OCR GCSE Mathematics: Probability scale and relative frequency (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 unit 11.01 (Basic probability and experiments, outcomes 11.01a to 11.01d) of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026). Nothing in this unit is Higher tier only, so every question is for both tiers. Learn the ideas first in the study guide or the revision notes. See the course hub, the checklist and the free diagnostics.

Questions

1. (a) For each event, say whether it is impossible, has evens chance or is certain, and give its probability. (i) A fair coin lands heads. (ii) A month chosen at random has more than 32 days. (iii) A fair dice shows a score less than 7. [3] (b) Explain why 1.3 cannot be the probability of an event. [1]

2. A bag holds 12 counters: 5 green, 4 yellow and 3 white. One counter is taken at random. (a) Find the probability that it is yellow. [1] (b) Find the probability that it is not green. [2]

3. A plastic cap is flipped 80 times. It lands open side up 36 times, open side down 31 times and on its side 13 times. (a) Find the relative frequency of landing on its side. [1] (b) Estimate how many times the cap lands on its side in 500 flips. [2] (c) Explain why the answer to part (b) is only an estimate. [1]

4. A cinema sells 320 tickets: 200 adult and the rest child. Of the adult tickets, 140 include popcorn. Of the child tickets, 90 include popcorn. (a) Draw a frequency tree and fill in every number. [3] (b) A ticket is chosen at random. Find the relative frequency of a ticket that includes popcorn, as a fraction in its simplest form. [2] (c) Which group, adult or child, has the greater relative frequency of popcorn? Show your working. [1]

5. A spinner has 5 equal sections and 1 of them is blue. It is spun several times. The table shows the cumulative results.

Spins	10	50	200	1000	5000
Blue	4	8	46	193	1011

(a) Work out the relative frequency of blue for 10 spins and for 5000 spins. [2] (b) Write down the theoretical probability of blue. Which of the five relative frequencies is closest to it, and why would you expect this? [2] (c) Jo says, “After 10 spins the relative frequency was 0.4, so the spinner is biased.” Comment. [1]

6. A fair 8-sided dice is numbered 1 to 8. (a) Find the probability of scoring more than 5. [1] (b) Find the probability of scoring a factor of 12. [2] (c) Find the probability of not scoring a square number. [1]

7. Ayesha flips a coin 800 times and it lands heads 436 times. In a separate run of 40 flips it lands heads 26 times. (a) Work out the relative frequency of heads for the 800 flips. [1] (b) Which of the two runs gives the more reliable estimate of P(heads)? Give a reason. [1] (c) Does the evidence show that the coin is not fair? Explain. [2] (d) Ben says, “A fair coin must alternate heads and tails.” Explain why he is wrong. [1]

8. Two fair dice are thrown and their scores summed. (a) Find the probability that the total is 4. [2] (b) Find the probability that the total is less than 5. [2] (c) Find the probability that the total is a multiple of 4. [2]

9. (a) The probability that a part from a machine is defective is 0.03. How many defective parts are expected in a batch of 2500? [2] (b) A fair dice is rolled 300 times. How many times is a score greater than 4 expected? [2]

10. At a school fair, a spinner with three colours is spun 400 times. It lands on red 148 times, blue 172 times and green 80 times. (a) Work out the relative frequency of each colour. [3] (b) Show that your relative frequencies add up to 1. [1] (c) Use your relative frequency for green to estimate how many greens in 1500 spins. [2] (d) The organiser says all three colours are equally likely. Use the results to comment. [1]

11. A stall game uses two fair dice. A player wins if the total is 11 or 12. (a) Find the probability of winning one game. [2] (b) In 600 games, how many wins are expected? [1] (c) In the last 600 games there were 71 wins. Work out the relative frequency of winning. [1] (d) Mira says the dice must be biased. Use your answers to comment on her claim. [2] (e) Work out the relative frequency of not winning in the last 600 games, and compare it with the theoretical probability of not winning. [2]

Answers

1. (a)(i) Heads is as likely as tails: evens, 0.5 [1]. (ii) No month has more than 31 days: impossible, 0 [1]. (iii) Every score from 1 to 6 is less than 7: certain, 1 [1]. [3] (b) A probability cannot be greater than 1, and 1.3 is greater than 1 [1]. [1] Examiner insight: give both the word and the number when asked for both, and for the explanation state the rule that probabilities lie between 0 and 1 rather than just saying “it is too big”.

2. (a) 4 yellow out of 12 counters: 4/12 = 1/3 [1]. [1] (b) Counters that are not green: 4 + 3 = 7 [1]. Probability = 7/12 [1]. [2] Examiner insight: writing the working “12 - 5 = 7” or “1 - 5/12” shows how you reached 7/12 and makes your method clear if the final fraction is mis-simplified.

3. (a) 13/80 = 0.1625 [1]. [1] (b) Expected number = 0.1625 × 500 [1] = 81.25, so about 81 [1]. [2] (c) The relative frequency comes from only 80 flips and would change in another experiment, so the number in 500 flips is expected on average and not guaranteed [1]. [1] Examiner insight: the explanation for an estimate should mention variation between experiments or the limited number of trials, not just “it is not exact”.

4. (a) Child tickets: 320 - 200 = 120 [1]. Adult without popcorn: 200 - 140 = 60 [1]. Child without popcorn: 120 - 90 = 30 [1]. Tree: 320 tickets, adult 200 (popcorn 140, no popcorn 60), child 120 (popcorn 90, no popcorn 30). [3] (b) Tickets with popcorn: 140 + 90 = 230 [1]. 230/320 = 23/32 [1]. [2] (c) Adult: 140/200 = 0.7. Child: 90/120 = 0.75. Child has the greater relative frequency [1]. [1] Examiner insight: in part (c) compare relative frequencies, not the counts, because the groups are different sizes. Show both values so the comparison is visible.

5. (a) 4/10 = 0.4 [1]. 1011/5000 = 0.2022 [1]. [2] (b) Theoretical probability = 1/5 = 0.2 [1]. The closest is 0.2022 from 5000 spins, because relative frequencies get closer to the theoretical probability as the number of trials increases [1]. [2] (c) 10 spins is a very small sample, so a relative frequency of 0.4 is not good evidence of bias; more spins are needed [1]. [1] Examiner insight: a good answer links the conclusion to the number of trials. Saying “the spinner is fair” or “biased” without mentioning sample size is a weak response.

6. (a) Scores 6, 7, 8: 3/8 [1]. [1] (b) Factors of 12 up to 8: 1, 2, 3, 4, 6, so 5 outcomes [1]. Probability = 5/8 [1]. [2] (c) Squares: 1 and 4, so 2 outcomes. Not a square: 6 outcomes. 6/8 = 3/4 [1]. [1] Examiner insight: list the favourable outcomes before writing the fraction.

7. (a) 436/800 = 0.545 [1]. [1] (b) The 800 flips, because more trials give a more reliable estimate [1]. [1] (c) For a fair coin P(heads) = 0.5, and 0.545 is a little above this [1]. The difference is small, so the evidence suggests the coin may be slightly biased towards heads but does not prove it; more flips are needed to be sure [1]. [2] (d) Each flip is random, and earlier results do not affect the next one, so runs of heads or tails can happen [1]. [1] Examiner insight: in “does the evidence show” questions, give a judgement that is hedged to match the data and name the next step, such as more trials, rather than a firm yes or no.

8. (a) Pairs (1,3) (2,2) (3,1) give total 4: 3 outcomes [1]. Probability = 3/36 = 1/12 [1]. [2] (b) Totals 2, 3 and 4: 1 + 2 + 3 = 6 outcomes [1]. 6/36 = 1/6 [1]. [2] (c) Totals 4, 8, 12: 3 + 5 + 1 = 9 outcomes [1]. 9/36 = 1/4 [1]. [2] Examiner insight: the denominator is 36 because the 36 pairs are equally likely. A grid of totals shown in your working is the clearest way to show you have counted every pair.

9. (a) Expected = 0.03 × 2500 [1] = 75 [1]. [2] (b) P(5 or 6) = 2/6 = 1/3 [1]. 1/3 × 300 = 100 [1]. [2] Examiner insight: write the product p × n so the method is visible.

10. (a) Red 148/400 = 0.37 [1]. Blue 172/400 = 0.43 [1]. Green 80/400 = 0.2 [1]. [3] (b) 0.37 + 0.43 + 0.2 = 1 [1]. [1] (c) 0.2 × 1500 [1] = 300 [1]. [2] (d) If all three were equally likely, each would have probability 1/3, so about 133 greens in 400 spins. Only 80 greens is well below this, so the results do not support the organiser’s claim [1]. [1] Examiner insight: compare the observed figure with what equally likely outcomes would give, and use numbers from the table in the comment.

11. (a) Totals 11 and 12: 2 + 1 = 3 pairs [1]. Probability = 3/36 = 1/12 [1]. [2] (b) 1/12 × 600 = 50 [1]. [1] (c) 71/600 = 0.118 (to 3 decimal places) [1]. [1] (d) About 50 wins are expected but there were 71, which is noticeably more [1]. This suggests the game may favour wins, but one run of 600 games does not prove the dice are biased; Mira would need to collect more results [1]. [2] (e) Games not won: 600 - 71 = 529 and 529/600 = 0.882 [1]. Theoretical P(not win) = 1 - 1/12 = 11/12 = 0.917, so the observed value is lower [1]. [2] Examiner insight: in a comment, quote the expected and observed values and stop short of claiming certainty.

Where marks are usually lost

• Writing an answer as a ratio or as “1 in 12” when the question asks for a probability.
• Counting the 11 totals of two dice as the sample space instead of 36 pairs.
• Comparing counts instead of relative frequencies when the groups have different sizes.
• Calling a coin or spinner biased on the basis of a short run of results.
• Treating the expected number as an exact count, or forgetting to say that it is an estimate.
• Using the wrong total on a frequency tree, such as dividing by 320 when the question is about the 120 child tickets.
• Not showing working for p × n, so your method is visible if a small slip occurs.

Next steps

• Re-read the study guide for any outcome you found hard.
• Use the revision notes for a final recap.
• Check the course hub and the checklist.
• Take one of the free 10-minute diagnostics.
• For help with any question you could not do, 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. Unit 11.01: Basic probability and experiments.