Practise › Questions › The binomial distribution
The binomial distribution questions
Count the successes in a fixed number of independent yes-or-no trials and the answer has a distribution you can write down. The binomial is the exam's favourite model, and the skill is checking it applies before you reach for the calculator.
8 original questions · 22 marks · the the binomial distribution notes · Statistics
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
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.
Worked 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 — the word that earns the mark is uniform, and the reason is equally likely. A total of two dice is the standard trap: 7 is more likely than 2, so that variable is not uniform.State the conditions under which a count X can be modelled by a binomial distribution B(n, p).
Worked 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.X ~ B(8, 0.25). Find P(X = 2) to 3 significant figures.
Worked 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. Three ingredients multiply in one line: the number of ways of placing the two successes, the probability of those successes, and the probability of the six failures. The exponents must sum to 8, which is the quickest check that the line is right.X ~ B(12, 0.2). Find P(X = 3) and P(X = 0), each to 3 significant figures.
Worked 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.X ~ B(20, 0.15). Write P(X ≥ 4) in a form a calculator's cumulative function accepts.
Worked 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.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.
Worked 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 earns the marks.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.
Worked 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. An answer has to name the condition and tie it to the context; 'the students are not independent' alone is too vague, so say why they are not.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.
Worked 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. Marks go for turning 'at least one' into its complement, for forming the inequality, for reversing it correctly when dividing by a negative logarithm, and for rounding up rather than to the nearest whole number. In codes 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. Rounding 20.6 down to 20 is the loss the question is built around, and quoting the two probabilities is the cheapest way to prove you have not made it.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise the binomial distribution one question at a time
The player marks nothing for you. It shows one question, waits, then shows the worked answer so you can mark yourself, and brings a question back sooner when it went badly.