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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- (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.
- (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.
- (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.
- (c) 500 — Method: the expected number of successes is the number of trials multiplied by the probability, so to find the number of trials, divide the expected number by the probability. Working: let n be the number of seeds planted. Then n multiplied by 0.6 must come to 300, so n = 300 ÷ 0.6 = 500. Answer: the grower should plant 500 seeds. The distractors: 180 comes from multiplying instead of dividing, 300 × 0.6 = 180, which answers how many of 300 seeds would germinate; 750 comes from dividing by the probability of not germinating, 300 ÷ 0.4 = 750; 120 comes from multiplying by that same 0.4, 300 × 0.4 = 120.
- (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.
- (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.
- (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.
- (c) 300 — Expected number = probability × number of trials = 0.15 × 2,000 = 300. Moving the decimal point one place too far, using 0.015 instead of 0.15, gives 2,000 × 0.015 = 30. Working out the expected number of letters that are NOT bills, using the complement 1 − 0.15 = 0.85, gives 2,000 × 0.85 = 1,700. Rounding 0.15 up to 0.2 before multiplying gives 2,000 × 0.2 = 400.
- (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.
- (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) £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.
- (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) 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) £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) 45 — Three of the ten numbers (8, 9 and 10) are greater than 7, so the probability is 3/10, and 150 × 3/10 = 45. Writing 105 is wrong because 150 × 7/10 = 105 uses the seven numbers that are NOT greater than 7 (1 to 7), the opposite of what is asked. Writing 60 is wrong because it counts 7, 8, 9 and 10 as four numbers greater than 7, wrongly including 7 itself: 150 × 4/10 = 60. Writing 50 is wrong because 150 ÷ 3 = 50 divides by the count of favourable numbers instead of multiplying by the correct fraction of the spinner. The expected number of spins landing on a number greater than 7 is 45.
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