Mathematics 9709 · syllabus 5.4
Cambridge A Level Mathematics 9709 — Discrete random variables topical past papers
Probability distribution tables, expectation and variance, and the binomial and geometric distributions as models for successes and trials.
- Questions
- 211
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 83
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What Discrete random variables covers
A discrete random variable X takes separate values, each with a probability, and the probabilities add up to 1. In section 5.4 of Paper 5 you build a probability distribution table from a situation, such as the number of sixes in a few throws, the larger score on two spinners or the number of red counters drawn, and calculate the expectation E(X) and the variance Var(X).
Two named models follow. The binomial distribution B(n, p) counts the successes in a fixed number n of independent trials, each with the same probability of success p. The geometric distribution Geo(p) counts the number of trials up to and including the first success. You use their probability formulae, the mean and variance of the binomial distribution and the mean of the geometric distribution, and you say when each model fits a situation, or why it does not.
The section builds on earlier Paper 5 work: binomial coefficients come from the permutations and combinations of section 5.2, and E(X) and Var(X) have the same shape as the mean and variance of a frequency distribution in section 5.1. The binomial distribution returns in section 5.5, where a normal distribution approximates it for large n, and in Paper 6, where it is the model for some hypothesis tests and can be approximated by a Poisson distribution.
In the syllabus
The headings this topic has in the Cambridge International 2026–2027 syllabus, as the board prints them.
- 5.4 Discrete random variables
What you should be able to do
- Draw up a probability distribution table and use the fact that the probabilities add up to 1
- Calculate E(X) and Var(X) from a table
- Recognise situations that fit B(n, p) or Geo(p), and state the conditions in context
- Calculate binomial probabilities, including 'at least' and 'at most' cases
- Calculate geometric probabilities, including P(X > r) = (1 − p)ʳ
- Use np, np(1 − p) and 1/p for means and variances
Key ideas
- E(X) = Σxp and Var(X) = Σx²p − (E(X))²
- X ~ B(n, p): P(X = r) = ⁿCr pʳ(1 − p)ⁿ⁻ʳ, with E(X) = np and Var(X) = np(1 − p)
- X ~ Geo(p): P(X = r) = p(1 − p)ʳ⁻¹ for r = 1, 2, 3, …, with E(X) = 1/p
- For Geo(p), P(X > r) = (1 − p)ʳ, the probability that the first r trials all fail
- A binomial model needs a fixed number of independent trials, each with two outcomes and the same probability of success
How Discrete random variables has been examined
From 2022 to 2026, Discrete random variables was on Paper 5 · Probability & Statistics 1 in 14 of 14 exam series, worth about 35% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 5 · Probability & Statistics 1 | 83 | 35% | 14 of 14 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 7.
Filed in Momentum under: Discrete Random Variables, The Binomial Distribution, Geometric Distribution.
A worked example
A player scores from a free throw with probability 0.3, independently of other throws. Find the probability that she scores at least 2 times in 10 throws, and the probability that her first score comes on her 4th throw.
- Let X be the number of scores in 10 throws, so X ~ B(10, 0.3).
- P(X ≥ 2) = 1 − P(X = 0) − P(X = 1).
- P(X = 0) = 0.7¹⁰ = 0.02825, and P(X = 1) = ¹⁰C₁ × 0.3 × 0.7⁹ = 0.12106.
- So P(X ≥ 2) = 1 − 0.02825 − 0.12106 = 0.85069.
- For the first score, let Y ~ Geo(0.3): P(Y = 4) = 0.7³ × 0.3 = 0.1029.
Answer: P(at least 2 scores) = 0.851; P(first score on the 4th throw) = 0.103
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Forgetting to subtract (E(X))² when finding the variance, which gives Σx²p instead of Var(X).
- Reading 'at least 3' as P(X > 3): it is P(X ≥ 3) = 1 − P(X ≤ 2).
- Using a binomial model when items are drawn without replacement from a small group, so the probability changes from one trial to the next.
- Counting only the failures before the first success for the geometric distribution: here X includes the trial on which the success happens.
Exam technique
- Check that a distribution table's probabilities add up to 1 before calculating E(X).
- For 'at least' questions, use the complement and write out the probabilities you are subtracting.
- State the distribution with its parameters, such as X ~ B(10, 0.3), before using it, and give answers to 3 significant figures.
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.