OCR GCSE Mathematics: Tree diagrams and probability laws (J560) – 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 OCR GCSE (9-1) Mathematics (J560), Version 2.1 (August 2026), content references 11.02d, 11.02e and 11.02f. Questions 1 to 8 are for both tiers. Questions 9 to 12 are marked (Higher tier only). The mark allocations are set by Marlbridge for practice. Read the study guide first, then use the revision notes as a quick reminder. Course pages: OCR GCSE Mathematics hub, checklist and the free diagnostics.

Questions

1. A bag holds blue, white and black discs only. One disc is taken at random. p(blue) = 0.3 and p(white) = 0.45.

(a) Find p(blue or white). [1] (b) Find p(black). [1]

2. A netball player takes two shots at goal. The probability that the first scores is 0.8 and the probability that the second scores is 0.6. The shots are independent.

(a) Copy and complete the tree diagram by writing the four route probabilities. Label scores as S and misses as M. [2] (b) Find the probability that exactly one shot scores. [2]

3. A box holds 12 chocolates: 5 dark and 7 milk. Two are chosen at random without replacement.

(a) Find the probability that both are dark. [2] (b) Find the probability that the two chocolates are of different types. [3]

4. In a cafe, the probability that a customer buys a drink is 0.55 and the probability that a customer buys a snack is 0.4. The probability that a customer buys both is 0.2. Find the probability that a customer buys a drink or a snack (or both), and the probability that they buy neither. [3]

5. A machine makes three parts one after another. Each part is faulty with probability 0.05, independently of the others. Give each answer to 3 significant figures.

(a) Find the probability that none of the three parts is faulty. [2] (b) Find the probability that at least one is faulty. [1] (c) Find the probability that exactly one is faulty. [2]

6. The probability that it is rainy on Monday is 0.3. If Monday is rainy, the probability that Tuesday is rainy is 0.6. If Monday is dry, the probability that Tuesday is rainy is 0.2.

(a) Draw a tree diagram and write all probabilities on it. [2] (b) Find the probability that it rains on exactly one of the two days. [3] (c) Find the probability that Tuesday is rainy. [2]

7. Zara buys two tickets from a raffle of 25 tickets. Four of the 25 tickets win a prize. Find the probability that at least one of her tickets wins. [4]

8. A learner driver takes up to three driving tests. The probability of passing on the first attempt is 0.45. After a failure, the probability of passing the next attempt is 0.6 for the second attempt and 0.75 for the third attempt.

(a) Find the probability that the learner passes on the second attempt. [2] (b) Find the probability that the learner passes within three attempts. [3] (c) Find the probability that the learner fails all three attempts. [1]

9. (Higher tier only) A fair ten-sided die is numbered 1 to 10 and rolled once.

(a) Find the probability that the score is prime, given that it is odd. [2] (b) Find the probability that the score is odd, given that it is prime. [2]

10. (Higher tier only) A commuter catches a train. p(train on time) = 0.7. If the train is on time, the probability of reaching work before 9 am is 0.9. If the train is not on time, the probability of reaching work before 9 am is 0.3.

(a) Find the probability that the train is on time and the commuter reaches work before 9 am. [2] (b) Find the probability that the commuter reaches work before 9 am. [2] (c) Given that the commuter reaches work before 9 am, find the probability that the train was on time. [2]

11. (Higher tier only) Events A, B, C and D satisfy: p(A) = 0.3, p(B) = 0.6, p(A and B) = 0.18, p(C) = 0.5, p(D) = 0.4 and p(C and D) = 0.15.

(a) Show that A and B are independent. [3] (b) Show that C and D are not independent. [2]

12. (Higher tier only) A bag holds 5 black and 4 white counters. Two counters are taken at random without replacement.

(a) Draw a tree diagram and find the probability that the two counters are the same colour. [4] (b) Given that the two counters are the same colour, find the probability that both are black. [3] (c) Show that “the first counter is black” and “the second counter is black” are not independent. [3]

Answers

1. (a) p(blue or white) = 0.3 + 0.45 = 0.75 [1] (b) p(black) = 1 - 0.75 = 0.25 [1] [2] Examiner insight: Blue, white and black cover every outcome and cannot overlap, so state why you may add or subtract from 1 rather than writing a bare number.

2. (a) SS: 0.8 × 0.6 = 0.48 [1]; SM: 0.8 × 0.4 = 0.32, MS: 0.2 × 0.6 = 0.12, MM: 0.2 × 0.4 = 0.08 [1] (check: they sum to 1) (b) Exactly one score is SM or MS: 0.32 + 0.12 [1] = 0.44 [1] [4] Examiner insight: Write both routes and show the addition so your method is clear.

3. (a) p(both dark) = 5/12 × 4/11 [1] = 20/132 = 5/33 [1] (b) Dark then milk: 5/12 × 7/11 = 35/132 [1]. Milk then dark: 7/12 × 5/11 = 35/132 [1]. Total 70/132 = 35/66 [1] [5] Examiner insight: “Different types” has two routes. Check you counted both orders.

4. p(drink or snack) = 0.55 + 0.4 - 0.2 [1] = 0.75 [1]. p(neither) = 1 - 0.75 = 0.25 [1] [3] Examiner insight: Adding 0.55 and 0.4 without taking away the 0.2 counts the customers who buy both twice, giving 0.95.

5. (a) p(none faulty) = 0.95 × 0.95 × 0.95 [1] = 0.857375 = 0.857 [1] (b) p(at least one faulty) = 1 - 0.857375 = 0.142625 = 0.143 [1] (c) One faulty can happen in three orders, each 0.05 × 0.95 × 0.95 [1]. Probability = 3 × 0.045125 = 0.135375 = 0.135 [1] [5] Examiner insight: Use the unrounded value from part (a) in part (b); working from 0.857 can shift the third significant figure.

6. (a) First stage: rain 0.3, dry 0.7 [1]. Second stage after rain: rain 0.6, dry 0.4. After dry: rain 0.2, dry 0.8 [1] (b) Rain then dry: 0.3 × 0.4 = 0.12 [1]. Dry then rain: 0.7 × 0.2 = 0.14 [1]. Total 0.26 [1] (c) Rain Tuesday via rain Monday: 0.3 × 0.6 = 0.18 [1]. Via dry Monday: 0.14. Total 0.32 [1] [7] Examiner insight: The two Tuesday branch pairs are different, so putting 0.6 and 0.4 on both Monday branches is a typical way to lose the later marks.

7. p(neither ticket wins) = 21/25 × 20/24 [2] = 420/600 = 0.7 [1]. p(at least one wins) = 1 - 0.7 = 0.3 [1] [4] Examiner insight: The denominator drops to 24 at the second ticket and the 21 non-winners drop to 20; keeping the numbers from the first stage is the usual error.

8. (a) Fail the first (1 - 0.45 = 0.55) then pass the second: 0.55 × 0.6 [1] = 0.33 [1] (b) Pass on third attempt: 0.55 × 0.4 × 0.75 = 0.165 [1]. Total = 0.45 + 0.33 + 0.165 [1] = 0.945 [1] (c) Fail all three: 0.55 × 0.4 × 0.25 = 0.055 [1] [6] Examiner insight: Part (c) gives a check, since 1 - 0.055 = 0.945; use it to confirm part (b).

9. (Higher tier only) (a) Odd scores are 1, 3, 5, 7, 9, so there are 5 outcomes [1]. Primes among them are 3, 5, 7, so p = 3/5 [1] (b) Prime scores are 2, 3, 5, 7, so the reduced group has 4 outcomes [1]. Odd ones are 3, 5, 7, so p = 3/4 [1] [4] Examiner insight: Name the reduced group of outcomes before counting. Using the total of 10 as the denominator shows the condition was ignored.

10. (Higher tier only) (a) p(on time and before 9) = 0.9 × 0.7 [1] = 0.63 [1] (b) Not on time and before 9: 0.3 × 0.3 = 0.09 [1]. p(before 9) = 0.63 + 0.09 = 0.72 [1] (c) p(on time given before 9) = 0.63 ÷ 0.72 [1] = 7/8 (0.875) [1] [6] Examiner insight: In part (c) the denominator is p(before 9), the event after “given”, not p(on time).

11. (Higher tier only) (a) p(A) × p(B) = 0.3 × 0.6 = 0.18 [1]. This equals p(A and B) = 0.18 [1]. So A and B are independent. [1] (b) p(C) × p(D) = 0.5 × 0.4 = 0.2 [1], but p(C and D) = 0.15, so they are not equal and C and D are not independent. [1] [5] Examiner insight: A “show that” needs the comparison written out and a closing statement; the product alone is not a conclusion.

12. (Higher tier only) (a) First stage: black 5/9, white 4/9. Second stage after black: black 4/8, white 4/8. After white: black 5/8, white 3/8 [1]. BB: 5/9 × 4/8 = 20/72 [1]. WW: 4/9 × 3/8 = 12/72 [1]. Same colour: 32/72 = 4/9 [1] (b) p(both black given same colour) = (20/72) ÷ (32/72) [2] = 5/8 [1] (c) p(second black given first black) = 4/8 = 1/2 [1]. p(second black) = 5/9 × 4/8 + 4/9 × 5/8 = 20/72 + 20/72 = 40/72 = 5/9 [1]. Since 1/2 ≠ 5/9, the events are not independent. [1] [10] Examiner insight: Compare p(second black given first black) with p(second black) from the whole tree, not with 5/9 alone, and finish with a statement that uses the test.

Where marks are usually lost

• Putting the same second-stage fractions on every branch of a without-replacement tree.
• Finding only one of the two orders for “one of each” or “different”.
• Adding p(A) and p(B) when the events overlap and forgetting p(A and B).
• Rounding the answer to part (a) and then using the rounded value in later parts.
• Using the total, rather than the group after “given”, as the denominator in conditional probability.
• Confusing p(A given B) with p(B given A).
• Writing “independent” without comparing p(A) × p(B) with p(A and B), or p(A given B) with p(A).
• Not checking that branches from one point sum to 1.

Next steps

• Revisit the study guide for any question you missed.
• Use the revision notes to rehearse the laws.
• Tick the unit off the checklist and visit the course hub.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR GCSE (9-1) Mathematics (J560), Version 2.1 (August 2026), for first assessment in 2017, published by OCR. Content references 11.02d, 11.02e and 11.02f.