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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- (d) 0.468 — The probability that Leo does not wake up before his alarm is 1 − 0.35 = 0.65, and the probability that his sister does not is 1 − 0.28 = 0.72. Since the two events are independent, multiply the complements: 0.65 × 0.72 = 0.468. A candidate who answers 0.63 has added the two given probabilities, 0.35 + 0.28, instead of finding and multiplying the complements. A candidate who answers 0.098 has multiplied the two given probabilities directly, 0.35 × 0.28, without taking complements first. A candidate who answers 0.532 has correctly reached 0.468 but then subtracted it from 1 again by mistake.
- (d) 21 — Method: 'yoga only' means yoga but not pilates, so subtract the number who do both from the total who do yoga. Working: 32 − 11 = 21. Answer: 21. Watch out: writing down 32, the total who do yoga, answers 'how many do yoga' rather than 'how many do yoga only' — it still includes the 11 who also do pilates. Adding the overlap instead of subtracting it, 32 + 11 = 43, moves in the wrong direction entirely. And writing down 11 gives the number who do both activities, which is the opposite of yoga only.
- (a) 4/15 — Method: first find how many pupils travel only by car, then write that as a fraction of the 60 pupils surveyed. Working: pupils who walk or cycle or both = 32 + 24 − 12 = 44. Only by car = 60 − 44 = 16. P(only by car) = 16/60 = 4/15. Answer: 4/15. Watch out: writing down 1/5 takes the overlap of 12 pupils on its own, 12/60, mistaking the group who do both for the group who travel only by car. Writing down 4/5 comes from 60 − 12 = 48, subtracting only the overlap from the total instead of the whole walk-or-cycle count, so cyclists and walkers who are not in the overlap are wrongly swept into the only-car group. And writing down 7/15 comes from 60 − 32 = 28, subtracting the walkers alone and forgetting the cyclists altogether.
- (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.
- (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.
- (a) 17/32 — Because the counter is replaced, each pick is independent with P(red) = 5/8 and P(blue) = 3/8 every time. Both red has probability 5/8 × 5/8 = 25/64, and both blue has probability 3/8 × 3/8 = 9/64. 'Same colour' means either of these, so add them: 25/64 + 9/64 = 34/64 = 17/32. Giving 25/64 finds only the probability of both counters being red, leaving out both counters being blue, which also counts as the same colour. Giving 15/64 finds the probability of one red and one blue counter in just one of the two possible orders — the opposite of what was asked, and only half of it. Giving 13/28 works out the probabilities as if the counter had NOT been replaced, using 5/8 × 4/7 and 3/8 × 2/7, even though the question states it was put back.
- (b) 10 — The number who use the pool or the sauna (or both) is the total minus those who use neither: 70 − 12 = 58. Since pool + sauna double-counts the overlap, n(P ∩ S) = 38 + 30 − 58 = 10. Adding the pool and sauna counts without subtracting the overlap at all gives 38 + 30 = 68, more members than are in the whole gym. Subtracting the sauna count from the union, 58 − 30 = 28, actually finds the number who use ONLY the pool, not both. Reporting the 'neither' count, 12, confuses it with the 'both' region — they describe opposite corners of the diagram.
- (b) 37/75 — There are 150 − 84 = 66 men. 84 − 50 = 34 women prefer weight training, and 40 men prefer weight training, so 34 + 40 = 74 people in total prefer weight training, out of 150: 74/150 = 37/75. Writing 4/15 is wrong because it only counts the men who prefer weight training (40/150, simplified), leaving out the 34 women. Writing 17/75 is wrong because it only counts the women who prefer weight training (34/150, simplified), leaving out the 40 men. Writing 37/42 is wrong because it uses the number of women (84) as the denominator instead of the whole gym (150) — 74/84 simplifies to 37/42, but that is not a probability out of the whole group. The probability is 37/75.
- (d) 2/3 — Method: the pupil picked is known to be a girl, so the sample space is the 45 girls and not all 80 pupils; divide the number of girls who walk by the number of girls. Working: 30 of the 45 girls walk to school, which gives 30/45. Dividing the numerator and the denominator by 15 gives 2/3. Answer: the probability is 2/3. The distractors: 3/8 is 30/80, dividing the girls who walk by every pupil in the group, which is the commonest slip on a conditional probability because it keeps the restricted numerator but the whole denominator; 1/3 is 15/45, counting the 15 girls who do not walk to school and so answering the opposite event inside the correct group; 11/20 is 44/80, adding the 30 girls and the 14 boys who walk and dividing by the whole group, which throws away the information that the pupil picked is a girl.
- (d) 3/10 — Method: list the full possibility space as pairs of coin and spinner results, then count how many pairs satisfy both conditions and divide by the size of the whole space. Working: the coin gives 2 outcomes and the spinner gives 5, so the full space has 2 × 5 = 10 equally likely pairs. The pairs with a tail and an odd number are (T,1), (T,3) and (T,5), which is 3 out of 10. Answer: 3/10. Watch out: writing down 1/2 uses only the coin's own chance of a tail and ignores that the spinner also has to land on an odd number. Writing down 3/5 uses only the spinner's chance of landing on an odd number and ignores the coin altogether. And writing down 1/10 counts just one matching outcome, such as (T,1), instead of all three pairs that satisfy both conditions.
- (b) 1/4 — There are 6 × 2 = 12 equally likely outcomes. The even-numbered cards are 2, 4 and 6, so there are 3 × 1 = 3 outcomes with an even card and tails, giving a probability of 3/12 = 1/4. Choosing 1/2 comes from working out only the probability of drawing an even card, 3/6, and forgetting to combine it with the coin landing on tails. Choosing 1/12 comes from treating only one specific outcome, such as card 6 with tails, as the only one that counts, instead of all three even cards paired with tails. Choosing 1/8 comes from doubling the coin stage when counting the total, using 6 × 2 × 2 = 24 outcomes instead of 6 × 2 = 12, and giving 3/24 = 1/8.
- (c) 0.15 — Method: for two independent events, multiply along the branches of the tree to find the probability of both outcomes happening together. Working: P(red and heads) = P(red) × P(heads) = 0.3 × 0.5 = 0.15. Answer: 0.15. Watch out: adding the two probabilities, 0.3 + 0.5 = 0.8, does not give the probability of both — probabilities along one path of a tree are multiplied, not added. Writing down 0.5 ignores the spinner altogether and gives only the coin's probability. And writing down 0.65 is the probability of red OR heads, which is 0.3 + 0.5 − 0.15 = 0.65, a different question from the one asked here.
- (a) 0.368 — Combining both samples, the spinner landed on red 34 + 58 = 92 times out of a total of 85 + 165 = 250 spins, so the best estimate of the probability is 92/250 = 0.368. Writing 0.400 is wrong because it uses only the first sample, 34/85 = 0.400, ignoring the extra 165 spins recorded afterwards. Writing 0.352 is wrong because it uses only the second sample, 58/165 = 0.352 (to 3 decimal places), ignoring the first 85 spins. Writing 0.376 is wrong because it averages the two separate estimates, (0.400 + 0.352) ÷ 2 = 0.376, instead of combining the actual numbers of reds and spins across both samples. The best estimate of the probability that the spinner lands on red, using all 250 spins, is 0.368.
- (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.
- (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.
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