OxfordAQA A Level Mathematics: Bernoulli and Binomial Distributions — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Bernoulli and Binomial Distributions study guide · Bernoulli and Binomial Distributions revision notes

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Section A

1. State the four conditions required for a binomial distribution to apply. [4]

2. State the mean and variance of a single Bernoulli trial with probability of success p. [2]

Section B

3. A biased coin has P(heads) = 0.35. It is tossed 6 times. Find the probability of exactly 4 heads. [4]

4. X ~ B(12, 0.25). Find the mean and variance of X. [3]

5. A bag contains 10 red and 5 blue balls. Five balls are drawn one at a time without replacement. Explain why the number of red balls drawn should not be modelled using a binomial distribution, and state what would need to change for the binomial model to apply. [4]

6. Derive the mean of a binomial distribution X ~ B(n, p) from the mean of a single Bernoulli trial, explaining your reasoning in full sentences rather than only quoting the formula. [4]

7. Describe how you would use a cumulative binomial table to find P(X ≤ 5) for X ~ B(15, 0.4), and explain why this approach is generally preferred over calculating each term of the formula separately. [4]

8. X ~ B(10, p) has variance 1.6. Find the two possible values of p. [5]

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Answers

1. A fixed number of trials, n [1]; each trial has only two possible outcomes [1]; the probability of success, p, is constant across all trials [1]; the trials are independent of each other [1].

2. Mean E(X) = p [1]. Variance Var(X) = p(1 − p) [1].

3. Confirm the conditions: n = 6, p = 0.35, trials independent [1]. P(X = 4) = ⁶C₄ × (0.35)⁴ × (0.65)² [1] ⁶C₄ = 15, (0.35)⁴ = 0.01500625, (0.65)² = 0.4225 [1] P(X = 4) = 15 × 0.01500625 × 0.4225 = 0.0951 (3 s.f.) [1].

4. Mean = np = 12 × 0.25 = 3 [1] [1]. Variance = np(1 − p) = 12 × 0.25 × 0.75 = 2.25 [1].

5. Drawing without replacement means that once a ball is removed, the probability of drawing a red ball on the next draw changes, since the total number of balls and the number of red balls remaining are both different [1] [1]. This violates the requirement that trials be independent and that p stays constant across all trials [1], so the binomial model does not strictly apply. For the binomial model to apply, each ball would need to be replaced (and the bag mixed) before the next draw, keeping the composition of the bag, and therefore p, constant across all five draws [1].

6. A binomial random variable X ~ B(n, p) is the sum of n independent, identically distributed Bernoulli trials [1]. The mean of a sum of independent random variables equals the sum of their individual means [1]. Since each individual Bernoulli trial has mean p [1], summing n of them gives a total mean of n × p = np [1]. The same reasoning extends to variance: because the trials are independent, the variance of the sum equals the sum of the individual variances, giving n lots of p(1 − p), or np(1 − p). (Stating this chain of reasoning, rather than only quoting the two final formulas, is what a strong answer on this sub-topic demonstrates.)

7. First confirm the binomial conditions hold for X ~ B(15, 0.4) [1]; then locate the row or column corresponding to n = 15 and p = 0.4 in the cumulative binomial table [1], and read off the tabulated value for P(X ≤ 5) directly [1]. This is preferred over calculating each term separately because summing six individual formula calculations (P(X = 0) through P(X = 5)) by hand is slower and carries more opportunity for arithmetic error than reading one value directly from a table that has already performed that summation [1].

8. Variance = np(1 − p), so 10 × p(1 − p) = 1.6 [1], giving p(1 − p) = 0.16 [1]. Expanding: p − p² = 0.16, so p² − p + 0.16 = 0 [1]. Using the quadratic formula: p = (1 ± √(1 − 0.64)) ÷ 2 = (1 ± 0.6) ÷ 2 [1]. So p = 0.8 or p = 0.2 [1]. Both values are valid solutions of the variance equation, since p and 1 − p play symmetric roles in the product p(1 − p) — swapping which outcome is labelled “success” turns one solution into the other. (A full answer should give both roots rather than stopping at the first one found.)

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Where marks are usually lost

• Applying the binomial formula without first checking all four conditions genuinely hold in the scenario described.
• Confusing single-trial Bernoulli mean/variance (p, p(1 − p)) with the full binomial values (np, np(1 − p)).
• Miscalculating or omitting the binomial coefficient ⁿCₓ.
• Giving only one root when a variance equation produces a quadratic in p, rather than both valid solutions.
• Describing dependence and non-constant probability as separate issues in a without-replacement scenario, rather than recognising that removing an item changes both at once.