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.Two ordinary fair dice are rolled and the two scores are multiplied together. Work out the probability that the product is odd.
- 2.150 people at a gym were asked whether they prefer weight training or cardio; each person chose exactly one. 84 of the people are women. 50 of the women prefer cardio. 40 of the men prefer weight training. Work out the probability that a person, chosen at random from the 150, prefers weight training. Give your answer as a fraction in its simplest form.
- 3.A bag contains beads that are exactly one of red, white or black. The probability of taking a red bead is 3/10 and the probability of taking a white bead is 1/4. Work out the probability of taking a bead that is red or white.
- 4.A survey of 90 people records which of two apps, X and Y, they use. 55 people use app X, 42 people use app Y, and 20 people use neither app. Work out the probability that a randomly chosen person uses both apps, giving your answer as a fraction in its simplest form.
- 5.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.
- 6.A machine makes 4000 light bulbs a day and runs 5 days a week. Two inspectors test bulbs from this machine. Inspector A tests 40 bulbs and finds 4 faulty. Inspector B tests 500 bulbs and finds 30 faulty. Using the better of the two estimates, work out how many faulty bulbs the machine is expected to make in one week.
- 7.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?
- 8.On the probability scale from 0 (impossible) to 1 (certain), evens is exactly halfway. An event is 3/10 more likely than evens. Work out the probability of that event, giving your answer as a decimal.
- 9.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.
- 10.A council surveyed 140 households about recycling. 85 of the households are in Zone A. 52 of the Zone A households recycle glass. 21 of the Zone B households do not recycle glass. Work out the probability that a household, chosen at random from the 140, recycles glass. Give your answer as a fraction in its simplest form.
- 11.A two-way table records how 180 students at a school travel: by bus or on foot, split by year group. There are 84 students in Year 11, of whom 38 travel by bus and the rest walk. The rest of the 180 students are in Year 10, and 42 of the Year 10 students travel by bus. Work out the probability that a randomly chosen Year 10 student walks to school. Give your answer as a fraction in its simplest form.
- 12.Priya spins a fair spinner with P(red) = 0.3, and separately flips a fair coin with P(heads) = 0.5. Using a tree diagram, work out the probability that she gets red AND heads.
- 13.In a survey of 200 households, 120 have a garden and 80 own a dog. 54 of the households have a garden and own a dog. Work out the probability that a household owns a dog given that it has a garden, and compare it with the probability that a household picked from the whole survey owns a dog.
- 14.At a school car park, each car is a saloon, an estate or a hatchback, and no car is more than one of these. Among 200 cars, the probability that a car chosen at random is a saloon is 0.28 and the probability that it is an estate is 0.37. Work out how many of the 200 cars are hatchbacks.
- 15.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.
Answer key
- (d) 1/4 — Method: a product is odd only when BOTH factors are odd, so list the ordered pairs where both scores are odd and divide by 36. Working: the odd scores on a dice are 1, 3 and 5, so there are 3 × 3 = 9 ordered pairs where both scores are odd, out of the 36 equally likely pairs, cancelling down to 1/4. Answer: 1/4. Watch out: writing down 3/4 finds the probability that AT LEAST ONE score is odd, 1 minus the probability both are even, which is a different, easier condition to meet than both being odd. Considering only the first dice's score and ignoring the second gives 1/2, since 3 of the first dice's 6 scores are odd — but the product also depends on what the second dice shows. And writing down 1/12 comes from counting only the pairs where the SAME odd number appears twice, (1, 1), (3, 3) and (5, 5), missing pairs like (1, 3) and (5, 1) where the two odd scores differ.
- (b) 37/75 — There are 150 − 84 = 66 men. 84 − 50 = 34 women prefer weight training, and 40 men prefer weight training, so 34 + 40 = 74 people in total prefer weight training, out of 150: 74/150 = 37/75. Writing 4/15 is wrong because it only counts the men who prefer weight training (40/150, simplified), leaving out the 34 women. Writing 17/75 is wrong because it only counts the women who prefer weight training (34/150, simplified), leaving out the 40 men. Writing 37/42 is wrong because it uses the number of women (84) as the denominator instead of the whole gym (150) — 74/84 simplifies to 37/42, but that is not a probability out of the whole group. The probability is 37/75.
- (c) 11/20 — 'Red or white' combines two mutually exclusive events, so add their probabilities: 3/10 = 6/20 and 1/4 = 5/20, giving 6/20 + 5/20 = 11/20. Multiplying the two probabilities instead of adding them, 3/10 × 1/4, gives 3/40, which would be the probability of red and white together, not red or white — and a bead can't be both colours. Subtracting the sum from 1, 1 − 11/20 = 9/20, gives the probability of the bead being black instead of red or white. Converting 1/4 as 4/20 instead of 5/20 (dividing 20 by 4 but forgetting to scale the numerator) gives 6/20 + 4/20 = 1/2.
- (b) 3/10 — The number who use at least one app is 90 − 20 = 70. Since 55 + 42 double-counts the overlap, n(X ∩ Y) = 55 + 42 − 70 = 27, so P(both) = 27/90 = 3/10. Forgetting to subtract the 20 who use neither, and using the full 90 as the union, gives 55 + 42 − 90 = 7, so 7/90. Reporting the probability of using X or Y (or both), 70/90 = 7/9, answers a different question about the union, not the overlap. Reporting the probability of using neither app, 20/90 = 2/9, is the complement of the union, not the intersection.
- (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.
- (a) 1200 — Method: take the estimate from the larger sample, because an unbiased relative frequency tends towards the true probability as the sample grows, then multiply by the number of bulbs made in a week. Working: Inspector B tested 500 bulbs, far more than Inspector A's 40, so use B's relative frequency: 30 ÷ 500 = 0.06. A week's production is 4000 × 5 = 20000 bulbs. The expected number of faulty bulbs is 20000 × 0.06 = 1200. Answer: about 1200 faulty bulbs a week. The distractors: 2000 uses Inspector A's estimate, 4 ÷ 40 = 0.1, giving 20000 × 0.1 = 2000, and so rests on a sample of only 40 bulbs; 1600 comes from averaging the two estimates of 0.1 and 0.06 to get 0.08, and 20000 × 0.08 = 1600, which gives the small sample equal weight with the large one; 240 uses the right estimate but stops at a single day, 4000 × 0.06 = 240.
- (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.
- (d) 0.80 — Evens is exactly halfway along the scale, at 0.5, and 3/10 written as a decimal is 0.3. A probability 3/10 higher than evens is 0.5 + 0.3 = 0.80. Giving 0.30 as the answer converts the increase but never adds it to the value of evens. Adding the increase to 1, the 'certain' end of the scale, instead of to evens gives 1 + 0.3 = 1.30, which cannot be a probability. Writing 3/10 as 0.03 instead of 0.3, a place-value slip, gives 0.5 + 0.03 = 0.53.
- (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%.
- (a) 43/70 — There are 140 − 85 = 55 Zone B households, and 55 − 21 = 34 of them recycle glass. In total, 52 + 34 = 86 households recycle glass, out of 140: 86/140 = 43/70. Writing 13/35 is wrong because 52/140 simplifies to 13/35, and 52 only counts Zone A, leaving out the 34 Zone B recyclers. Writing 17/70 is wrong because 34/140 simplifies to 17/70, and 34 only counts Zone B, leaving out the 52 Zone A recyclers. Writing 73/140 is wrong because it adds the 52 Zone A recyclers to the 21 Zone B households that do NOT recycle, mixing up two different groups instead of adding the two recycling groups. The probability is 43/70.
- (c) 9/16 — There are 180 students in total and 84 are in Year 11, so Year 10 has 180 − 84 = 96 students. Of those 96, 42 travel by bus, so 96 − 42 = 54 walk. P(Year 10 student walks) = 54/96 = 9/16. Using the whole school of 180 as the denominator instead of just the 96 Year 10 students gives 54/180 = 3/10. Using the bus count, 42, as if it were the number who walk gives 42/96 = 7/16, the wrong branch of the Year 10 row. Working out the probability for Year 11 instead of Year 10 — 46 walkers out of 84 — gives 46/84 = 23/42.
- (c) 0.15 — Method: for two independent events, multiply along the branches of the tree to find the probability of both outcomes happening together. Working: P(red and heads) = P(red) × P(heads) = 0.3 × 0.5 = 0.15. Answer: 0.15. Watch out: adding the two probabilities, 0.3 + 0.5 = 0.8, does not give the probability of both — probabilities along one path of a tree are multiplied, not added. Writing down 0.5 ignores the spinner altogether and gives only the coin's probability. And writing down 0.65 is the probability of red OR heads, which is 0.3 + 0.5 − 0.15 = 0.65, a different question from the one asked here.
- (a) 0.45, different from 0.4 for all the households — Method: work out the probability inside the restricted group of garden owners, then work out the probability across the whole survey, and compare the two. Working: 54 of the 120 households with a garden own a dog, so the conditional probability is 54 divided by 120, which is 0.45. Across the whole survey 80 of the 200 households own a dog, which is 0.4. Since 0.45 is not 0.4, having a garden changes the chance of owning a dog and the two events are not independent. Answer: 0.45, different from 0.4 for all the households. The distractors: 0.27 is 54/200, dividing the households with both by the whole survey instead of by the 120 with a garden; 0.675 is 54/80, the probability that a household has a garden given that it owns a dog, which is the condition and the event the wrong way round; 0.4 is 80/200, the probability of owning a dog with the garden information never used, which is why that route also reports no difference.
- (b) 70 — Saloon, estate and hatchback are exhaustive, so their probabilities sum to 1: the probability of a hatchback is 1 − 0.28 − 0.37 = 0.35. The number of hatchbacks is 0.35 × 200 = 70. Treating the SUM of the other two probabilities, 0.28 + 0.37 = 0.65, as the probability of a hatchback instead of its complement gives 0.65 × 200 = 130. Multiplying the correct probability, 0.35, by 100 instead of the 200 cars actually surveyed gives 35. Averaging the two given probabilities, (0.28 + 0.37) ÷ 2 = 0.325, instead of subtracting them from 1, and then multiplying by 200 gives 65.
- (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.
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