OCR A Level Mathematics A: Probability: Venn and tree diagrams, conditional probability and modelling (H240) – 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 from your board or school.

These questions cover section 2.03 Probability of the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. Questions 1 to 5 use Stage 1 content (2.03a, 2.03b), apart from part (c) of question 3, which is conditional probability and so A Level only. Questions 6 to 11 need A Level only content (2.03c to 2.03e). Learn the methods first from the study guide and the revision notes. Course hub: /boards/ocr/a-level/mathematics/. Checklist: /checklists/ocr/a-level/mathematics/.

Questions

1. A and B are two events that cannot occur together. P(A) = 0.45 and P(B) = 0.3.

(a) Find P(A ∪ B). [1] (b) Find P(A’ ∩ B’). [1] (c) Explain why A and B are not independent. [1]

2. A random variable X has P(X = x) = k/x for x = 1, 2, 4, and takes no other values.

(a) Show that k = 4/7. [2] (b) Find P(X > 1). [1] (c) Two independent values of X are observed. Find the probability that their sum is 5. [3]

3. A fair spinner numbered 1 to 5 and a fair spinner numbered 1 to 3 are each spun once. The two numbers are multiplied to give the score.

(a) Find the probability that the score is even. [2] (b) Find the probability that the score is greater than 8. [2] (c) Given that the score is even, find the probability that it is greater than 8. [2]

4. Of 78 stallholders at a market, 41 sell food (F), 33 sell crafts (C) and 15 sell neither.

(a) Find the number who sell both. [2] (b) One stallholder is chosen at random. Find the probability that they sell food only. [1] (c) Find the probability that they sell exactly one of the two types of goods. [2]

5. A phone app crashes on start-up with probability 0.08 and, independently, on logging out with probability 0.05.

(a) Find the probability that it does not crash on either occasion. [2] (b) Find the probability that it crashes exactly once. [2]

6. A bag holds 5 chocolate, 4 mint and 3 toffee sweets. Two are drawn one after the other and not put back.

(a) Find the probability that both sweets have the same flavour. [3] (b) Find the probability that the second is toffee, given that the first is chocolate. [1] (c) Find the probability that at least one sweet is mint. [2]

7. A survey of 180 residents recorded age group and car ownership.

	Under 30	30 to 59	60 and over
Owns a car	20	62	38
No car	28	18	14

A resident is chosen at random.

(a) Find the probability that the resident owns a car, given that they are 60 or over. [2] (b) Find the probability that the resident is aged 30 to 59, given that they do not own a car. [2] (c) Determine whether “owns a car” and “under 30” are independent. [3]

8. P(A) = 0.6, P(B | A) = 0.25 and P(B | A’) = 0.5.

(a) Find P(A ∩ B). [1] (b) Find P(B). [2] (c) Find P(A ∪ B). [2]

9. An integer is chosen at random from 1 to 40 inclusive. Event A is “the number is a multiple of 4” and event B is “the number is a perfect square”.

(a) Find P(A | B). [2] (b) Find P(B | A). [2]

10. P(A) = 0.52, P(B) = 0.46 and P(A’ ∩ B’) = 0.18.

(a) Find P(A ∩ B). [2] (b) Find P(A | B). [2] (c) Show that A and B are not independent. [2] (d) Find P(B’ | A’). [2]

11. A transport survey uses a model in which each of four students chosen from one form travels by bus with probability 0.35, independently of the others.

(a) Find the probability that at least one of the four travels by bus. [2] (b) State two assumptions made by the model. [2] (c) The four students are close friends who often travel together. Explain how this affects your answer to (a). [2]

Answers

1. (a) P(A ∪ B) = 0.45 + 0.3 = 0.75 [1] (b) P(A’ ∩ B’) = 1 − 0.75 = 0.25 [1] (c) Mutually exclusive means P(A ∩ B) = 0, but P(A) × P(B) = 0.135, which is not 0, so the product rule fails. [1] [3] Examiner insight: For an explanation, quote the numbers: stating that P(A ∩ B) = 0 differs from the product 0.135 is more convincing than the general claim that “exclusive events are not independent”.

2. (a) The probabilities sum to 1: k + k/2 + k/4 = 7k/4 [1], so 7k/4 = 1 and k = 4/7 [1]. [2] (b) P(X > 1) = 1 − 4/7 = 3/7 [1] (c) P(X = 1) = 4/7, P(X = 2) = 2/7, P(X = 4) = 1/7. The pairs summing to 5 are (1, 4) and (4, 1). [1] Probability = 2 × 4/7 × 1/7 [1] = 8/49. [1] [3] Examiner insight: In a “show that” question, write the line where the probabilities are added and set equal to 1 before the result, because the given value cannot be the starting point of your working.

3. (a) There are 15 equally likely outcomes. Odd products need both numbers odd: 3 × 2 = 6 outcomes. [1] Even = 9 outcomes, so P = 9/15 = 3/5. [1] [2] (b) Products above 8 are 9, 10, 12, 15, from (3, 3), (5, 2), (4, 3), (5, 3). [1] P = 4/15. [1] [2] (c) Of those four, 10 and 12 are even. [1] P = 2/9, since there are 9 even outcomes in total. 2/9 [1] [2] Examiner insight: A conditional question restricts the sample space. Using 2/15 (a share of all outcomes) instead of 2/9 is the typical slip, so state the new total before dividing.

4. (a) Selling at least one type: 78 − 15 = 63. [1] Both = 41 + 33 − 63 = 11. [1] [2] (b) Food only = 41 − 11 = 30, so P = 30/78 = 5/13. [1] (c) Crafts only = 33 − 11 = 22. Exactly one = 30 + 22 = 52 [1], so P = 52/78 = 2/3. [1] [2] Examiner insight: Drawing the Venn diagram with all four regions totalling 78 lets you check your work and shows method even if one subtraction slips.

5. (a) P(no crash) = 0.92 × 0.95 [1] = 0.874 [1] [2] (b) P(exactly once) = 0.08 × 0.95 + 0.92 × 0.05 [1] = 0.076 + 0.046 = 0.122 [1] [2] Examiner insight: “Exactly once” has two routes. Showing both products before adding gives a clear method, and a missing second route is easy to see and costly.

6. (a) P(same) = 5/12 × 4/11 + 4/12 × 3/11 + 3/12 × 2/11 [1] = 20/132 + 12/132 + 6/132 [1] = 38/132 = 19/66 [1] [3] (b) After one chocolate is removed, 11 sweets remain and 3 are toffee, so P = 3/11. [1] (c) P(no mint) = 8/12 × 7/11 = 14/33 [1], so P(at least one mint) = 1 − 14/33 = 19/33. [1] [2] Examiner insight: The second-stage denominator must be 11. Labelling each branch with its fraction on a quick tree makes that visible and protects the method.

7. (a) Restrict to the 60-and-over column, which has 38 + 14 = 52 residents. [1] P = 38/52 = 19/26. [1] [2] (b) The “no car” row totals 28 + 18 + 14 = 60. [1] P = 18/60 = 3/10. [1] [2] (c) P(car) = 120/180 = 2/3 [1]. P(car | under 30) = 20/48 = 5/12 [1]. These differ, so the events are not independent. [1] [3] Examiner insight: An independence conclusion needs both probabilities written down and a statement in words. A bare “not independent” has nothing to credit.

8. (a) P(A ∩ B) = 0.6 × 0.25 = 0.15 [1] (b) P(A’ ∩ B) = 0.4 × 0.5 = 0.2 [1]. P(B) = 0.15 + 0.2 = 0.35. [1] [2] (c) P(A ∪ B) = 0.6 + 0.35 − 0.15 [1] = 0.8 [1] [2] Examiner insight: Part (b) needs the A’ branch as well. Using only P(B | A) is a common shortcut that answers a different question.

9. (a) Squares: 1, 4, 9, 16, 25, 36, so 6 outcomes. [1] Those that are multiples of 4: 4, 16, 36, so P(A | B) = 3/6 = 1/2. [1] [2] (b) Multiples of 4: 4, 8, …, 40, so 10 outcomes. [1] P(B | A) = 3/10. 3/10 [1] [2] Examiner insight: The two answers differ, which is the point. Write the “given” set and its size first so the denominator is not chosen by habit.

10. (a) P(A ∪ B) = 1 − 0.18 = 0.82 [1]. P(A ∩ B) = 0.52 + 0.46 − 0.82 = 0.16. [1] [2] (b) P(A | B) = 0.16 / 0.46 [1] = 8/23. [1] [2] (c) If independent, P(A ∩ B) would equal 0.52 × 0.46 = 0.2392. [1] This is not 0.16, so A and B are not independent. [1] [2] (d) P(A’) = 0.48. P(B’ | A’) = 0.18 / 0.48 [1] = 3/8. [1] [2] Examiner insight: A “show that” for non-independence is complete only when the product is evaluated and compared with the intersection. Say what the two values are.

11. (a) P(at least one) = 1 − 0.65⁴ [1] = 1 − 0.1785 = 0.821 (3 s.f.) [1] [2] (b) Each student has the same probability, 0.35, of travelling by bus. [1] The students’ choices are independent of each other. [1] [2] (c) Friends travelling together makes the students’ choices positively linked, so independence fails. [1] Bus users tend to appear together, so a group with no bus user becomes more likely and P(at least one) would be expected to be smaller than 0.821. [1] [2] Examiner insight: Link the comment to the context and give a direction. “Not realistic” alone has no content, while “less likely than 0.821 because choices cluster” does.

Where marks are usually lost

• Writing “independent” or “not independent” with no product test shown.
• Treating mutually exclusive events as independent, or the reverse.
• Using the full total as the denominator when the question says “given that”.
• Using P(A) + P(B) for a union when the two events overlap.
• Forgetting that the second stage of a without-replacement tree has a smaller total.
• Computing only one route for “exactly one” or “different colours”.
• Rounding part-way through, so a later equality test fails by a small amount.
• Giving a modelling comment with no named assumption or direction of effect.

Next steps

• Revise with the revision notes and the study guide.
• Return to the course hub and tick off topics on the checklist.
• Find weak spots with the free diagnostics.
• Book a free trial class for help with any question you could not finish.

Official syllabus

OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR. Section 2.03 Probability.