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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- (b) 0.0309 — Method: P(both defective | at least one defective) = P(both defective) ÷ P(at least one defective). Find each using independence: P(both) = 0.06², P(at least one) = 1 − P(neither) = 1 − 0.94². Working: P(both) = 0.06² = 0.0036. P(neither) = 0.94² = 0.8836, so P(at least one) = 1 − 0.8836 = 0.1164. P(both | at least one) = 0.0036 ÷ 0.1164 = 0.0309 (3 s.f.). Answer: 0.0309. Watch out: leaving the answer as 0.0036 gives P(both defective) itself, not the probability once you already know at least one is defective — you still need to divide by P(at least one defective). Giving 0.0600 answers with the single-component defect rate, ignoring the condition altogether. And 0.5000 assumes that 'at least one' makes the outcomes 'exactly one defective' and 'both defective' equally likely, which is not how these probabilities combine.
- (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) £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.
- (a) 6.75% — Finishing, retiring and being disqualified are exhaustive, so the three percentages sum to 100%: 100% − 68.5% − 24.75% = 6.75%. Adding the two given percentages instead of subtracting them from 100% gives 68.5% + 24.75% = 93.25%, the combined probability of finishing or retiring, not of being disqualified. Subtracting only the retiring percentage from 100% and forgetting the finishing percentage gives 100% − 24.75% = 75.25%. Subtracting only the finishing percentage and forgetting the retiring percentage gives 100% − 68.5% = 31.50%.
- (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) 24/125 — On the fiction branch, 160 − 112 = 48 books were returned late. None of the non-fiction books were late, so the total number of late books is 48, out of 250 books altogether: 48/250 = 24/125. Writing 3/10 is wrong because it divides the 48 late fiction books by the fiction total (160) instead of the whole library total (250). Writing 56/125 is wrong because 112/250 simplifies to 56/125, and 112 is the number of fiction books returned ON TIME, not late. Writing 9/25 is wrong because 90/250 simplifies to 9/25, and 90 is simply the number of non-fiction books, which has nothing to do with late returns. The probability is 24/125.
- (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.
- (d) 1/4 — Method: a product is odd only when BOTH factors are odd, so list the ordered pairs where both scores are odd and divide by 36. Working: the odd scores on a dice are 1, 3 and 5, so there are 3 × 3 = 9 ordered pairs where both scores are odd, out of the 36 equally likely pairs, cancelling down to 1/4. Answer: 1/4. Watch out: writing down 3/4 finds the probability that AT LEAST ONE score is odd, 1 minus the probability both are even, which is a different, easier condition to meet than both being odd. Considering only the first dice's score and ignoring the second gives 1/2, since 3 of the first dice's 6 scores are odd — but the product also depends on what the second dice shows. And writing down 1/12 comes from counting only the pairs where the SAME odd number appears twice, (1, 1), (3, 3) and (5, 5), missing pairs like (1, 3) and (5, 1) where the two odd scores differ.
- (a) 43/70 — There are 140 − 85 = 55 Zone B households, and 55 − 21 = 34 of them recycle glass. In total, 52 + 34 = 86 households recycle glass, out of 140: 86/140 = 43/70. Writing 13/35 is wrong because 52/140 simplifies to 13/35, and 52 only counts Zone A, leaving out the 34 Zone B recyclers. Writing 17/70 is wrong because 34/140 simplifies to 17/70, and 34 only counts Zone B, leaving out the 52 Zone A recyclers. Writing 73/140 is wrong because it adds the 52 Zone A recyclers to the 21 Zone B households that do NOT recycle, mixing up two different groups instead of adding the two recycling groups. The probability is 43/70.
- (c) 1/10 — On the adult branch, 210 − 189 = 21 appointments were missed. There are 300 − 210 = 90 child appointments, and 90 − 81 = 9 of those were missed. In total, 21 + 9 = 30 appointments were missed, out of 300: 30/300 = 1/10. Writing 7/100 is wrong because 21/300 simplifies to 7/100, and 21 only counts the adult branch, leaving out the 9 missed child appointments. Writing 3/100 is wrong because 9/300 simplifies to 3/100, and 9 only counts the child branch, leaving out the 21 missed adult appointments. Writing 1/9 is wrong because it divides the 30 missed appointments by the 270 that were attended (300 − 30) instead of by the whole 300 booked. The probability is 1/10.
- (b) 12/19 — Method: turn the percentages into expected frequencies out of 1000, total the faulty pens, then divide supplier Y's faulty pens by that total, because the pen picked is known to be faulty. Working: supplier X provided 700 pens and 2% of them are faulty, which is 14 pens. Supplier Y provided 300 pens and 8% of them are faulty, which is 24 pens. Altogether 38 pens are faulty, so the probability is 24/38, and dividing the numerator and the denominator by 2 gives 12/19. Answer: the probability is 12/19. The distractors: 3/10 is supplier Y's share of the stock, the answer before the faulty information is used at all; 3/125 is 24/1000, the probability that a pen is from supplier Y and faulty, which stops at the joint probability and never divides by the probability of a fault; 4/5 is 8 divided by 2 + 8, comparing the two fault rates as though the suppliers provided equal numbers of pens, so the 70 to 30 split is thrown away.
- (b) 0.55 — Method: it is easier to find the probability that Ffion wins NEITHER stage, then subtract that from 1. Working: P(lose stage 1) = 1 − 0.4 = 0.6, and P(lose stage 2) = 1 − 0.25 = 0.75. P(neither) = 0.6 × 0.75 = 0.45. P(at least one) = 1 − 0.45 = 0.55. Answer: 0.55. Watch out: adding the two win probabilities, 0.4 + 0.25 = 0.65, treats winning both as impossible and overcounts — that is not how independent probabilities combine. Multiplying the two win probabilities, 0.4 × 0.25 = 0.1, gives the probability of winning BOTH stages, not at least one. And stopping at 0.45, the probability of winning neither stage, forgets the final step of subtracting from 1.
- (c) 11/20 — 'Red or white' combines two mutually exclusive events, so add their probabilities: 3/10 = 6/20 and 1/4 = 5/20, giving 6/20 + 5/20 = 11/20. Multiplying the two probabilities instead of adding them, 3/10 × 1/4, gives 3/40, which would be the probability of red and white together, not red or white — and a bead can't be both colours. Subtracting the sum from 1, 1 − 11/20 = 9/20, gives the probability of the bead being black instead of red or white. Converting 1/4 as 4/20 instead of 5/20 (dividing 20 by 4 but forgetting to scale the numerator) gives 6/20 + 4/20 = 1/2.
- (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.
- (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.
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