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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Answer key: Probability worksheet — GCSE Higher
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- (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.
- (c) 9/16 — Method: two steps. Total the patients who had a flu jab, since the patient picked is known to be one of them, then divide the over 65s who had a jab by that total. Working: 90 patients over 65 and 70 patients aged 65 or under had a jab, so 160 patients had one. The over 65s give 90/160, and dividing the numerator and the denominator by 10 gives 9/16. Answer: the probability is 9/16. The distractors: 7/16 is 70/160, the probability that the patient picked is aged 65 or under, which is the other part of the same restricted group; 3/4 is 90/120, the probability that a patient had a jab given that they are over 65, which is the condition and the event the wrong way round and needs the 120 patients over 65; 9/40 is 90/400, dividing by every patient on the list instead of by the 160 who had a jab.
- (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) 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.
- (c) Red — Theoretical probability is 1/3 ≈ 0.333 for each colour. Red's relative frequency is 38/90 ≈ 0.422, above 1/3, so red is over-represented. Blue's relative frequency is 26/90 ≈ 0.289, below 1/3, so blue is under-represented, not over. Green's relative frequency is also 26/90 ≈ 0.289, below 1/3 for the same reason. Since red's relative frequency clearly exceeds 1/3, it is not true that none of the colours are over-represented.
- (b) 38% — Method: the two swimming percentages are quoted inside different age groups, so weight each one by the size of its group and add the two results. Working: the under 18s are 45% of the members and 60% of them swim, giving 0.45 × 60 = 27% of all the members. The members aged 18 or over are 55% of the members and 20% of them swim, giving 0.55 × 20 = 11% of all the members. Adding these gives 38%. Answer: 38% of the members swim each week. The distractors: 80% comes from adding 60% and 20% straight off, treating two rates quoted inside different groups as though they could be added; 40% is the mean of 60% and 20%, which would be right only if the two age groups were the same size, and they are not; 42% comes from pairing each swimming rate with the wrong age group, working out 0.45 × 20 added to 0.55 × 60.
- (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.
- (d) 3/16 — Method: work out each draw's own probability first, then multiply them together since the two draws are independent. Working: there are 8 tickets in all, 3 of them blue, so P(blue) = 3/8. P(spinner number greater than 2) = 2/4 = 1/2, since 3 and 4 qualify. Multiplying gives 3/8 × 1/2, which comes to 3/16. Answer: 3/16. Watch out: writing down 3/8 stops after the first draw and never brings in the spinner at all. Writing down 1/2 does the opposite, using only the spinner and ignoring the ticket draw. And writing down 5/16 uses 5/8, the probability of a RED ticket, instead of 3/8 for blue — reading the wrong colour off the raffle.
- (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) 500 — Method: when a dice is known to be fair, the theoretical probability is the best thing to work from, and the more trials there are the closer the results tend to it. Working: for a fair dice the probability of a six is 1/6, so the expected number of sixes in 3000 rolls is 3000 × 1 ÷ 6 = 500. The class experiment gave a relative frequency of 14/60, but 60 trials is far too few to overturn a known theoretical value, and the school's 3000 rolls will tend towards 1/6 in any case. Answer: about 500 sixes. The distractors: 700 comes from using the class relative frequency instead of the theory, 3000 × 14 ÷ 60 = 700; 600 comes from splitting the difference between the two, since 1/6 is about 0.167 and 14/60 is about 0.233, whose mean is 0.2, and 3000 × 0.2 = 600; 2500 uses 5/6 instead of 1/6 and counts the rolls expected not to be a six.
- (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.
- (a) Game B, by £8 — Game A's expected total is 40 × 0.2 × £5 = £40. Game B's expected total is 40 × 0.1 × £12 = £48. Game B is higher, by £48 − £40 = £8. Writing 'Game A, by £8' is wrong because it has the right difference but the wrong game — Game A's total (£40) is actually LOWER than Game B's, not higher. Writing 'Game B, by £48' is wrong because £48 is Game B's whole expected total, not the DIFFERENCE between the two games. Writing 'Game A, by £40' is wrong in the same way, using Game A's whole total as if it were the margin, and naming the wrong game as the winner. Game B gives the higher expected total, by £8.
- (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.
- (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.
- (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.
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