IGCSE Mathematics: Probability — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Probability revision notes

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Questions

1. A fair six-sided dice is rolled once. State the probability that the score is a prime number. [1]

2. A biased coin is tossed 200 times and lands on heads 118 times.

(a) Calculate the relative frequency of heads. [1] (b) The coin is tossed a further 50 times. Use the relative frequency from part (a) to estimate the expected number of heads in these 50 tosses. [2]

3. (Extended) A bag contains 5 red counters and 3 blue counters. Two counters are taken from the bag at random, one after the other, without replacement.

(a) Draw and label a tree diagram to show the possible outcomes and their probabilities. [3] (b) Calculate the probability that both counters are red. [2] (c) Calculate the probability that the two counters are different colours. [3]

4. (Extended) For two events A and B, P(A) = 0.3 and P(A and B) = 0.15. Find P(B | A), the probability of B given that A has occurred. [2]

5. (Extended) In a group of 40 students, 24 study French, 18 study Spanish, and 10 study both. A student is picked at random from those who study French. Find the probability that this student also studies Spanish. [3]

6. A biased dice has P(rolling a 6) = 0.15.

(a) Find the probability of not rolling a 6. [1] (b) The dice is rolled 300 times. Estimate the number of times a 6 is not rolled. [2]

7. A fair coin is tossed twice. Using a tree diagram approach, find the probability of getting exactly one head. [3]

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Answers

1. Prime scores on a dice are 2, 3, 5 — three out of six outcomes. P(prime) = 3/6 = ½ [1].

2. (a) Relative frequency = 118 ÷ 200 = 0.59 [1]. (b) Expected heads = 0.59 × 50 [1] = 29.5 [1].

3. (a) First pick: P(red) = 5/8, P(blue) = 3/8 [1]. Second pick (no replacement): from red first, P(red) = 4/7, P(blue) = 3/7; from blue first, P(red) = 5/7, P(blue) = 2/7 [2] — a fully labelled two-stage tree diagram showing these four branches. (b) P(both red) = 5/8 × 4/7 [1] = 20/56 = 5/14 [1]. (c) P(different colours) = P(R then B) + P(B then R) = (5/8 × 3/7) + (3/8 × 5/7) [2] = 15/56 + 15/56 = 30/56 = 15/28 [1].

4. P(B | A) = P(A and B) ÷ P(A) [1] = 0.15 ÷ 0.3 = 0.5 [1].

5. Of the 24 who study French, 10 also study Spanish [1]. P(Spanish | French) = 10 ÷ 24 [1] = 5/12 [1].

6. (a) P(not 6) = 1 − P(6) = 1 − 0.15 = 0.85 [1], since the probabilities of an outcome and its complement always sum to 1. (b) Expected number of “not 6” rolls = 0.85 × 300 [1] = 255 [1].

7. Each toss is independent, so the probability stays 0.5 for heads and 0.5 for tails on both branches, unlike a without-replacement problem [1]. Exactly one head happens via two paths: heads-then-tails or tails-then-heads [1]. P(exactly one head) = (0.5 × 0.5) + (0.5 × 0.5) = 0.25 + 0.25 = 0.5 [1].

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Where marks are usually lost

• Confusing theoretical probability (counting outcomes) with relative frequency (an experimental estimate from data) — they are related but not automatically equal, and a question about one is not answered with the other.
• Multiplying the wrong branch probabilities together on a tree diagram, or forgetting that probabilities without replacement change on the second branch.
• Adding two combined-event probabilities when they should be multiplied (independent/sequential events), or the reverse when combining two mutually exclusive outcomes.
• Not simplifying a fraction at the end, or giving a probability greater than 1 or as a ratio instead of a fraction/decimal.
• In conditional probability, dividing by the wrong total — P(B | A) means restricting attention to the cases where A has already happened, not the whole sample space.

Examiner report insight

• On a Venn diagram, shading the intersection (A n B) by default is a common habit – check what the question actually asked for before shading, since the intersection is only one of many possible regions.
• Set-notation questions built from n(...) (a numerical count of elements) need the actual number of elements evaluated, not just a region shaded on a diagram – practise both skills separately.
• Elements that belong only to the universal set – outside every named subset – are still part of the total and are easy to forget when completing a Venn diagram.
• The complement A’ is everything in the universal set U that is not in A – this includes the part of any other set B that lies outside A (B minus the A∩B overlap) and the region outside both A and B, but it excludes all of A, including the A∩B overlap itself (since A∩B is part of A). A common error wrongly includes A∩B in A’.

Source: Cambridge International, 0580 Mathematics Principal Examiner Report, June 2024 series, Papers 12, 13, 21, 22, 23, 31 (verified 2026-09-02).

Approaching probability questions

Before multiplying or adding anything, decide what kind of question is in front of you: a single event (count outcomes), a combined event across two or more stages (build a tree and multiply along branches), or a conditional probability (restrict to a smaller sample space first). Getting this classification right first is what determines whether the next step should be a multiplication, an addition, or a division – most errors on this topic come not from arithmetic mistakes but from applying the wrong one of these three operations to a correctly understood situation.