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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- (a) 20 — Dice A is fair, so its expected number of sixes is 150 × 1/6 = 25. Dice B has P(6) = 0.3, so its expected number of sixes is 150 × 0.3 = 45. The difference is 45 − 25 = 20. Adding the two expected values instead of subtracting them gives 25 + 45 = 70. Reporting Dice B's expected sixes on their own, without comparing to Dice A, gives 45. Using the fair probability 1/6 for Dice B as well as Dice A ignores the bias altogether, giving 150 × 1/6 = 25 for both dice and a difference of 0.
- (c) 325 — Method: first find the relative frequency of NOT landing on red from the 200 spins, then scale that up to 500 spins. Working: non-red results = 50 + 80 = 130, out of 200 spins, so P(not red) = 130 ÷ 200 = 0.65. Expected non-red results in 500 spins = 500 × 0.65 = 325. Answer: 325. Watch out: writing down 175 finds the expected number of RED results instead, 70 ÷ 200 × 500 = 175, answering the opposite of what was asked. Writing down 250 assumes landing red or not landing red must be a fair 50-50 split, but the spinner is biased and the actual results do not split evenly. And writing down 130 stops after finding how many of the 200 spins were non-red and forgets to scale that figure up to the 500 spins asked for.
- (d) 3/16 — Method: work out each draw's own probability first, then multiply them together since the two draws are independent. Working: there are 8 tickets in all, 3 of them blue, so P(blue) = 3/8. P(spinner number greater than 2) = 2/4 = 1/2, since 3 and 4 qualify. Multiplying gives 3/8 × 1/2, which comes to 3/16. Answer: 3/16. Watch out: writing down 3/8 stops after the first draw and never brings in the spinner at all. Writing down 1/2 does the opposite, using only the spinner and ignoring the ticket draw. And writing down 5/16 uses 5/8, the probability of a RED ticket, instead of 3/8 for blue — reading the wrong colour off the raffle.
- (b) 50% — The total number of customers who bought a cake is 54 + 21 = 75, combining both hot-drink and non-hot-drink customers. As a percentage of all 150 customers, this is (75 ÷ 150) × 100 = 50%. Choosing 36% comes from only counting the hot-drink customers who bought a cake, (54 ÷ 150) × 100 = 36%, and forgetting the 21 non-hot-drink customers who also bought a cake. Choosing 14% comes from only counting the non-hot-drink customers who bought a cake, (21 ÷ 150) × 100 = 14%, and forgetting the 54 hot-drink customers who also bought a cake. Choosing 60% comes from dividing by the hot-drink total of 90 instead of the grand total of 150, (54 ÷ 90) × 100 = 60%.
- (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) 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.
- (c) 15/23 — Method: find P(rough and delayed) and the overall P(delayed) using the tree, then divide. Working: P(rough and delayed) = 0.2 × 0.75 = 0.15. P(calm and delayed) = 0.8 × 0.1 = 0.08. P(delayed) = 0.15 + 0.08 = 0.23. P(rough | delayed) = 0.15 ÷ 0.23 = 15/23. Answer: 15/23. Watch out: leaving the answer as 0.15 (3/20) gives P(rough and delayed) itself, without dividing by the overall probability that a crossing is delayed. Giving 0.75 (3/4) is the probability you were told to start with — that a crossing is delayed GIVEN the sea is rough — which is the reverse of what's being asked. And 0.2 (1/5) is just the original probability that the sea is rough, before you take the fact that the crossing was delayed into account.
- (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.
- (b) 38% — Method: the two swimming percentages are quoted inside different age groups, so weight each one by the size of its group and add the two results. Working: the under 18s are 45% of the members and 60% of them swim, giving 0.45 × 60 = 27% of all the members. The members aged 18 or over are 55% of the members and 20% of them swim, giving 0.55 × 20 = 11% of all the members. Adding these gives 38%. Answer: 38% of the members swim each week. The distractors: 80% comes from adding 60% and 20% straight off, treating two rates quoted inside different groups as though they could be added; 40% is the mean of 60% and 20%, which would be right only if the two age groups were the same size, and they are not; 42% comes from pairing each swimming rate with the wrong age group, working out 0.45 × 20 added to 0.55 × 60.
- (d) 36/91 — Method: P(both red | same colour) = P(both red) ÷ P(same colour), where P(same colour) = P(both red) + P(both green). Working: P(both red) = 9/20 × 8/19 = 72/380 = 18/95. P(both green) = 11/20 × 10/19 = 110/380 = 11/38. P(same colour) = 18/95 + 11/38 = 36/190 + 55/190 = 91/190. P(both red | same colour) = (36/190) ÷ (91/190) = 36/91. Answer: 36/91. Watch out: stopping at 18/95 gives P(both red) itself, without dividing by the probability that the colours matched at all. Working out 55/91 finds the same-colour probability for green instead of red — check which colour's count you are putting on top. And 9/20 is just the chance the first ball drawn is red, which ignores the second draw and the without-replacement condition completely.
- (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.
- (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) 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) £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.
- (d) 56 — Method: count the trials over the whole period first, then multiply the number of trials by the probability. Working: 4 weeks is 4 × 7 = 28 days, and at 25 trains a day that is 25 × 28 = 700 trains. The expected number of late trains is 700 × 0.08 = 56. Answer: about 56 late trains over the 4 weeks. The distractors: 2 is the expected number for a single day, 25 × 0.08 = 2, with the 28 days never brought in; 14 uses one week instead of four, 25 × 7 × 0.08 = 14; 644 is 700 − 56 and counts the trains expected to be on time.
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