Questions on the binomial distribution

Question 1

A fair twelve-sided die is rolled once and X is the score shown. Name the distribution of X, and state the probability it assigns to each value.

Accepted Answer

Every one of the twelve scores is equally likely, so X has a discrete uniform distribution on 1 to 12, with P(X = x) = 1/12 for each value. B1 for naming the discrete uniform distribution, B1 for 1/12 on every value. The specification asks you to recognise and name this distribution, not to compute with it. A total of two dice is the standard trap: 7 is more likely than 2, so that variable is not uniform.

Question 2

State the conditions under which a count X can be modelled by a binomial distribution B(n, p).

Accepted Answer

A fixed number of trials n, each trial with exactly two outcomes, a constant probability of success p, and independence between trials. B1 for a fixed number of trials with two outcomes, B1 for a constant probability of success, B1 for independence. All four are needed, and independence is the one exam questions attack.

Question 3

X ~ B(8, 0.25). Find P(X = 2) to 3 significant figures.

Accepted Answer

P(X = 2) = ⁸C₂ × 0.25² × 0.75⁶ = 28 × 0.0625 × 0.1780 = 0.311. M1 for the complete product, A1 for 0.311. The exponents must sum to 8, which is the quickest check that the line is right.

Question 4

A discrete random variable X has P(X = x) = kx for x = 1, 2, 3, 4, where k is a constant. Find the value of k and the value of P(X ≥ 3).

Accepted Answer

The four probabilities have to total 1, so k(1 + 2 + 3 + 4) = 10k = 1 and k = 0.1. M1 for setting the total to 1, A1 for k = 0.1. Then P(X ≥ 3) = P(X = 3) + P(X = 4) = 0.3 + 0.4 = 0.7. A1 for 0.7. Every discrete probability distribution satisfies that sum, which is what turns one unknown constant into an equation, and it is also the check to make at the end: 0.1 + 0.2 + 0.3 + 0.4 = 1.

Question 5

X ~ B(12, 0.2). Find P(X = 3) and P(X = 0), each to 3 significant figures.

Accepted Answer

P(X = 3) = ¹²C₃ × 0.2³ × 0.8⁹ = 220 × 0.008 × 0.1342 = 0.236. P(X = 0) = 0.8¹² = 0.0687, with no coefficient because there is only one way to fail every time. M1 for the product, A1 for 0.236, B1 for 0.0687. Sense check: np = 2.4, so a value of 3 should be common, and it is. The mode is 2, at 0.283, with 3 the next most likely outcome.

Question 6

X ~ B(20, 0.15). Write P(X ≥ 4) in a form a calculator's cumulative function accepts.

Accepted Answer

P(X ≥ 4) = 1 − P(X ≤ 3). B1 for the complement, B1 for the boundary dropping to 3. Calculators only sum upwards from zero, so every 'at least' converts to one minus an 'at most', with the boundary dropped by one. Writing 1 − P(X ≤ 4) instead quietly removes the case X = 4 from the answer, and that off-by-one is the main hazard of the topic.

Question 7

A machine produces faulty parts with probability 0.1, independently. In a box of 10 parts, find the probability of at most one faulty part, to 3 significant figures.

Accepted Answer

X ~ B(10, 0.1), and P(X ≤ 1) = P(X = 0) + P(X = 1) = 0.9¹⁰ + 10 × 0.1 × 0.9⁹ = 0.349 + 0.387 = 0.736. M1 for adding the two point probabilities, A1 for 0.349 and 0.387, A1 for 0.736. With a boundary this low, summing the two point probabilities by hand shows the working more clearly than the calculator's cumulative function does, and either route is acceptable.

Question 8

Twenty students in the same class are each asked whether they attended a party at the weekend, and X counts the yes answers. Give a reason why B(20, p) may be a poor model.

Accepted Answer

Friends go to parties together, so one student answering yes makes their friends more likely to answer yes too. The trials are not independent, and independence is a binomial condition. B1 for naming independence, B1 for tying it to friends attending together.

Question 9

X ~ B(15, 0.35). Find P(X > 7) and P(3 ≤ X ≤ 6), each to 3 significant figures.

Accepted Answer

Both have to be rewritten in the ≤ form a calculator's cumulative function accepts. For the first, X > 7 means X ≥ 8, so P(X > 7) = 1 − P(X ≤ 7) = 1 − 0.8868 = 0.113. M1 for the complement with its boundary at 7, A1 for 0.113. For the second, P(3 ≤ X ≤ 6) = P(X ≤ 6) − P(X ≤ 2) = 0.7548 − 0.0617 = 0.693, where the lower boundary drops to 2 so that X = 3 stays in the range. M1 for a difference of two cumulative values, A1 for 0.693. Subtracting P(X ≤ 3) instead would strip X = 3 out of a range that includes it, and would return 0.582.

Question 10

In a town, 12% of adults are left-handed. A random sample of 20 adults is taken and X is the number of them who are left-handed. Find P(X = 2) and P(X ≥ 3), each to 3 significant figures.

Accepted Answer

Take the sample as 20 independent trials with a constant probability, so X ~ B(20, 0.12). Then P(X = 2) = ²⁰C₂ × 0.12² × 0.88¹⁸ = 190 × 0.0144 × 0.1002 = 0.274. M1 for the complete product, A1 for 0.274. For the tail, P(X ≥ 3) = 1 − P(X ≤ 2) = 1 − 0.5631 = 0.437. B1 for 0.437.

Question 11

A gardener plants 50 bulbs taken from a single packet and counts how many of them flower. Give one reason why B(50, p) may be a reasonable model here, and one reason why it may not.

Accepted Answer

In its favour: the number of bulbs is fixed at 50 and each bulb either flowers or does not. B1 for a fixed number of trials with two outcomes. The bulbs come from one packet, so they are of one variety and one age and a constant probability p of flowering is plausible. B1 for a constant p tied to the packet. Against it: the bulbs are planted in the same bed and share the soil, the drainage and the weather, so a cold spell or a waterlogged corner would stop several of them together and the trials are not independent. B1 for independence failing, with a reason drawn from the context.

Question 12

A spinner lands on red with probability 0.2 on each spin, independently. It is spun n times. Find the smallest value of n for which the probability of at least one red exceeds 0.99.

Accepted Answer

Let X ~ B(n, 0.2). Then P(X ≥ 1) = 1 − P(X = 0) = 1 − 0.8ⁿ, so the requirement is 1 − 0.8ⁿ > 0.99, that is 0.8ⁿ < 0.01. Take logarithms of both sides: n log 0.8 < log 0.01. Since log 0.8 is negative, the inequality reverses on dividing, giving n > log 0.01/log 0.8 = 20.6. So the smallest whole number is n = 21. Confirm it: 1 − 0.8²⁰ = 0.9885, which fails, and 1 − 0.8²¹ = 0.9908, which passes. That is M1 for the complement, A1 for 0.8ⁿ < 0.01, M1 for taking logarithms, A1 for reversing the inequality, A1 for n = 21.

Question 13

X ~ B(20, p), where 0 < p < 0.5, and P(X = 0) = 0.0115 correct to 3 significant figures. Find the value of p, and hence find P(X ≥ 3) to 3 significant figures.

Accepted Answer

There is exactly one way to fail every trial, so P(X = 0) = (1 − p)²⁰ and no coefficient appears. M1 for that expression. Setting it equal to 0.0115 gives (1 − p)²⁰ = 0.0115, so 1 − p = 0.0115^1/20 = 0.800. M1 for taking the 20th root, or for taking logarithms and dividing by 20, A1 for 0.800. Hence p = 0.2, which lies in the range given. A1. With X ~ B(20, 0.2), P(X ≥ 3) = 1 − P(X ≤ 2) = 1 − 0.2061 = 0.794. M1 for the complement with its boundary at 2, A1 for 0.794. Check the value forwards before finishing: 0.8²⁰ = 0.011529, which is 0.0115 to 3 significant figures, as the question gave. The 20th root delivers 1 − p and not p, so the value to write down is 0.2 rather than 0.800, and the stated range 0 < p < 0.5 confirms which of the two the question is asking for.

Question 14

A fair six-sided die is rolled 4 times. X is the number of rolls showing a score of 5 or more. State the distribution of X and find P(X = 2). Y is the highest score shown in the four rolls. Explain why Y does not have a binomial distribution, and find P(Y ≤ 4).

Accepted Answer

A single roll shows 5 or more with probability 2/6 = 1/3, the four rolls are independent and that probability is the same every time, so X ~ B(4, 1/3). B1 for the distribution with both parameters. Then P(X = 2) = ⁴C₂ × (1/3)² × (2/3)² = 6 × (1/9) × (4/9) = 24/81 = 8/27 = 0.296. M1 for the product, A1 for 8/27. Y is not binomial because it counts nothing: it records the largest score seen, and takes the values 1 to 6 rather than a number of successes from 0 to 4. B1 for Y not being a count of successes. For the last part, the highest score is at most 4 exactly when every one of the four rolls is at most 4, and a single roll is at most 4 with probability 4/6 = 2/3, so P(Y ≤ 4) = (2/3)⁴ = 16/81 = 0.198. M1 for turning the condition on the maximum into a condition on all four rolls, A1 for 16/81.

Question 15

Components are faulty with probability 0.04, independently of one another, and are packed into boxes of 25. A box is rejected if it contains more than one faulty component. Find the probability that a box is rejected, and the probability that exactly one of 10 boxes is rejected.

Accepted Answer

Let X be the number of faulty components in a box, so X ~ B(25, 0.04). M1 for the first model. A box is rejected when X ≥ 2, so P(rejected) = 1 − P(X ≤ 1) = 1 − 0.7358 = 0.2642. M1 for the complement, A1 for 0.264. Now let Y be the number of the 10 boxes that are rejected. The boxes are filled independently and each is rejected with the same probability 0.2642, so Y ~ B(10, 0.2642). M1 for the second binomial with that probability. Then P(Y = 1) = ¹⁰C₁ × 0.2642 × 0.7358⁹ = 0.167. A1 for 0.167. Carry the unrounded 0.26419 into the second stage: entering 0.26 returns 0.173 and entering 0.3 returns 0.121.