Printable · GCSE Foundation · ages 14-16
Probability worksheet — GCSE Foundation
Fifteen questions across the probability statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Answer key: Probability worksheet — GCSE Foundation
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- (b) 0.49 — Adding all four pupils' flips gives 40 + 25 + 20 + 15 = 100 flips in total, and adding their heads gives 24 + 10 + 9 + 6 = 49 heads in total, so the combined estimate is 49/100 = 0.49. Writing 0.46 is wrong because it averages the four pupils' individual rates (0.60, 0.40, 0.45 and 0.40) as if they all came from the same number of flips, which they do not — this ignores that Sam and Priti flipped far more times than Leo and Fatima. Writing 0.60 is wrong because it only uses Sam's own result (24/40 = 0.60), ignoring the other three pupils. Writing 0.40 is wrong because it only combines Priti and Fatima's results (16/40 = 0.40), leaving out Sam and Leo entirely. The combined estimate from all four pupils is 0.49.
- (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.
- (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.
- (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.
- (c) 1/3 — Method: list the scores that satisfy the condition, count them and write that count over the total number of equally likely scores, then simplify. Working: the scores greater than 4 are 5 and 6, so 2 of the 6 equally likely scores qualify, giving 2/6. Answer: 2/6 cancels to 1/3. The distractors: 2/3 comes from giving the probability that the score is not greater than 4, the complement rather than the event asked for; 1/6 comes from giving the probability of one particular qualifying score; 1/2 comes from reading 'greater than 4' as '4 or more' and counting 4, 5 and 6, which gives 3/6.
- (b) 0.15 — Method: success and failure are the only two outcomes of the task, so they form an exhaustive set and their probabilities add to 1; subtract the given probability from 1. Working: writing 1 as 1.00 so that both numbers have two decimal places gives 1.00 − 0.85; exchanging once, the hundredths give 10 − 5 = 5 and the tenths, now 9, give 9 − 8 = 1. Answer: 0.15, close to the left-hand end of the 0 to 1 scale because the machine nearly always succeeds. The distractors: 0.85 comes from giving back the probability of success instead of the probability of failure; 0.25 comes from taking each decimal column from 10 on its own in 1.00 − 0.85, writing 10 − 5 = 5 in the hundredths and 10 − 8 = 2 in the tenths instead of reducing the tenths to 9 after the exchange; 0.5 comes from assuming that success and failure must be equally likely because there are only two outcomes.
- (a) 753 — The estimate from 2000 spins is the most reliable, since it comes from the largest sample size, so the best estimate of the probability is 0.251. Over a further 3000 spins, the expected number landing on green is 3000 × 0.251 = 753. Writing 1050 is wrong because 3000 × 0.350 = 1050 uses the estimate from only 20 spins, the LEAST reliable of the three. Writing 870 is wrong because 3000 × 0.290 = 870 uses the estimate from 200 spins rather than the more reliable 2000-spin estimate. Writing 750 is wrong because 3000 × 0.25 = 750 ignores the recorded data completely and simply assumes each of the 4 colours is equally likely. The best estimate is 753 expected green spins.
- (c) No — 7/30 is the relative frequency; theory stays 1/6. — The theoretical probability of rolling a 6 on an ordinary dice is fixed at 1/6, worked out from the number of equally likely outcomes, and does not change however the dice is actually rolled. The relative frequency from this trial is 7/30, found from what happened in these particular 30 rolls. Since 7/30 and 1/6 are different numbers, the correct statement is 'No — 7/30 is the relative frequency; theory stays 1/6.' Assuming the two values must always match because they describe the same event gives 'Yes — relative frequency always equals theory.' Believing that an observed result redefines the theoretical probability gives 'Yes — the theoretical probability has now become 7/30.' Refusing to work out either value at all gives 'Neither can be found — 30 rolls is too few to tell', which ignores that both numbers CAN be calculated from the information given.
- (c) The green bag (7/24) — Method: fractions can only be ordered once they share a denominator, so rewrite all four over the lowest common denominator and compare the numerators. Working: the lowest common denominator of 12, 8, 24 and 3 is 24, and scaling gives 5/12 = 10/24, 3/8 = 9/24, 7/24 stays as it is, and 1/3 = 8/24; the numerators are then 10, 9, 7 and 8. Answer: the smallest numerator is 7, so the green bag, with 7/24, is the least likely and sits furthest to the left on the 0 to 1 scale. The distractors: the red bag (5/12) comes from finding the largest of the four probabilities instead of the smallest; the blue bag (3/8) comes from scaling 3/8 by changing only the denominator to 24, which turns it into 3/24 and makes it look the smallest; the yellow bag (1/3) comes from comparing numerators alone and assuming the fraction with the numerator 1 must be the smallest.
- (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.
- (a) 1/29 — The probability that Priya's ticket wins is 6/30. Since her ticket is not returned, there are now only 5 winning tickets left out of 29 tickets in total, so the probability that Tom's ticket also wins is 5/29. Multiplying these, 6/30 × 5/29 = 30/870 = 1/29. A candidate who answers 1/25 has treated Priya's ticket as returned, using 6/30 twice. A candidate who answers 1/30 has correctly reduced the winning tickets to 5 for Tom but forgotten to reduce the total number of tickets, using 5/30 instead of 5/29. A candidate who answers 11/59 has added the numerators and added the denominators, (6+5)/(30+29), instead of multiplying.
- (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) 15/64 — Method: because the first marble is put back, the second draw is made from the same 8 marbles, so the two draws are independent and their probabilities are multiplied. Working: the bag holds 8 marbles both times, so yellow first has probability 5/8 and blue second has probability 3/8. Multiplying fractions gives 5 × 3 = 15 on the top and 8 × 8 = 64 on the bottom. Answer: the probability is 15/64. The distractors: 9/64 comes from working out the probability that BOTH marbles are blue, 3/8 × 3/8, instead of one of each colour in the stated order; 15/16 comes from multiplying the numerators but adding the denominators; 5/8 comes from writing down the first draw only and never combining it with the second.
- (d) 7/15 — Both socks are the same colour either if both are red or if both are blue. The probability both are red is 4/6 × 3/5 = 12/30. The probability both are blue is 2/6 × 1/5 = 2/30. Adding these gives 12/30 + 2/30 = 14/30 = 7/15. Choosing 2/5 comes from only working out the 'both red' path, 12/30, and forgetting the 'both blue' path also counts. Choosing 5/9 comes from treating the draws as if the first sock were replaced, using 4/6 × 4/6 + 2/6 × 2/6 = 20/36 = 5/9, instead of reducing the totals for the second draw. Choosing 7/18 comes from reducing the number of socks removed but not the number left to choose from, using 4/6 × 3/6 + 2/6 × 1/6 = 14/36 = 7/18, instead of 5 remaining socks for the second draw.
- (b) (48/52) × (47/51) — Method: for two deals one after the other with nothing put back, multiply the probability of the first by the probability of the second worked out from the cards that are left. Working: 52 − 4 = 48 cards are not aces, so the first card is not an ace with probability 48/52. One card has now gone and it was not an ace, so 51 cards remain and 47 of them are not aces, giving 47/51. Answer: the probability is (48/52) × (47/51). The distractors: (48/52) × (48/52) comes from leaving the pack at 52 cards for the second deal, which is only true if the first card is replaced; (4/52) × (3/51) comes from working out the probability that both cards ARE aces instead of neither; (4/52) × (4/51) comes from the same misreading with the ace count left at 4 while the total is reduced, adjusting only half of the second fraction.
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