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Binomial Distribution - IB Mathematics Applications and Interpretation

  1. What is the expected number of times that a sample of 8 people would contain exactly 2 left-handed people?
    Answer: 19 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.
  2. Which of the following scenarios is best modelled by a binomial distribution?
    Answer: The number of students in a class of 25 who pass a test, where each student has an independent 80% chance of passing. When you encounter probability distribution questions, you need to check whether the scenario meets the specific conditions for each distribution type. For binomial distributions, you need exactly four conditions: a fixed number of independent trials, each with only two possible outcomes, where the probability of success remains constant across all trials. Option D satisfies all binomial conditions perfectly. You have a fixed number of trials (25 students), each trial is independent (one student's performance doesn't affect another's), there are exactly two outcomes (pass or fail), and the probability remains constant (80% for each student). This gives you X∼B(25,0.8)X \sim B(25, 0.8). Option A fails the independence requirement because drawing cards without replacement changes the probability with each draw. After drawing one red card, there are fewer red cards remaining, so subsequent probabilities shift. Option B describes a geometric distribution, not binomial. You're looking for the number of trials needed to get the first success, which has no fixed number of trials and focuses on "time until first success" rather than "number of successes in fixed trials." Option C violates the independence condition since rainy days affect the probability of subsequent rainy days. This describes a dependent process where probabilities change based on previous outcomes. Remember this key strategy: For binomial distribution questions, mentally check off all four conditions like a checklist. Independence is the most commonly violated condition in wrong answers, so pay special attention to whether each trial affects the others.
  3. What is the probability, to three significant figures, that the student gets at most 3 questions correct?
    Answer: 0.461 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.
  4. 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?
    Answer: 0.457 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.
  5. What is the probability that the agent makes more than their expected number of sales in a randomly chosen hour?
    Answer: 0.447 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.
  6. What is the probability, to three significant figures, that the inspector finds at least two defective chips?
    Answer: 0.190 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.
  7. What is the probability, to three significant figures, that the student passes the test?
    Answer: 0.212 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.
  8. 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%?
    Answer: 8 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(0.4)^n. We need to solve (0.4)n(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.
  9. Which statement is the most accurate interpretation of finding 10 defective items in the new batch?
    Answer: This is a statistically unlikely outcome, which may suggest the defect rate has increased. 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.
  10. 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?
    Answer: 0.126 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.
  11. 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?
    Answer: 0.200 Let XAX_A be the number of defects from Machine A, and XBX_B be the number of defects from Machine B. We have XAX_A ~ B(15, 0.03) and XBX_B ~ B(10, 0.05). We want to find P(XAX_A=0 AND XBX_B=1). Since the machines are independent, this is P(XAX_A=0) × P(XBX_B=1). Calculate each probability: P(XAX_A=0) = binompdf(15, 0.03, 0) ≈ 0.63326. P(XBX_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.
  12. Assuming the company's claim is true, what are the mean and standard deviation, respectively, for the number of seeds that germinate?
    Answer: 138 and 3.32 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.
  13. If a retailer buys 20 packs, what is the probability, to three significant figures, that exactly 3 of them are substandard?
    Answer: 0.155 This is a two-step binomial problem. First, find the probability (psp_s) that a single pack is substandard. Let Y be the number of defective batteries in a pack of 10, so Y ~ B(10, 0.05). A pack is substandard if Y ≥ 2. psp_s = P(Y ≥ 2) = 1 - P(Y ≤ 1) = 1 - binomcdf(10, 0.05, 1) ≈ 1 - 0.91386 = 0.08614. Now, let X be the number of substandard packs out of 20. X ~ B(20, 0.08614). We need to find P(X=3). P(X=3) = binompdf(20, 0.08614, 3) ≈ 0.155.
  14. 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?
    Answer: 0.286 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.
  15. For which value of p will the variance of X be the largest?
    Answer: 0.5 The variance of a binomial distribution is given by the formula Var(X) = np(1-p). In this case, n=40, so Var(X) = 40p(1-p). To find the value of p that maximizes this expression, we can consider the quadratic function f(p) = p(1-p) = p - p2p^2. This is a downward-opening parabola with roots at p=0 and p=1. The maximum value occurs at the vertex, which is halfway between the roots, at p = 0.5. Therefore, the variance is largest when the probability of success is 0.5.
  16. What is the probability of obtaining exactly 5 successes in this experiment, correct to three significant figures?
    Answer: 0.190 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.
  17. What is the most likely number of successful free throws she will make?
    Answer: 14 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.
  18. Given this information, what is the probability of obtaining at least two successes, P(X ≥ 2)?
    Answer: 0.800 First, we must find p. We are given P(X=0) = 0.05. The formula for P(X=0) is (1−p)n(1-p)^n. So, (1−p)16(1-p)^{16} = 0.05. To solve for p, we take the 16th root: 1-p = (0.05)1/16(0.05)^{1/16} ≈ 0.8285. This gives p ≈ 1 - 0.8285 = 0.1715. Now we need to find P(X ≥ 2). This is 1 - P(X ≤ 1) = 1 - [P(X=0) + P(X=1)]. We already know P(X=0) = 0.05. We calculate P(X=1) = binompdf(16, 0.1715, 1) ≈ 0.1500. So, P(X ≥ 2) = 1 - (0.05 + 0.1500) = 1 - 0.2000 = 0.800.
  19. Find the smallest number of specialty coffees, k, she should have ingredients for, to be at least 95% certain she can meet the demand.
    Answer: 26 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.

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