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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- (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.
- (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) 1276 — Method: an unbiased relative frequency tends towards the theoretical probability as the number of trials increases, so use the record resting on the most trials, then multiply by the number of new trials. Working: the three records rest on 50, 200 and 1000 drops, so the most reliable is the one after 1000 drops, namely 0.638, and the run is indeed settling as the trials increase. The expected number of point up landings in 2000 further drops is 2000 × 0.638 = 1276. Answer: about 1276 times. The distractors: 1440 uses the earliest record, which rests on only 50 drops, giving 2000 × 0.720 = 1440; 1330 uses the middle record, treating 200 drops as a safe compromise when 1000 drops is better still, giving 2000 × 0.665 = 1330; 1348 comes from averaging the three records, since 0.720 + 0.665 + 0.638 = 2.023 and 2.023 ÷ 3 = 0.674, then 2000 × 0.674 = 1348, which gives the 50 drop record the same weight as the 1000 drop record.
- (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.
- (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%.
- (d) 12/25 — The group holds 40 of the 250 tickets, so for any one prize the probability the group wins it is 40/250 = 4/25. There are 3 prizes and the group has the same chance at each one, so the expected number won is 3 × 4/25 = 12/25. Writing 4/25 is wrong because it is the chance of winning just ONE prize, without multiplying by the 3 prizes available. Writing 4/75 is wrong because it divides by the 3 prizes instead of multiplying (4/25 ÷ 3 = 4/75), which would mean the group did worse the more prizes were on offer. Writing 64/15625 is wrong because it multiplies the single-prize probability by itself three times, (4/25)³, as though all three prizes had to be won together, instead of adding up the expected number across the three separate prizes. The expected number of prizes won by the group is 12/25.
- (a) £124.80 — On the cake branch, 150 − 100 = 50 cakes were bought by children. Adding the 54 biscuits bought by children gives 50 + 54 = 104 items sold to children in total, and at £1.20 each that raises 104 × £1.20 = £124.80. Writing £60.00 is wrong because 50 × £1.20 = £60.00 only counts the cake sales to children and leaves out the 54 biscuits. Writing £163.20 is wrong because it uses the ADULT sales instead of children's: 100 cake adults plus 90 − 54 = 36 biscuit adults gives 136 × £1.20 = £163.20. Writing £136.80 is wrong because it finds the cake children's number by subtracting the wrong branch (150 − 90 = 60 instead of 150 − 100 = 50), giving 60 + 54 = 114 items and 114 × £1.20 = £136.80. The total raised from sales to children is £124.80.
- (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.
- (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) £100 — Over 250 games, the expected total winnings are 250 × (1/5) × £12 = £600, since a player wins on 1 of the 5 equally likely sections. The total cost of playing is 250 × £2 = £500. The players' expected profit is the winnings minus the cost: £600 − £500 = £100. Writing £500 is wrong because that is only the total cost of playing, without any winnings included. Writing £600 is wrong because that is only the total expected winnings, without subtracting what was paid to play. Writing £2,500 is wrong because it assumes a win on every single game (250 × £12 = £3,000) instead of using the 1-in-5 probability, then subtracts the cost: £3,000 − £500 = £2,500. The players' expected profit over the 250 games is £100.
- (b) 80 — To find the number of attempts needed, divide the target number of successes by the probability of success: 60 ÷ 0.75 = 80. Writing 45 is wrong because 60 × 0.75 = 45 multiplies instead of dividing — that is the number of successes expected from 60 attempts, not the number of attempts needed for 60 successes. Writing 240 is wrong because 60 ÷ 0.25 = 240 uses 0.25, the probability of MISSING, instead of 0.75, the probability of scoring. Writing 90 is wrong because it comes from misremembering 0.75 as 2/3 and dividing by that instead: 60 ÷ (2/3) = 90. She needs to attempt 80 free throws.
- (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.
- (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.
- (b) 22.8% — Relative frequency as a percentage is the faulty count divided by the total, then multiplied by 100: 33 ÷ 145 × 100 = 22.76, which rounds to 22.8%. Giving 33.0% as the answer uses the frequency, 33, directly as a percentage without dividing by the total 145 at all. Rounding 22.76 down to 22.7% instead of up applies the wrong rounding direction at the first decimal place. Finding the relative frequency of the bulbs that were NOT faulty first: 145 − 33 = 112, and 112 ÷ 145 × 100 = 77.24, answers the opposite question and rounds to 77.2%.
- (b) £13.50 — The total cost of Nadia's 25 tickets is 25 × £1.50 = £37.50. The expected number of winning tickets is 25 × 0.12 = 3, so the expected prize money is 3 × £8 = £24.00. Nadia's expected loss is the cost minus the expected prize money: £37.50 − £24.00 = £13.50. A candidate who answers £24.00 has given the expected prize money and mistaken it for the loss. A candidate who answers £37.50 has given the total cost of the tickets, forgetting to subtract the expected prize money. A candidate who answers £34.50 has subtracted the expected number of wins, 3, from the cost instead of first converting it to prize money by multiplying by £8.
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