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) 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) 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) 57/100 — Pooling both trials: total heads = 24 + 33 = 57, total flips = 40 + 60 = 100, so the combined relative frequency is 57/100, which is already in its simplest form since 57 and 100 share no common factor. Averaging the two separate relative frequencies instead, (24/40 + 33/60) ÷ 2 = (0.6 + 0.55) ÷ 2 = 0.575 = 23/40, treats the two trials as equally weighted even though Ben made more flips, which is not correct. Using only Leah's data gives 24/40 = 3/5. Using only Ben's data gives 33/60 = 11/20.
- (d) 3400 — Combining all three greenhouses gives 200 + 150 + 250 = 600 seeds planted in total, and 172 + 126 + 212 = 510 germinated, so the combined estimate of the germination probability is 510/600 = 0.85. Out of a new batch of 4000 seeds, the expected number to germinate is 4000 × 0.85 = 3400. Writing 3440 is wrong because it uses only Greenhouse 1's rate, 172/200 = 0.86, instead of the combined rate from all three: 4000 × 0.86 = 3440. Writing 3360 is wrong because it uses only Greenhouse 2's rate, 126/150 = 0.84: 4000 × 0.84 = 3360. Writing 510 is wrong because that is the total number that germinated in the ORIGINAL trial, not scaled up to the new batch of 4000 seeds at all. The best estimate is 3400 seeds.
- (b) 3/5 — The question asks about the girls only, so use the girls' total of 70 as the denominator: 42 out of 70 girls walk, giving 42/70 = 3/5. Choosing 21/65 comes from using the whole survey of 130 pupils as the denominator instead of just the 70 girls, 42/130 = 21/65. Choosing 33/70 comes from using the boys' walking count, 33, over the girls' total of 70, mixing up the two rows of the table. Choosing 2/5 comes from using the number of girls who CYCLE, 70 − 42 = 28, instead of the number who walk, giving 28/70 = 2/5.
- (b) 39 — Method: put the counts into a two-way table and fill each missing cell by subtracting along a row or down a column. Working: the number of female members is 150 − 80 = 70. The pool column holds 66 members and 35 of them are male, so the number of female pool users is 66 − 35 = 31. Subtracting along the female row, 70 − 31 = 39 female members use the gym. Answer: 39 female members use the gym. The distractors: 45 comes from subtracting along the male row instead, 80 − 35 = 45, which counts male gym users; 31 is the female pool cell, written down one step before the gym cell; 84 is 150 − 66 and counts every gym user, male and female together.
- (c) 100 — There are 3 even numbers on a fair dice (2, 4 and 6), so the probability of landing on an even number is 3/6 = 1/2, and 300 × 1/2 = 150. The probability of landing on a six is 1/6, so 300 × 1/6 = 50. The dice is expected to land on an even number 150 − 50 = 100 more times than on a six. Writing 50 is wrong because that is just the expected number of sixes on its own, without comparing it to the expected number of evens. Writing 150 is wrong because that is just the expected number of evens on its own, without subtracting the sixes. Writing 200 is wrong because it adds the two expected frequencies together (150 + 50 = 200) instead of finding the difference between them. The dice is expected to land on an even number 100 more times than on a six.
- (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) 30 — The numbers less than 4 are 1, 2 and 3, so the probability of that event is 3/6, and the expected count in 90 rolls is 90 × 3/6 = 45. The probability of rolling a 6 is 1/6, and the expected count is 90 × 1/6 = 15. The difference between the two expected counts is 45 − 15 = 30. A candidate who answers 45 has given the expected count for 'less than 4' only, forgetting to subtract the other expected count. A candidate who answers 15 has given the expected count for '6' only. A candidate who answers 36 has used a dice with 5 possible numbers instead of 6, giving 90 × 3/5 = 54 and 90 × 1/5 = 18, a difference of 36.
- (d) 136 — 15% of 160 = 0.15 × 160 = 24 patients reported side effects, so 160 − 24 = 136 did not. Stopping after finding the number who reported side effects, 24, answers the wrong question — it is not the number who did NOT report them. Misreading '15%' as a raw count of 15 patients, rather than a percentage, gives 160 − 15 = 145. Subtracting 15% of 160 twice, 160 − 24 − 24 = 112, double-counts the side-effect group.
- (b) 12/19 — Method: turn the percentages into expected frequencies out of 1000, total the faulty pens, then divide supplier Y's faulty pens by that total, because the pen picked is known to be faulty. Working: supplier X provided 700 pens and 2% of them are faulty, which is 14 pens. Supplier Y provided 300 pens and 8% of them are faulty, which is 24 pens. Altogether 38 pens are faulty, so the probability is 24/38, and dividing the numerator and the denominator by 2 gives 12/19. Answer: the probability is 12/19. The distractors: 3/10 is supplier Y's share of the stock, the answer before the faulty information is used at all; 3/125 is 24/1000, the probability that a pen is from supplier Y and faulty, which stops at the joint probability and never divides by the probability of a fault; 4/5 is 8 divided by 2 + 8, comparing the two fault rates as though the suppliers provided equal numbers of pens, so the 70 to 30 split is thrown away.
- (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.
- (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.
- (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.
- (a) 75% — 'Percentage of the women' restricts the group to the 80 women, of whom 60 attend yoga: 60/80 = 0.75 = 75%. Dividing by the number of men (200 − 80 = 120) instead of the number of women gives 60/120 = 0.5 = 50%. Dividing by all 200 members instead of just the 80 women gives 60/200 = 0.3 = 30%. Using the 20 women who do NOT attend yoga (80 − 60) as the numerator instead of the 60 who do gives 20/80 = 0.25 = 25%.
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