OCR GCSE Mathematics: Sample spaces, systematic listing, product rule, Venn diagrams and sets (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 follow sections 11.02a to 11.02c of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026): sample spaces, systematic listing, the product rule, and Venn diagrams and sets. Questions marked “(Higher tier only)” use Higher tier content; the rest are for both tiers. Learn the methods from the study guide and the revision notes. The OCR GCSE Mathematics hub and checklist help with planning.

Questions

1. A fair spinner has three equal sections: red, blue and green. It is spun and a fair coin is flipped. List all the possible outcomes and find the probability of blue and tails. [3]

2. The letters M, N and O are written in a row, each used once. List all the possible orders. One order is chosen at random. Find the probability that N is not first. [3]

3. Two fair four-sided dice, each numbered 1 to 4, are rolled. The score is the product of the two numbers.

(a) Draw a table showing all the scores. [2] (b) Find the probability that the score is at least 6. [2] (c) Find the probability that the score is 4. [1]

4. A pizza has two different toppings chosen from ham, olives, peppers and onion. List all the possible pairs of toppings. Find the probability that a pair chosen at random includes olives. [4]

5. (Higher tier only) A locker code has one letter from A to E, then two digits from 0 to 9, then one symbol from three available symbols. Letters, digits and symbols may be repeated.

(a) How many different codes are there? [2] (b) A code is guessed at random. Find the probability that it is correct. [1] (c) How many codes have two different digits? [2]

6. (Higher tier only) Eight runners take part in a race. Medals are given for first, second and third places.

(a) In how many different ways can the three medals be given out? [2] (b) Three particular runners are friends. Find the probability that they win the three medals between them, in any order. [2]

7. (a) A = {x | x is an odd number, 3 < x < 12}. List the members of A. [1] (b) C = {x | x is a factor of 24}. List the members of C. [2] (c) B = {2, 4, 6, 8, 10}. Write B using set-builder notation. [1]

8. 50 pupils are asked about pets. 22 have a dog, 19 have a cat and 8 have both.

(a) Draw a two-circle Venn diagram and write in the number for each region, including the number outside both circles. [4] (b) A pupil is chosen at random. Find the probability that they have a dog only. [1] (c) Find the probability that they have neither pet. [1]

9. The whole numbers 1 to 30 are sorted using A = {x | x is a multiple of 4} and B = {x | x is a factor of 24}.

(a) Draw a Venn diagram showing which numbers are in A only, in both sets, in B only, and how many are in neither. [3] (b) A number is chosen at random from 1 to 30. Find the probability that it is in B but not in A. [2]

10. (Higher tier only) A club has 84 members. 50 are adults and 34 are juniors. 46 members prefer swimming and the rest prefer gym. 21 of the juniors prefer swimming.

(a) Complete a two-way table showing adults and juniors against swimming and gym. [3] (b) A member who prefers swimming is chosen at random. Find the probability that they are a junior. [1] (c) An adult member is chosen at random. Find the probability that they prefer swimming. [1]

11. (Higher tier only) At a leisure centre, 80 people were asked about the facilities they used. 52 used the pool, 41 used the gym and 9 used neither.

(a) Work out how many used both the pool and the gym. [2] (b) Show all the information on a Venn diagram. [1] (c) One person who used the pool is chosen. Find the probability that they also used the gym. [2] (d) One person who did not use the gym is chosen. Find the probability that they used the pool. [2]

12. A fair five-sided spinner is numbered 1 to 5. A fair four-sided spinner is numbered 1 to 4. Both are spun and the numbers are added.

(a) Draw a table of all the possible totals. [2] (b) Find the probability that the total is 6. [1] (c) Find the probability that the total is a prime number. [2] (d) (Higher tier only) The five-sided spinner is now spun three times. How many sequences of three numbers are possible, and what is the probability that all three numbers are different? [3]

Answers

1. Outcomes: red-heads, red-tails, blue-heads, blue-tails, green-heads, green-tails [1]. There are 6 outcomes in total [1]. Only blue-tails is favourable, so the probability is 1/6 [1]. [3] Examiner insight: Listing in a fixed order, colour first then coin, makes it easy to see you have all six.

2. MNO, MON, NMO, NOM, OMN, ONM [1]. N is first in 2 of the 6, so N is not first in 4 of the 6 [1]. The probability is 4/6 = 2/3 [1]. [3] Examiner insight: A written systematic list shows your method even if the final fraction slips.

3. (a) The table is below, with one die across the top and the other down the side [1] with all 16 products correct [1]. [2]

	1	2	3	4
1	1	2	3	4
2	2	4	6	8
3	3	6	9	12
4	4	8	12	16

(b) The scores of at least 6 are 6, 8, 6, 9, 12, 8, 12 and 16, which is 8 of the 16 cells [1]. The probability is 8/16 = 1/2 [1]. [2] (c) The score 4 appears 3 times, so the probability is 3/16 [1]. [1] Examiner insight: A score such as 4 occurs in more than one cell, so count cells, not score values.

4. The pairs are ham and olives, ham and peppers, ham and onion, olives and peppers, olives and onion, peppers and onion [1] with no pair missing or repeated [1]. Three of the six pairs include olives [1]. The probability is 3/6 = 1/2 [1]. [4] Examiner insight: Olives and ham is the same pair as ham and olives, so listing twelve pairs is a common error.

5. (a) 5 × 10 × 10 × 3 [1] = 1500 [1]. [2] (b) The probability is 1/1500 [1]. [1] (c) The first digit can be any of 10 and the second any of the 9 remaining: 5 × 10 × 9 × 3 [1] = 1350 [1]. [2] Examiner insight: Showing the number of choices at each stage gives clear working; the 9 in part (c) shows you allowed for different digits.

6. (a) 8 × 7 × 6 [1] = 336 [1]. [2] (b) The three friends can be arranged in 3 × 2 × 1 = 6 ways [1]. The probability is 6/336 = 1/56 [1]. [2] Examiner insight: Part (b) needs the favourable count as well as the total; write both before simplifying.

7. (a) A = {5, 7, 9, 11} [1]. [1] (b) The factors of 24 are 1 and 24, 2 and 12, 3 and 8, 4 and 6 [1]. C = {1, 2, 3, 4, 6, 8, 12, 24} [1]. [2] (c) For example B = {x | x is an even number, 2 ≤ x ≤ 10} [1]. [1] Examiner insight: A rule that gives exactly the five members is the aim; state what type of number x is and make the end values clear.

8. (a) Both = 8 [1]. Dog only = 22 − 8 = 14 [1]. Cat only = 19 − 8 = 11 [1]. Neither = 50 − (14 + 8 + 11) = 17 [1]. [4]

Dog only: 14 | Both: 8 | Cat only: 11 | Neither: 17   (total 50)

(b) The probability is 14/50 = 7/25 [1]. [1] (c) The probability is 17/50 [1]. [1] Examiner insight: Subtract the overlap from each set total, and check the four regions add up to 50.

9. (a) The multiples of 4 up to 30 are 4, 8, 12, 16, 20, 24, 28. The factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. Both: 4, 8, 12, 24 [1]. A only: 16, 20, 28. B only: 1, 2, 3, 6 [1]. Neither: 30 − (3 + 4 + 4) = 19 numbers [1]. [3] (b) B but not A means B only, which has 4 numbers [1]. The probability is 4/30 = 2/15 [1]. [2] Examiner insight: “B but not A” means B only, not all of B.

10. (a) Juniors who prefer swimming = 21, so adults who prefer swimming = 46 − 21 = 25 [1]. Adults who prefer gym = 50 − 25 = 25 [1]. Juniors who prefer gym = 34 − 21 = 13 [1]. [3]

	Swimming	Gym	Total
Adults	25	25	50
Juniors	21	13	34
Total	46	38	84

(b) The group is the 46 who prefer swimming, and 21 of them are juniors, so the probability is 21/46 [1]. [1] (c) The group is the 50 adults, and 25 prefer swimming, so the probability is 25/50 = 1/2 [1]. [1] Examiner insight: For “given that”, the denominator is the row or column named in the condition, not 84.

11. (a) At least one facility: 80 − 9 = 71 [1]. Both = 52 + 41 − 71 = 22 [1]. [2] (b) Pool only = 30, both = 22, gym only = 19, neither = 9; total 80 [1]. [1] (c) The group is the 52 pool users, of whom 22 used the gym, so the denominator is 52 [1] and the probability is 22/52 = 11/26 [1]. [2] (d) The group is those who did not use the gym: 80 − 41 = 39 [1]. Of these, 30 used the pool, so the probability is 30/39 = 10/13 [1]. [2] Examiner insight: When no diagram is given, drawing your own labelled diagram organises the data and shows your reading of it.

12. (a) The table has 5 × 4 = 20 cells [1] with all totals correct [1]. [2]

	1	2	3	4
1	2	3	4	5
2	3	4	5	6
3	4	5	6	7
4	5	6	7	8
5	6	7	8	9

(b) The total 6 appears 4 times: 4/20 = 1/5 [1]. [1] (c) The prime totals are 2, 3, 5 and 7. They occur 1, 2, 4 and 3 times, which is 10 cells [1]. The probability is 10/20 = 1/2 [1]. [2] (d) Higher tier only. There are 5 × 5 × 5 = 125 sequences [1]. Sequences with all different numbers: 5 × 4 × 3 = 60 [1]. The probability is 60/125 = 12/25 [1]. [3] Examiner insight: Check the grid has 20 cells before counting favourable ones.

Where marks are usually lost

• Treating pairs in different orders as one outcome, or as two when order does not matter.
• Counting score values instead of cells, or using an incomplete table.
• Not simplifying the final fraction.
• Subtracting the overlap from the group total instead of each set total.
• Forgetting the number outside both circles.
• Using the whole group as the denominator for a “given that” probability.
• Keeping the same number of choices at every stage when choices cannot repeat.
• Writing set-builder notation without saying what type of number x is.

Next steps

• Re-read the study guide and revision notes.
• Visit the hub and checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR GCSE (9-1) Mathematics J560, Version 2.1 (August 2026), first assessment 2017, published by OCR.