Printable · GCSE Foundation · ages 14-16
Expected outcomes and fairness worksheet — GCSE Foundation
Fifteen questions on "expected outcomes and fairness" — DfE statement P2. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Answer key: Expected outcomes and fairness worksheet — GCSE Foundation
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- (d) 1/20 — Mia has 25 of the 500 tickets, so the probability she wins is 25/500 = 1/20, dividing the top and the bottom by 25. Writing 19/20 is wrong because that is the probability she does NOT win (1 − 1/20 = 19/20), the opposite of what is asked. Writing 1/25 is wrong because it puts 1 over the number of tickets Mia holds, as though her 25 tickets were the whole raffle — the denominator has to be the 500 tickets sold, not her own share. Writing 1/10 is wrong because it treats the raffle as having only 250 tickets instead of the actual 500: 25/250 = 1/10. The probability that Mia wins is 1/20.
- (b) 10 — The probability that the spinner lands on green is its angle out of the whole circle: 60° ÷ 360° = 1/6. Expected number of times on green = 1/6 × 60 = 10. Assuming the three colours are equally likely because there are three sectors, regardless of their different angles, gives 60 ÷ 3 = 20. Using red's angle of 180° instead of green's 60° gives a probability of 1/2, so 60 × 1/2 = 30. Treating green's angle, 60°, as a percentage instead of finding its fraction of 360° gives 60 × 0.60 = 36.
- (b) 120 — Purple's probability is its angle out of the full circle: 144° ÷ 360° = 2/5. Expected number on purple = 2/5 × 300 = 120. Assuming the three colours are equally likely, ignoring their different angles, gives 300 ÷ 3 = 100. Using orange's angle, 90°, instead of purple's 144° gives a probability of 1/4, so 300 × 1/4 = 75. Treating 144 as a percentage instead of finding its fraction of 360° gives 300 × 0.144 = 43.2, which rounds to 43.
- (a) 60 — Method: because the counter goes back each time, every draw is the same experiment, so the expected number of reds is the number of draws multiplied by P(red). Working: there are 8 red counters out of 20, so P(red) = 8 ÷ 20 = 0.4. Over 150 draws the expected number of reds is 150 × 0.4 = 60. Answer: about 60 red counters would be expected. The distractors: 90 uses P(blue) by mistake, 150 × 12 ÷ 20 = 90; 100 comes from writing the probability from the red to blue ratio of 8 to 12, giving 150 × 8 ÷ 12 = 100; 75 comes from treating red and blue as equally likely because there are two colours, which gives 150 ÷ 2 = 75.
- (a) £150 — Each game, the expected payout is 0.1 × £20 = £2, so the fête's expected profit per game is the £3 charged minus the £2 expected payout, £1. Over 150 games, that is 150 × £1 = £150. Writing £300 is wrong because 150 × £2 = £300 is the total expected PAYOUT, not the profit — it has not been subtracted from the entry fees. Writing £450 is wrong because 150 × £3 = £450 is the total money taken in entry fees, without accounting for what is expected to be paid out in prizes. Writing £1 is wrong because that is only the expected profit for ONE game — it has not been scaled up to all 150 games. The fête's expected profit is £150.
- (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.
- (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.
- (a) 50 — To find the number of shots needed for an expected 12 hits, divide the number of hits wanted by the probability of a hit: 12 ÷ 0.24 = 50. Multiplying the number of hits by the probability instead of dividing gives 12 × 0.24 = 2.88, which rounds to 3 shots. Rounding 0.24 to 0.25 before dividing gives 12 ÷ 0.25 = 48. Using the probability of missing, 1 − 0.24 = 0.76, instead of the probability of hitting, gives 12 ÷ 0.76 = 15.79, which rounds to 16.
- (a) 80 — Method: list the outcomes that count as a success, write the probability from them, then multiply by the number of rolls. Working: the scores of 3 or more are 3, 4, 5 and 6, which is 4 of the 6 equally likely scores, so the probability is 4/6, which cancels to 2/3. Over 120 rolls the expected number is 120 × 2 ÷ 3 = 80. Answer: about 80 of the rolls would be expected to give 3 or more. The distractors: 60 comes from reading a score of 3 or more as a score above 3 and counting only 4, 5 and 6, giving 120 × 3 ÷ 6 = 60; 40 is the expected number of rolls that are not 3 or more, 120 × 2 ÷ 6 = 40; 20 is 120 ÷ 6 and is the expected count for one single score.
- (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.
- (b) £100 — Method: find the expected number of wins, turn that into the expected pay out, then compare it with what the games cost. Working: the expected number of wins is 200 × 0.15 = 30. Each win pays £10, so the expected pay out is 30 × 10 = 300 pounds. Playing 200 times at £2 a go costs 200 × 2 = 400 pounds. The expected loss is 400 − 300 = 100 pounds. Answer: Amir should expect to be about £100 down. The distractors: £300 is the expected winnings on their own, with the cost of playing never taken off; £400 is the total cost of playing, with the winnings never taken off; £700 comes from adding the two totals, 400 + 300 = 700, instead of subtracting one from the other.
- (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.
- (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.
- (c) 120 — 27 of the 45 sweets are orange (45 − 18 = 27), so the probability of taking an orange sweet is 27/45 = 3/5, and 200 × 3/5 = 120. Writing 80 is wrong because 200 × 18/45 = 80 uses the LEMON sweets' fraction instead of orange. Writing 182 is wrong because it takes the 18 lemon sweets away from the 200 repeats (200 − 18 = 182), applying the ‘the rest are orange’ subtraction to the number of goes instead of to the 45 sweets in the bag. Writing 27 is wrong because it is simply the number of orange sweets in the bag — it has not been scaled up to account for the 200 repeats. The expected number of times an orange sweet is taken is 120.
- (b) 60 — The probability of landing on purple in a single spin is 2/6. The expected number of times it lands on purple in 180 spins is 180 × 2/6 = 60. A candidate who answers 90 has used 3/6 instead of 2/6, miscounting the purple sections as 3. A candidate who answers 30 has used 1/6 instead of 2/6, forgetting one of the two purple sections. A candidate who answers 120 has used the probability of NOT landing on purple, 4/6, by mistake.
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