Printable · GCSE Higher · ages 14-16
Probability worksheet — GCSE Higher
Fifteen questions across the probability statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Probability worksheet — GCSE Higher
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- 1.A fair six-sided dice is rolled 150 times. The table shows how many times each number came up: 1 came up 22 times, 2 came up 27 times, 3 came up 24 times, 4 came up 34 times, 5 came up 21 times and 6 came up 22 times. The theoretical probability of each number is 1/6. Which number is most over-represented compared with its theoretical probability?
- 2.A factory finds that the probability a randomly chosen light bulb is defective is 0.035. In a batch of 4,000 bulbs, work out how many bulbs you would expect to work correctly.
- 3.A raffle sells 400 tickets at 50p each. There is one prize of £45. Aisha buys 8 tickets. Work out how much money Aisha should expect to lose from playing, giving your answer in pounds.
- 4.A box contains 9 red balls and 11 green balls. Two balls are taken out at random, one after the other, without being replaced. Given that both balls taken out are the same colour, work out the probability that both balls are red.
- 5.A fair coin is flipped again and again. After the first 10 flips there have been 7 heads. After 1000 flips there have been 528 heads. Which statement best describes what these results show?
- 6.At a fun run, each runner finishes, retires or is disqualified, and cannot do more than one of these. The probability that a runner finishes is 68.5% and the probability that a runner retires is 24.75%. Work out the probability, as a percentage, that a runner is disqualified.
- 7.Two goalkeepers face penalty kicks. Elin saves 34% of the penalties she faces. Noah saves 11/32 of the penalties he faces. Which statement correctly compares them?
- 8.A Venn diagram shows two sets, P and Q, inside a universal set. n(P) = 34, n(Q) = 27, n(P ∩ Q) = 11, and n(ξ) = 90, where ξ is the universal set. Work out n((P ∪ Q)′), the number of elements in neither P nor Q.
- 9.The probability that Leo wakes up before his alarm is 0.35. The probability that his younger sister wakes up before her alarm, independently of Leo, is 0.28. Work out the probability that neither of them wakes up before their alarm.
- 10.A clinic recorded 300 booked appointments using a frequency tree. The first branch splits them into 210 adult appointments and the rest child appointments. Of the adult appointments, 189 were attended and the rest were missed. Of the child appointments, 81 were attended. Work out the probability that a booked appointment, chosen at random from the 300, was missed. Give your answer as a fraction in its simplest form.
- 11.A basketball player takes two free throws, and the throws are independent. The probability of scoring on each throw is 0.6, and the probability of missing is 0.4. Work out the probability that she scores exactly one of the two throws.
- 12.A charity fundraiser runs a game using a spinner with 5 equal sections numbered 1 to 5. A player wins a £4 prize if the spinner lands on 5, and wins nothing otherwise. It costs £1 to play, and the game is played 100 times during the fundraiser. Decide which statement correctly describes the game.
- 13.At a school fête, a tombola stall costs £1.50 to play. The probability of winning is 0.2, and the prize is worth £6. Work out the stall's expected profit, on average, from each game played.
- 14.A call centre finds that 40% of its calls are about billing. 35% of the calls about billing are dealt with in under five minutes. The centre takes 500 calls on Monday. Work out how many of Monday's calls you would expect to be about billing and dealt with in under five minutes.
- 15.A phone network sends automatic text alerts to customers. On average, 1,500 alerts are sent each day, and the probability that a customer replies 'STOP' to an alert is 0.18. Work out how many replies of 'STOP' the network should expect over a 30-day month.
Answer key
- (b) 4 — With 150 rolls and probability 1/6 for each number, the expected count is 150 ÷ 6 = 25. Comparing each actual count with 25: 1 is 22 (3 below), 2 is 27 (2 above), 3 is 24 (1 below), 4 is 34 (9 above), 5 is 21 (4 below) and 6 is 22 (3 below). Number 4 is furthest above its expected count, so it is the most over-represented. Number 2 is also above its expected count, but by only 2, far less than 4's 9. Number 3's count of 24 is below the expected 25, so it is under-represented, not over. Number 6's count of 22 is also below the expected 25, so it too is under-represented.
- (a) 3,860 — The probability a bulb works correctly is the complement of being defective: 1 − 0.035 = 0.965. Expected number working correctly = 0.965 × 4,000 = 3,860. Using the probability of being defective instead of its complement gives 4,000 × 0.035 = 140, the expected number of DEFECTIVE bulbs, not working ones. Shifting the decimal point in the complement, using 0.0965 instead of 0.965, gives 4,000 × 0.0965 = 386. Assuming every bulb works, ignoring the 0.035 probability altogether, gives the full batch of 4,000.
- (c) £3.10 — Aisha's tickets cost 8 × 50p = £4.00. Her expected winnings are (8/400) × £45 = £0.90, since she holds 8 of the 400 tickets. Her expected loss is the cost minus the expected winnings: £4.00 − £0.90 = £3.10. Writing £4.00 is wrong because it is only the cost of her tickets, with no account taken of the expected winnings she might get back. Writing £0.90 is wrong because that is her expected WINNINGS, not her loss — the cost has not been subtracted. Writing £3.89 is wrong because it uses 1 ticket instead of her actual 8 tickets when working out the expected winnings: (1/400) × £45 = £0.1125, giving £4.00 − £0.11 = £3.89. Aisha should expect to lose £3.10.
- (d) 36/91 — Method: P(both red | same colour) = P(both red) ÷ P(same colour), where P(same colour) = P(both red) + P(both green). Working: P(both red) = 9/20 × 8/19 = 72/380 = 18/95. P(both green) = 11/20 × 10/19 = 110/380 = 11/38. P(same colour) = 18/95 + 11/38 = 36/190 + 55/190 = 91/190. P(both red | same colour) = (36/190) ÷ (91/190) = 36/91. Answer: 36/91. Watch out: stopping at 18/95 gives P(both red) itself, without dividing by the probability that the colours matched at all. Working out 55/91 finds the same-colour probability for green instead of red — check which colour's count you are putting on top. And 9/20 is just the chance the first ball drawn is red, which ignores the second draw and the without-replacement condition completely.
- (b) The relative frequency is settling near 0.5 — Method: turn each result into a relative frequency before comparing them, because it is the relative frequency, and not the difference between the two counts, that tends towards the theoretical probability. Working: after 10 flips the relative frequency of a head is 7 ÷ 10 = 0.7, which is a long way from 0.5. After 1000 flips it is 528 ÷ 1000 = 0.528, which is much closer to 0.5. Meanwhile the gap between the two counts has grown rather than shrunk: it was 7 − 3 = 4 after 10 flips and is 528 − 472 = 56 after 1000 flips. Answer: the relative frequency is settling near 0.5, which is what an unbiased experiment does as the sample grows. The distractors: saying the counts are levelling out is the usual form of this idea and the figures contradict it, since the gap went from 4 to 56; saying the coin is biased treats 28 extra heads in 1000 flips as proof, when 0.528 sits close to 0.5 and a fair coin gives results like this often; saying the next flip is more likely to be a tail is the gambler's fallacy, since each flip stays at 1/2 whatever came before.
- (a) 6.75% — Finishing, retiring and being disqualified are exhaustive, so the three percentages sum to 100%: 100% − 68.5% − 24.75% = 6.75%. Adding the two given percentages instead of subtracting them from 100% gives 68.5% + 24.75% = 93.25%, the combined probability of finishing or retiring, not of being disqualified. Subtracting only the retiring percentage from 100% and forgetting the finishing percentage gives 100% − 24.75% = 75.25%. Subtracting only the finishing percentage and forgetting the retiring percentage gives 100% − 68.5% = 31.50%.
- (c) Noah — 11/32 = 0.34375, above 34%. — Converting 11/32 to a decimal gives 11 ÷ 32 = 0.34375, which is greater than 34% (0.34), so Noah has the better save rate: 'Noah — 11/32 = 0.34375, above 34%.' Comparing the raw numbers 34 and 11 directly, without converting the fraction to the same form, gives 'Elin — 34 is bigger than 11.' Treating a larger denominator as meaning a bigger value, rather than smaller equal shares, gives 'Noah — 32 is a bigger denominator.' Rounding 34.375% to 34% to the nearest whole percent hides the difference and gives 'Equal — both round to 34% to the nearest percent.'
- (b) 40 — n(P ∪ Q) = n(P) + n(Q) − n(P ∩ Q) = 34 + 27 − 11 = 50. The complement is everyone outside both sets: n((P ∪ Q)′) = 90 − 50 = 40. Adding P and Q without subtracting the overlap gives 34 + 27 = 61, so 90 − 61 = 29 double-subtracts the 11 who are in both. Reporting n(P ∪ Q) itself, 50, forgets to take the complement at all. Subtracting only n(P) from the universal set, 90 − 34 = 56, ignores set Q altogether.
- (d) 0.468 — The probability that Leo does not wake up before his alarm is 1 − 0.35 = 0.65, and the probability that his sister does not is 1 − 0.28 = 0.72. Since the two events are independent, multiply the complements: 0.65 × 0.72 = 0.468. A candidate who answers 0.63 has added the two given probabilities, 0.35 + 0.28, instead of finding and multiplying the complements. A candidate who answers 0.098 has multiplied the two given probabilities directly, 0.35 × 0.28, without taking complements first. A candidate who answers 0.532 has correctly reached 0.468 but then subtracted it from 1 again by mistake.
- (c) 1/10 — On the adult branch, 210 − 189 = 21 appointments were missed. There are 300 − 210 = 90 child appointments, and 90 − 81 = 9 of those were missed. In total, 21 + 9 = 30 appointments were missed, out of 300: 30/300 = 1/10. Writing 7/100 is wrong because 21/300 simplifies to 7/100, and 21 only counts the adult branch, leaving out the 9 missed child appointments. Writing 3/100 is wrong because 9/300 simplifies to 3/100, and 9 only counts the child branch, leaving out the 21 missed adult appointments. Writing 1/9 is wrong because it divides the 30 missed appointments by the 270 that were attended (300 − 30) instead of by the whole 300 booked. The probability is 1/10.
- (a) 0.48 — There are two ways to score exactly one throw: scoring on the first and missing the second, 0.6 × 0.4 = 0.24, or missing the first and scoring the second, 0.4 × 0.6 = 0.24. Adding these gives 0.24 + 0.24 = 0.48. Choosing 0.24 comes from working out only one of the two paths and forgetting the other one also gives exactly one score. Choosing 0.36 comes from working out the probability of scoring BOTH throws, 0.6 × 0.6 = 0.36, instead of exactly one. Choosing 0.84 comes from working out the probability of scoring AT LEAST one throw, 1 − 0.4 × 0.4 = 0.84, instead of exactly one.
- (b) Organiser favoured — expected pay-out is under £1 — The expected pay-out per game is the prize times the probability of winning: £4 × 1/5 = £0.80. The expected income per game is the £1 entry fee, which the organiser collects regardless of the result. Since £0.80 is less than £1, the game favours the organiser, because the expected pay-out is under £1. The claim that the game favours the player, because the pay-out is over £1, is wrong on both counts — the pay-out is not over £1, and it is the organiser who benefits. The claim that the organiser is favoured because the pay-out is over £1 reaches the right side but the wrong reason: £0.80 is under £1, not over it. The claim that the two expected amounts are equal is also wrong: £0.80 and £1 are different amounts, so the game is not fair to both sides.
- (b) £0.30 profit for the stall — The stall keeps the £1.50 entry fee whatever happens, and expects to pay out prize × probability of winning = £6 × 0.2 = £1.20 on average. So its expected profit per game is £1.50 − £1.20 = £0.30. Reporting the expected pay-out of £1.20 itself as the profit forgets that the stall also keeps the entry fee. Assuming the player always wins gives an expected cost of £6 − £1.50 = £4.50, treated as a loss for the stall. Using the probability of NOT winning, 0.8, to find the expected pay-out gives £6 × 0.8 = £4.80, and £1.50 − £4.80 = −£3.30, a £3.30 loss.
- (a) 70 — Method: two linked steps. Find the expected number of billing calls first, then take the 35% of those, because the 35% is quoted for billing calls only. Working: 40% of 500 is 200 billing calls. 35% of 200 is 70 calls. Answer: you would expect 70 calls. The distractors: 200 stops after the first step and gives the billing calls, forgetting that only some of them are dealt with quickly; 175 is 35% of 500, applying the quick response rate to every call the centre takes rather than to the billing calls only; 375 comes from adding 40% and 35% to get 75% and taking 75% of 500, which treats two stages of one journey as separate outcomes to be added.
- (b) 8,100 — First find the total number of alerts sent in the month: 1,500 × 30 = 45,000. Then apply the probability of a 'STOP' reply: 45,000 × 0.18 = 8,100. Stopping after finding only one day's expected replies, 1,500 × 0.18 = 270, forgets to scale up to the whole month. Multiplying the number of days by the probability instead of by the daily total of alerts gives 30 × 0.18 = 5.4, which rounds to 5. Shifting the decimal point in the probability, using 0.018 instead of 0.18, gives 45,000 × 0.018 = 810.
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