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.
Probability worksheet — GCSE Foundation
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- 1.The probability that a spinner lands on blue is 2/5. Write this probability as a percentage.
- 2.A charity raffle has three types of ticket: winning, near-miss and losing, and every ticket is exactly one of these. The probability that a ticket is winning is 1/8 and the probability that it is a near-miss is 1/4. Work out the probability that a ticket is losing.
- 3.Four pupils each flip the same coin a number of times and record the results. Sam flips it 40 times and gets 24 heads. Priti flips it 25 times and gets 10 heads. Leo flips it 20 times and gets 9 heads. Fatima flips it 15 times and gets 6 heads. Combine all four pupils' results to estimate the probability that the coin lands on heads.
- 4.A spinner is divided into sectors of 144°, 90° and 126°, coloured purple, orange and grey in that order. The spinner is spun 300 times. Work out how many times you would expect it to land on purple.
- 5.A charity raffle sells 250 tickets. A group of friends buy 40 of the tickets between them. Three prizes are awarded, each to a different randomly drawn ticket, and no ticket can win more than one prize. Work out the expected number of prizes won by the group. Give your answer as a fraction in its simplest form.
- 6.A two-way table records 200 members of a gym. 80 of the members are women and the rest are men. 60 of the women attend yoga classes. Work out the percentage of the women who attend yoga classes.
- 7.A cinema sold 150 tickets for a Saturday screening. A frequency tree records the sales: 90 of the tickets were bought by adults and the rest by children. Of the adult tickets, 54 were for the 3D showing and the rest were for the standard showing. Work out how many adult tickets were for the standard showing.
- 8.A fair six-sided dice is rolled once. Write down the probability of rolling a 6.
- 9.A two-way table records how 180 students at a school travel: by bus or on foot, split by year group. There are 84 students in Year 11, of whom 38 travel by bus and the rest walk. The rest of the 180 students are in Year 10, and 42 of the Year 10 students travel by bus. Work out the probability that a randomly chosen Year 10 student walks to school. Give your answer as a fraction in its simplest form.
- 10.Two cards are dealt one after the other from an ordinary pack of 52 playing cards. The first card is not put back before the second is dealt. The pack contains 4 aces. Work out the probability that neither card is an ace. Give your answer as a product of two fractions.
- 11.A garden centre tests 250 packets of seeds for germination, recording the results on a frequency tree. 68% of the 250 packets germinated successfully, and the rest did not. Work out how many packets germinated successfully.
- 12.At a school fête, a tombola stall costs £1.50 to play. The probability of winning is 0.2, and the prize is worth £6. Work out the stall's expected profit, on average, from each game played.
- 13.In a game, Freddie says the probability of winning is 0.45, the probability of drawing is 0.3 and the probability of losing is 0.35, and that all three of his probabilities are correct. Which statement about Freddie's probabilities is correct?
- 14.A frequency tree records 320 calls to a helpline. The calls first split into calls about billing and calls about technical support. Of the technical support calls, the probability that a call was resolved on the first contact is 0.75. If 180 technical support calls were resolved on the first contact, work out how many calls were about billing.
- 15.180 people entered a fun run. A frequency tree splits them into 108 adults and the rest children. It then splits each of those groups into those who finished and those who did not. 99 of the adults finished, and 153 of the 180 people finished altogether. Work out how many children did not finish.
Answer key
- (b) 40% — Method: a fraction becomes a percentage by scaling it so that its denominator is 100, or equivalently by dividing the numerator by the denominator and multiplying the decimal by 100. Working: 5 × 20 = 100, so the numerator is scaled by 20 as well, 2 × 20 = 40, giving 40/100; the same figure comes from 2 ÷ 5 = 0.4 and 0.4 × 100 = 40. Answer: 40%. The distractors: 20% comes from dividing 100 by the denominator alone, 100 ÷ 5 = 20, which is the percentage for 1/5 and never uses the numerator 2; 4% comes from finding the decimal 0.4 correctly and then writing the 4 down as the percentage instead of multiplying by 100; 2% comes from scaling only the denominator up to 100 and leaving the numerator as it was, giving 2/100.
- (a) 5/8 — Winning, near-miss and losing are exhaustive, so the three probabilities sum to 1. Writing 1/4 as 2/8 so every fraction has the same denominator, 1 − 1/8 − 2/8 = 8/8 − 1/8 − 2/8 = 5/8. Subtracting only the winning probability and forgetting the near-miss probability gives 1 − 1/8 = 7/8. Subtracting only the near-miss probability and forgetting the winning probability gives 1 − 1/4 = 3/4. Adding the two given probabilities and stopping there gives 1/8 + 2/8 = 3/8, the probability that a ticket is winning or a near-miss, not the probability that it is losing.
- (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) 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) 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) 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%.
- (c) 36 — On the adult branch there are 90 tickets in total, and 54 of them are for the 3D showing, so the standard-showing branch is 90 − 54 = 36. Subtracting 54 from the overall total of 150 gives 96, but 150 is the total for ALL tickets, not just the adult branch, so 96 is wrong. Writing 60 is wrong because that is the number of CHILD tickets (150 − 90 = 60), not adult standard tickets. Writing 54 again is wrong because that is the number of adult 3D tickets, not the number of adult standard tickets. The adult standard-showing branch is 36.
- (a) 1/6 — Method: for equally likely outcomes, write the number of favourable outcomes over the total number of outcomes. Working: the dice has one face showing 6, and there are 6 equally likely faces in total, so the probability is 1 over 6. Answer: 1/6, which is close to the left-hand end of the 0 to 1 scale because the outcome is unlikely. The distractors: 5/6 comes from giving the probability of not rolling a 6; 1/2 comes from treating 'a 6' and 'not a 6' as two equally likely outcomes; 1/5 comes from writing the one favourable face over the 5 faces that are not a 6 instead of over all 6 faces.
- (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.
- (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.
- (c) 170 — 68% of the 250 packets germinated: 250 × 0.68 = 170. Using the complement, the 32% that did NOT germinate, gives 250 × 0.32 = 80. Shifting the decimal point, using 0.068 instead of 0.68, gives 250 × 0.068 = 17. Rounding 68% up to 70% before multiplying gives 250 × 0.70 = 175.
- (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.
- (a) No — the three probabilities sum to 1.10, over 1. — Winning, drawing and losing are exhaustive and mutually exclusive, so their probabilities must sum to exactly 1. Adding Freddie's three values gives 0.45 + 0.3 + 0.35 = 1.10, which is more than 1, so his probabilities cannot all be correct: 'No — the three probabilities sum to 1.10, over 1.' Checking only that each value lies between 0 and 1 accepts them as 'Yes — each probability lies between 0 and 1' without ever adding the three together. Judging by which outcome sounds most likely leads to 'Yes — winning has the highest single probability', which never checks the total either. Noting that a runner cannot win, draw and lose at once, and treating that alone as enough, gives 'Yes — the three outcomes are mutually exclusive' — but mutually exclusive outcomes that are also exhaustive must still sum to 1, and 1.10 does not.
- (c) 80 — Since 180 calls are 0.75 of all the technical support calls, the technical support total is 180 ÷ 0.75 = 240. The billing calls make up the rest of the 320 calls, so 320 − 240 = 80. Choosing 240 comes from stopping after finding the technical support total and forgetting the question asks for the billing calls, which are the rest. Choosing 185 comes from multiplying 180 × 0.75 = 135 instead of dividing, then working out 320 − 135 = 185. Choosing 140 comes from using 180 directly as the whole technical support total, ignoring the probability altogether, then working out 320 − 180 = 140.
- (d) 18 — Method: a frequency tree can be filled from either end, because a pair of branches adds up to the number above it and the matching end branches add up to the total for that outcome. Working: the children number 180 − 108 = 72. Of the 153 finishers, 99 were adults, so the children who finished number 153 − 99 = 54. Along the children's branch, 72 − 54 = 18 children did not finish. Answer: 18 children did not finish. The distractors: 27 is 180 − 153 and counts everyone who did not finish, adults included; 9 is 108 − 99 and counts only the adults who did not finish; 54 is the number of children who did finish, which is the branch next to the one asked for.
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