Mathematics 9709 · syllabus 5.3
Cambridge A Level Mathematics 9709 — Probability topical past papers
Probabilities from equally likely outcomes, tree diagrams and counting, with mutually exclusive and independent events and conditional probability.
- Questions
- 155
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 58
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What Probability covers
Section 5.3 of Paper 5 covers the rules of probability in simple settings. You find probabilities by listing equally likely outcomes, such as the totals from two dice, by counting with permutations and combinations, or by multiplying along the branches of a tree diagram and adding the branches you need. The notation includes P(A ∪ B), P(A ∩ B), P(A | B) and A′ for the event that A does not happen.
Two definitions matter. Events are mutually exclusive if they cannot both happen, so P(A ∩ B) = 0; they are independent if one happening does not change the probability of the other, so P(A ∩ B) = P(A) × P(B), which is the test you use to decide independence. Conditional probability P(A | B) = P(A ∩ B)/P(B) is the probability of A given that B has happened; on a tree diagram it is often asked in reverse, as the probability of a first-stage outcome given the second.
Questions are mainly numerical and use no algebra beyond Paper 1. You are not expected to quote the addition rule P(A ∪ B) = P(A) + P(B) − P(A ∩ B), so when events overlap a Venn diagram or a direct count of outcomes is usually the clearest route. Probability underlies the rest of the statistics in 9709: discrete random variables and their distributions in section 5.4, and then the binomial, geometric, normal and Poisson models.
In the syllabus
The headings this topic has in the Cambridge International 2026–2027 syllabus, as the board prints them.
- 5.3 Probability
What you should be able to do
- Find probabilities by listing equally likely outcomes or drawing a sample space diagram
- Use counting methods to find the probabilities of selections
- Draw and use tree diagrams, including for selections without replacement
- Decide whether two events are independent by comparing P(A ∩ B) with P(A) × P(B)
- Calculate conditional probabilities, including the probability of a first outcome given a second
- Tell mutually exclusive events from independent ones
Key ideas
- P(A′) = 1 − P(A)
- Mutually exclusive events: P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B)
- Independent events: P(A ∩ B) = P(A) × P(B)
- Conditional probability: P(A | B) = P(A ∩ B) / P(B)
- Without replacement, the probabilities on later branches change because the totals change
How Probability has been examined
From 2022 to 2026, Probability was on Paper 5 · Probability & Statistics 1 in 14 of 14 exam series, worth about 24% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 5 · Probability & Statistics 1 | 58 | 24% | 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: Probability.
A worked example
A bag contains 5 red counters and 3 blue counters. Two counters are taken at random without replacement. Find the probability that the second counter is blue, and the probability that the first counter was red given that the second is blue.
- P(red then blue) = 5/8 × 3/7 = 15/56.
- P(blue then blue) = 3/8 × 2/7 = 6/56.
- P(second is blue) = 15/56 + 6/56 = 21/56 = 3/8.
- P(first red | second blue) = P(red then blue) ÷ P(second blue) = (15/56) ÷ (21/56).
- This simplifies to 15/21 = 5/7.
Answer: P(second blue) = 3/8; P(first red | second blue) = 5/7
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Treating mutually exclusive and independent as the same: exclusive events with non-zero probabilities are never independent, because P(A ∩ B) = 0 while P(A) × P(B) is not.
- Dividing by the wrong probability in a conditional: P(A | B) has P(B), the event that is given, on the bottom.
- Keeping the same probabilities on the second branches of a tree when items are taken without replacement.
- Arguing that events 'seem unrelated' when asked to show independence: the numbers P(A ∩ B) and P(A) × P(B) must be compared.
Exam technique
- Draw the tree diagram or sample space and label it fully before calculating; most of the marks come from the right structure.
- For 'given that' questions, write P(A | B) = P(A ∩ B)/P(B) and work out the top and the bottom separately.
- Leave probabilities as exact fractions where they are simple; otherwise give them 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.