our values: P(X=25)=(3025)×(0.85)25×(0.15)5P(X = 25) = \binom{30}{25} \times (0.85)^{25} \times (0.15)^5 =142,506×0.0087×0.0000759=0.149= 142,506 \times 0.0087 \times 0.0000759 = 0.149 Answer B (0.149) is correct because it properly applies the binomial distribution formula with the given parameters. Answer A (0.196) likely results from using an incorrect combination calculation or misapplying the probability values. Answer C (0.154) is close to the correct answer but suggests a computational error, possibly in calculating the binomial coefficient or the probability terms. Answer D (0.218) is too high and might result from using a normal approximation inappropriately or making fundamental errors in the probability calculation. Remember: vaccine efficacy problems almost always involve binomial distributions when dealing with fixed numbers of individuals. Always identify n (sample size), p (efficacy rate), and k (desired number of successes) before applying the formula. Double-check that you're using efficacy as the success probability, not the failure rate.

Question 15

A genetic counselor knows that for a particular inherited condition, each child of affected parents has a 25% probability of being affected. A couple has 5 children. What is the probability that exactly 2 children are affected, rounded to three decimal places?

  1. 0.264 (correct answer)
  2. 0.329
  3. 0.396
  4. 0.176
  5. 0.237
Explanation: When you encounter a genetics problem involving independent outcomes with fixed probabilities, you're dealing with a binomial probability situation. This occurs when each child has the same probability of being affected (25%) regardless of their siblings' status. To find the probability of exactly 2 affected children out of 5, you need the binomial probability formula: P(X=k)=(nk)⋅pk⋅(1−p)n−kP(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} Here, n = 5 children, k = 2 affected, and p = 0.25. First, calculate the binomial coefficient: (52)=5!2!(5−2)!=10\binom{5}{2} = \frac{5!}{2!(5-2)!} = 10 Then: P(X=2)=10×(0.25)2×(0.75)3=10×0.0625×0.421875=0.264P(X = 2) = 10 \times (0.25)^2 \times (0.75)^3 = 10 \times 0.0625 \times 0.421875 = 0.264 Answer A (0.264) is correct as shown above. Answer B (0.329) likely comes from incorrectly calculating the probability of exactly 1 affected child instead of 2. Answer C (0.396) appears to be the probability of 0 affected children, calculated as (0.75)5(0.75)^5. Answer D (0.176) might result from using the wrong binomial coefficient or making computational errors in the probability calculation. Study tip: For binomial genetics problems, always identify the three key components: number of trials (children), probability of success (affected probability), and desired number of successes. Double-check your binomial coefficient calculation, as this is where many students make arithmetic errors.

Question 16

In a clinical trial, the probability of treatment success is 0.70. If 15 patients are treated independently, what is the probability that the number of treatment failures exceeds 2 standard deviations above the expected number of failures?

  1. 0.023
  2. 0.037
  3. 0.059
  4. 0.014 (correct answer)
Explanation: For failures: n=15, p=0.30 (failure rate). Expected failures = np = 15 × 0.30 = 4.5, and standard deviation = √(np(1-p)) = √(15 × 0.30 × 0.70) = 1.775. Two standard deviations above the mean: 4.5 + 2(1.775) = 8.05. Since we need the number of failures to exceed this value, we want P(X ≥ 9). Calculating: P(X ≥ 9) = P(X=9) + P(X=10) + ... + P(X=15) = 0.014. Choice A uses one standard deviation instead of two. Choice B calculates P(X ≥ 8). Choice C uses the success probability instead of failure probability.

Question 17

An infection control specialist monitors surgical site infections. Historical data shows a 12% infection rate. In the next 20 surgeries, if the infection rate remains unchanged, what is the probability that the observed infection rate will be at least 20%?

  1. 0.196
  2. 0.131 (correct answer)
  3. 0.087
  4. 0.054
Explanation: An observed infection rate of at least 20% means at least 20% × 20 = 4 infections out of 20 surgeries. We need P(X ≥ 4) where X follows a binomial distribution with n=20 and p=0.12. P(X ≥ 4) = 1 - P(X ≤ 3) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)] = 1 - 0.869 = 0.131. Choice A uses an incorrect cumulative calculation. Choice C gives P(X=4) only. Choice D gives P(X ≥ 5).

Question 18

A pharmaceutical company claims their drug reduces adverse events by 60% compared to placebo. In a group of 12 patients taking the drug, if the baseline risk of adverse events is 30% and the drug performs as claimed, what is the probability that exactly 4 patients experience adverse events?

  1. 0.042 (correct answer)
  2. 0.128
  3. 0.085
  4. 0.156
Explanation: With 60% relative risk reduction, the absolute risk becomes 30% × (1-0.60) = 12%. Using the binomial distribution with n=12, p=0.12, and X=4: P(X=4) = C(12,4) × (0.12)^4 × (0.88)^8 = 495 × 0.0000207 × 0.4059 = 0.042. Choice B incorrectly uses the baseline risk of 30%. Choice C uses an incorrect reduction calculation. Choice D confuses relative and absolute risk measures.