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.
Answer key: Probability worksheet — GCSE Higher
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- (a) 1/12 — The probability that the first dice shows a 6 is 1/6. The probability that the second dice shows an even number, 2, 4 or 6, is 3/6 = 1/2. Since the two dice are independent, multiply the probabilities: 1/6 × 1/2 = 1/12. A candidate who answers 1/6 has considered only the first dice and forgotten the condition on the second dice. A candidate who answers 1/2 has considered only the second dice and forgotten the condition on the first dice. A candidate who answers 1/36 has treated 'an even number' as a single specific value rather than three possible values, using 1/6 × 1/6.
- (c) Noah — 11/32 = 0.34375, above 34%. — Converting 11/32 to a decimal gives 11 ÷ 32 = 0.34375, which is greater than 34% (0.34), so Noah has the better save rate: 'Noah — 11/32 = 0.34375, above 34%.' Comparing the raw numbers 34 and 11 directly, without converting the fraction to the same form, gives 'Elin — 34 is bigger than 11.' Treating a larger denominator as meaning a bigger value, rather than smaller equal shares, gives 'Noah — 32 is a bigger denominator.' Rounding 34.375% to 34% to the nearest whole percent hides the difference and gives 'Equal — both round to 34% to the nearest percent.'
- (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) 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) 5/12 — There are 36 equally likely ordered pairs. Listing the pairs where the red score is smaller than the blue score gives 15 outcomes, so the probability is 15/36 = 5/12. Choosing 1/2 comes from assuming the 36 outcomes split evenly between 'red smaller' and 'red bigger', which ignores the 6 outcomes where the two dice show the same score. Choosing 7/12 comes from including the outcomes where the two scores are equal, 15 + 6 = 21, giving 21/36 = 7/12, instead of using 'less than' strictly. Choosing 5/36 comes from counting only the outcomes where the two scores differ by exactly 1, such as (1, 2) and (2, 3), which is 5 outcomes, and missing the pairs that differ by more than 1.
- (b) 0.355 — Method: use the law of total probability across the two Monday branches: P(rain Tue) = P(rain Mon) × P(rain Tue | rain Mon) + P(no rain Mon) × P(rain Tue | no rain Mon). Working: P(no rain Mon) = 1 − 0.3 = 0.7. P(rain Tue) = (0.3 × 0.6) + (0.7 × 0.25) = 0.18 + 0.175 = 0.355. Answer: 0.355. Watch out: using only the rain-Monday branch (0.3 × 0.6) or only the no-rain-Monday branch (0.7 × 0.25) accounts for just one of the two ways Tuesday can turn out rainy — both branches must be added. And swapping which weekday-probability multiplies which branch (0.7 with the rain branch, 0.3 with the no-rain branch) uses the right numbers on the wrong branches.
- (d) 5/12 — The possible totals are 2 to 12. The prime totals are 2, 3, 5, 7 and 11, with 1, 2, 4, 6 and 2 ways to make them, giving 1 + 2 + 4 + 6 + 2 = 15 outcomes out of 36, so the probability is 15/36 = 5/12. Choosing 7/18 comes from forgetting that a total of 2 counts as prime, giving 14/36 = 7/18 instead. Choosing 19/36 comes from wrongly including a total of 9 as a prime number, adding its 4 ways to the correct 15 to get 19/36. Choosing 13/36 comes from forgetting the 2 ways of making a total of 11, giving 13/36 instead of 15/36.
- (a) 0.48 — There are two ways to score exactly one throw: scoring on the first and missing the second, 0.6 × 0.4 = 0.24, or missing the first and scoring the second, 0.4 × 0.6 = 0.24. Adding these gives 0.24 + 0.24 = 0.48. Choosing 0.24 comes from working out only one of the two paths and forgetting the other one also gives exactly one score. Choosing 0.36 comes from working out the probability of scoring BOTH throws, 0.6 × 0.6 = 0.36, instead of exactly one. Choosing 0.84 comes from working out the probability of scoring AT LEAST one throw, 1 − 0.4 × 0.4 = 0.84, instead of exactly one.
- (c) £3.10 — Aisha's tickets cost 8 × 50p = £4.00. Her expected winnings are (8/400) × £45 = £0.90, since she holds 8 of the 400 tickets. Her expected loss is the cost minus the expected winnings: £4.00 − £0.90 = £3.10. Writing £4.00 is wrong because it is only the cost of her tickets, with no account taken of the expected winnings she might get back. Writing £0.90 is wrong because that is her expected WINNINGS, not her loss — the cost has not been subtracted. Writing £3.89 is wrong because it uses 1 ticket instead of her actual 8 tickets when working out the expected winnings: (1/400) × £45 = £0.1125, giving £4.00 − £0.11 = £3.89. Aisha should expect to lose £3.10.
- (b) 40 — n(P ∪ Q) = n(P) + n(Q) − n(P ∩ Q) = 34 + 27 − 11 = 50. The complement is everyone outside both sets: n((P ∪ Q)′) = 90 − 50 = 40. Adding P and Q without subtracting the overlap gives 34 + 27 = 61, so 90 − 61 = 29 double-subtracts the 11 who are in both. Reporting n(P ∪ Q) itself, 50, forgets to take the complement at all. Subtracting only n(P) from the universal set, 90 − 34 = 56, ignores set Q altogether.
- (b) 12/25 — List every outcome as an ordered pair (first dice, second dice) out of the 25 equally likely outcomes. The pairs with a difference of exactly 2 are (1, 3), (3, 1), (2, 4), (4, 2), (3, 5) and (5, 3), the pairs with a difference of 3 are (1, 4), (4, 1), (2, 5) and (5, 2), and the pairs with a difference of 4 are (1, 5) and (5, 1), giving 6 + 4 + 2 = 12 outcomes with a difference of at least 2, so the probability is 12/25. Choosing 13/25 comes from finding the probability that the two scores differ by LESS than 2 instead, using the remaining 13 outcomes, the opposite of what was asked. Choosing 10/25 comes from forgetting the two outcomes where the difference is 4, (1, 5) and (5, 1), and adding only the difference-2 and difference-3 outcomes, 6 + 4 = 10 out of 25. Choosing 6/25 comes from counting only the pairs with a difference of exactly 2, forgetting that differences of 3 and 4 also count as at least 2.
- (a) 1276 — Method: an unbiased relative frequency tends towards the theoretical probability as the number of trials increases, so use the record resting on the most trials, then multiply by the number of new trials. Working: the three records rest on 50, 200 and 1000 drops, so the most reliable is the one after 1000 drops, namely 0.638, and the run is indeed settling as the trials increase. The expected number of point up landings in 2000 further drops is 2000 × 0.638 = 1276. Answer: about 1276 times. The distractors: 1440 uses the earliest record, which rests on only 50 drops, giving 2000 × 0.720 = 1440; 1330 uses the middle record, treating 200 drops as a safe compromise when 1000 drops is better still, giving 2000 × 0.665 = 1330; 1348 comes from averaging the three records, since 0.720 + 0.665 + 0.638 = 2.023 and 2.023 ÷ 3 = 0.674, then 2000 × 0.674 = 1348, which gives the 50 drop record the same weight as the 1000 drop record.
- (c) 9/25 — The frequency tree already shows the swim-and-cycle branch directly: 54 out of 150, which simplifies to 9/25. Adding both cycling branches together, 54 + 18 = 72, gives the total number of cyclists, so 72/150 = 12/25, not just those who also swim. Using the non-swimmers' cycling figure, 18, gives 18/150 = 3/25, the probability of cycling WITHOUT swimming. Using the swimmers' total of 90, before splitting by cycling, gives 90/150 = 3/5, the probability of swimming on its own.
- (d) 1/12 — Method: list the full possibility space of sandwich-and-drink pairs, then divide the one matching pair by the size of the whole space. Working: there are 3 × 4 = 12 equally likely sandwich-and-drink pairs, and exactly one of them is egg and water. Answer: 1/12. Watch out: writing down 1/7 comes from adding the two counts, 3 + 4 = 7, instead of multiplying them to build the possibility space. Writing down 1/3 uses only the chance of choosing egg out of 3 sandwiches and ignores the drink altogether. And writing down 1/4 uses only the chance of choosing water out of 4 drinks and ignores the sandwich altogether.
- (d) 5/11 — Method: the pen already taken was black, so update the contents of the box before working out the second probability. Working: the box held 12 pens and one black pen has gone, so 11 pens remain. None of the blue pens has been taken, so all 5 are still there, and the probability is 5/11, which will not cancel. Answer: the probability is 5/11. The distractors: 5/12 uses the box as it was at the start, which is only correct if the first pen is put back; 4/11 takes one off the blue count as well as the total, as though the pen removed had been blue; 6/11 gives the probability that the second pen is black, carrying on with the colour of the first pen instead of the colour asked for.
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