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The Binomial Distribution — Answers

Answer sheet with worked solutions for the Year 13 statistics worksheet.

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Success TutoringWorksheetsAnswer sheetYear 13 · Mathematics · Statistics

The Binomial Distribution — Answers

Part A: Bernoulli trials and binomial situations

A Bernoulli trial has exactly two outcomes, success (probability p) and failure (probability 1 − p). For one trial, E(X) = p and Var(X) = p(1 − p).

X is binomial, written X ~ Bin(n, p), when it counts the successes in a fixed number n of independent Bernoulli trials, each with the same probability p of success.

Worked example. A basketball player makes 70% of free throws and takes 10 shots, independently. The number of shots made is X ~ Bin(10, 0.7). The number of shots taken until the first miss is not binomial, because n is not fixed.

1.Which situation can be modelled by a binomial distribution?
  1. AThe number of sixes when a fair die is rolled 12 times
  2. BThe number of rolls of a die needed to get the first six
  3. CThe total of the numbers shown when a die is rolled 12 times
  4. DThe time until the next customer arrives

Answer: The number of sixes when a fair die is rolled 12 times

Rolling a die 12 times gives a fixed number of independent trials, each a success (six) or failure with the same p = . The number of rolls until the first six has no fixed n, the total is not a count of successes, and time is continuous.

2.Drawing 5 cards without replacement from a standard 52-card deck and counting the hearts gives a binomial distribution.

Answer: False

Without replacement the trials are not independent: the probability of a heart changes after each draw (13/52, then 12/51 or 13/51, …). It would be binomial only if each card were replaced.

3.A Bernoulli trial has probability of success p = 0.3. Find the mean and variance of the number of successes in one trial.

Answer: Mean 0.3, variance 0.21

  1. For a Bernoulli variable, E(X) = p = 0.3
  2. Var(X) = p(1 − p) = 0.3 × 0.7 = 0.21

Part B: Binomial probabilities

If X ~ Bin(n, p), then P(X = k) = [sup n]C[sup k] p[sup k] (1 − p)[sup n − k] for k = 0, 1, 2, …, n. The nCk counts the orders in which k successes can occur among the n trials.

For at least one, use the complement: P(X ≥ 1) = 1 − P(X = 0) = 1 − (1 − p)n. For at most k, add P(X = 0) up to P(X = k).

Worked example. A fair coin is tossed 4 times. P(exactly 3 heads) = 4C3 (0.5)3 (0.5)1 = 4 × = 0.25, and P(at least one head) = 1 − (0.5)4 = 1 − 0.0625 = 0.9375.

4.X ~ Bin(5, 0.5). Find P(X = 2).

Answer: 0.3125

  1. P(X = 2) = 5C2 (0.5)2 (0.5)3 = 10 × = = 0.3125
5.X ~ Bin(10, 0.2). Find P(X = 3), correct to 4 decimal places.

Answer: 0.2013

  1. P(X = 3) = 10C3 (0.2)3 (0.8)7
  2. = 120 × 0.008 × 0.2097152
  3. = 0.2013266… ≈ 0.2013
6.X ~ Bin(6, 0.3). Find P(X = 0) and P(X ≥ 1), correct to 4 decimal places.

Answer: P(X = 0) = 0.1176, P(X ≥ 1) = 0.8824

  1. P(X = 0) = (0.7)6 = 0.117649 ≈ 0.1176
  2. P(X ≥ 1) = 1 − P(X = 0) = 1 − 0.117649 = 0.882351 ≈ 0.8824
7.X ~ Bin(4, 0.6). Find P(X ≤ 1).

Answer: 0.1792

  1. P(X = 0) = (0.4)4 = 0.0256
  2. P(X = 1) = 4C1 (0.6)(0.4)3 = 4 × 0.6 × 0.064 = 0.1536
  3. P(X ≤ 1) = 0.0256 + 0.1536 = 0.1792
8.X ~ Bin(3, 0.4). Write the full probability distribution of X and check that the probabilities add to 1.

Answer: P(X = 0) = 0.216, P(X = 1) = 0.432, P(X = 2) = 0.288, P(X = 3) = 0.064

  1. P(X = 0) = (0.6)3 = 0.216
  2. P(X = 1) = 3C1 (0.4)(0.6)2 = 3 × 0.4 × 0.36 = 0.432
  3. P(X = 2) = 3C2 (0.4)2(0.6) = 3 × 0.16 × 0.6 = 0.288
  4. P(X = 3) = (0.4)3 = 0.064
  5. Check: 0.216 + 0.432 + 0.288 + 0.064 = 1
9.A multiple-choice quiz has 8 questions, each with 4 options. A student guesses every answer. Find, correct to 4 decimal places, (a) the probability of exactly 4 correct answers and (b) the probability of at least one correct answer.

Answer: (a) 0.0865 (b) 0.8999

  1. X ~ Bin(8, 0.25)
  2. (a) P(X = 4) = 8C4 (0.25)4 (0.75)4 = 70 × 0.00390625 × 0.31640625 = 0.086517… ≈ 0.0865
  3. (b) P(X ≥ 1) = 1 − (0.75)8 = 1 − 0.1001129… = 0.899887… ≈ 0.8999
10.A machine produces faulty items with probability 0.1, independently. What is the smallest number of items that must be checked for the probability of finding at least one faulty item to exceed 0.95?

Answer: 29 items

  1. P(at least one faulty) = 1 − 0.9n > 0.95, so 0.9n < 0.05
  2. n > = = 28.43… (the inequality flips because ln 0.9 is negative)
  3. Smallest whole number: n = 29. Check: 0.928 ≈ 0.0523 (too big) and 0.929 ≈ 0.0471 (less than 0.05).

Part C: Mean, variance and standard deviation

If X ~ Bin(n, p), then E(X) = np, Var(X) = np(1 − p) and SD(X) = √(np(1 − p)). The mean is the long-run average number of successes in n trials.

Worked example. A fair die is rolled 30 times and X is the number of sixes. E(X) = 30 × = 5, Var(X) = 30 × × = ≈ 4.17 and SD ≈ 2.04.

11.X ~ Bin(20, 0.15). Find the mean, the variance and the standard deviation of X (standard deviation to 2 decimal places).

Answer: Mean 3, variance 2.55, SD ≈ 1.60

  1. E(X) = np = 20 × 0.15 = 3
  2. Var(X) = np(1 − p) = 20 × 0.15 × 0.85 = 2.55
  3. SD = √2.55 = 1.5968… ≈ 1.60
12.A binomial random variable has mean 6 and variance 4.2. Find n and p.

Answer: n = 20, p = 0.3

  1. np = 6 and np(1 − p) = 4.2
  2. Divide: 1 − p = = 0.7, so p = 0.3
  3. n = = 20
13.A test has 25 multiple-choice questions, each with 5 options. A student guesses every answer. How many questions should the student expect to get right?

Answer: 5

  1. X ~ Bin(25, 0.2)
  2. E(X) = np = 25 × 0.2 = 5 questions
14.For a fixed n, the variance np(1 − p) of a binomial distribution is largest when p = (answer space).

Answer: 0.5

p(1 − p) is a quadratic in p with its maximum at p = 0.5, where p(1 − p) = 0.25.

15.Match each distribution to its mean.
  • Bin(10, 0.5)
  • Bin(40, 0.25)
  • Bin(12, 0.75)
  • Bin(50, 0.04)
  • 9
  • 5
  • 2
  • 10

Answer: Bin(10, 0.5) → 5; Bin(40, 0.25) → 10; Bin(12, 0.75) → 9; Bin(50, 0.04) → 2

Part D: Extended problem

A seed company states that 85% of its seeds germinate. A packet contains 12 seeds, and each seed germinates independently. Let X be the number of seeds in a packet that germinate.

(a) Find the probability that all 12 seeds germinate.

(b) Find the probability that exactly 10 seeds germinate.

(c) Find the probability that at least 10 seeds germinate.

(d) Find the mean and standard deviation of X (standard deviation to 2 decimal places).

(e) The company guarantees that at least 9 seeds in every packet will germinate. Find the probability that a packet fails this guarantee, and interpret your answer.

(f) A gardener buys 5 packets. Find the probability that all 5 packets meet the guarantee.

16.Answer parts (a) to (f), showing full working. Give probabilities correct to 4 decimal places.[11 marks]

Model answer: X ~ Bin(12, 0.85). (a) P(X = 12) = 0.8512 = 0.142241… ≈ 0.1422. (b) P(X = 10) = 12C10 (0.85)10 (0.15)2 = 66 × 0.196874… × 0.0225 = 0.292358… ≈ 0.2924. (c) P(X = 11) = 12 × 0.8511 × 0.15 = 0.301217…, so P(X ≥ 10) = 0.292358… + 0.301217… + 0.142241… = 0.735818… ≈ 0.7358. (d) E(X) = 12 × 0.85 = 10.2 seeds; SD = √(12 × 0.85 × 0.15) = √1.53 ≈ 1.24 seeds. (e) P(X = 9) = 12C9 (0.85)9 (0.15)3 = 0.171975…, so P(X ≥ 9) = 0.735818… + 0.171975… = 0.907793… and P(X < 9) = 1 − 0.907793… ≈ 0.0922. About 9% of packets fail the guarantee. (f) The packets are independent, so P(all 5 meet the guarantee) = 0.907793…5 ≈ 0.6165.

Success criteria:

  • (a) 0.1422 (1 mark)
  • (b) correct use of 12C10 with p = 0.85 to get 0.2924 (2 marks)
  • (c) adds P(X = 10), P(X = 11) and P(X = 12) to get 0.7358 (2 marks)
  • (d) mean 10.2 and standard deviation ≈ 1.24 (2 marks)
  • (e) finds P(X ≥ 9) and uses the complement to get 0.0922, with an interpretation (2 marks)
  • (f) raises P(X ≥ 9) to the power 5 to get 0.6165 (2 marks)
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