Use binomial moments and infer valid parameters
Calculate the centre and spread of a success count, and reject inferred parameters that cannot define a binomial model.
Section 5.4 · Discrete random variables
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.4. A probability table needs non-negative probabilities summing to one. E(X) is a probability-weighted mean; Var(X) = E(X²) − [E(X)]², with squared units. A binomial model B(n,p) counts successes in a fixed non-negative integer number n of independent two-outcome trials with a constant success probability p. A geometric model Geo(p) counts the trial of the first success, starting at 1, with independent trials and a constant 0 < p ≤ 1. Formula proofs are not required. The expectation of the geometric distribution is required; its variance is not required by section 5.4. Normal approximations belong to section 5.5 and are not used here. Exact model probabilities do not establish real-world independence or constant rates without evidence. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Define B(n,p) and its assumptions.
- Distinguish variance from standard deviation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the mean of B(12,1/4)?
What you will learn
- Use mean np and variance np(1 − p).
- Check probability bounds and integer trial counts when inferring n and p.
Mean and variance describe different features
For X ~ B(n,p), E(X) = np and Var(X) = np(1 − p). The standard deviation is √[np(1 − p)], not the variance. For B(12,1/4), the mean is 3, the variance is 9/4 and the standard deviation is 3/2. If X counts successes, variance has squared count units while the mean and standard deviation use count units.
The mean is a long-run average across repeated experiments, not a guarantee that each experiment has np successes. With n = 12 and p = 1/4, several count values are possible despite the mean being 3. The formulas describe the stated model; observed data need separate evidence before the binomial assumptions can be accepted.
An algebraic solution must define a valid model
If an exact model mean μ > 0 and variance v are supplied, v/μ = 1 − p. Hence p = 1 − v/μ, followed by n = μ/p when p > 0. Exact mean 8 and variance 6 give p = 1/4 and n = 32, a valid positive integer. Substitute both values back into the moment formulas.
Exact mean 4 with variance 5 is impossible because a binomial variance cannot exceed its mean. Exact mean 4 with variance 2.5 gives p = 3/8 but n = 32/3, which is not an integer, so it also cannot be an exact binomial model. Estimated sample moments need a separate fitting discussion; do not round an invalid exact trial count to manufacture a solution. Zero-mean or endpoint models need direct consideration rather than division by zero.
Pause and explain
Exact mean 4 and variance 2.5 imply p = 3/8 and n = 32/3. Is this a valid binomial model?
Worked example
A binomial variable has exact mean 8 and variance 6. Find n and p.
Show the worked solution
- Divide variance by mean: 6/8 = 1 − p.
- Thus p = 1/4 and n = 8/(1/4) = 32.
- Check 32(1/4) = 8 and 32(1/4)(3/4) = 6, with an integer n and valid p.
Answer n = 32; p = 1/4.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
X ~ B(12,1/4). Find the mean, variance and standard deviation.
Give me a hint
Use np, then np(1 − p), then the square root.
Compare my reasoning
- Mean = 12 × 1/4 = 3.
- Variance = 12(1/4)(3/4) = 9/4.
- Standard deviation = √(9/4) = 3/2.
Mean 3; variance 9/4; standard deviation 3/2.
Look for these in your work
- Separate centre, variance and standard deviation.
- Retain exact values and the appropriate units.
A binomial variable has exact mean 8 and variance 6. Infer n and p and verify them.
Give me a hint
Divide variance by the positive mean to find 1 − p.
Compare my reasoning
- 1 − p = 6/8 = 3/4, giving p = 1/4.
- n = 8/(1/4) = 32, an integer.
- The supplied moments are recovered as 32/4 = 8 and 32(1/4)(3/4) = 6.
n = 32; p = 1/4.
Look for these in your work
- Check the integer trial count and probability bounds.
- Substitute back into both moment equations.
Assess these exact binomial claims: mean 4, variance 5; and mean 4, variance 2.5.
Give me a hint
Test both the variance bound and the integrality of an inferred n.
Compare my reasoning
- The first has variance greater than mean, which would force a negative p.
- The second gives p = 1 − 2.5/4 = 3/8, then n = 4/(3/8) = 32/3.
- Neither defines a valid exact binomial model: the first fails probability bounds and the second fails integer n.
Neither is valid; variance 5 exceeds mean 4, and variance 2.5 gives n = 32/3.
Look for these in your work
- Check every model parameter, not only p.
- Do not round to repair exact moment claims.
Common mistakes
- Take a square root to obtain standard deviation.
- The number of trials must be a non-negative integer.
- Do not divide by a zero mean or round exact inferred parameters.
What can you explain now?
Can exact binomial mean 4 and variance 5 coexist?
Compare with the explanation
No; np(1 − p) ≤ np for valid p.
The variance-to-mean ratio cannot exceed one under a valid binomial model.
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Make it stick
Calculate np and np(1 − p), distinguish standard deviation, and check that inferred parameters define a genuine fixed-trial model.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.