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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- (c) The green bag (7/24) — Method: fractions can only be ordered once they share a denominator, so rewrite all four over the lowest common denominator and compare the numerators. Working: the lowest common denominator of 12, 8, 24 and 3 is 24, and scaling gives 5/12 = 10/24, 3/8 = 9/24, 7/24 stays as it is, and 1/3 = 8/24; the numerators are then 10, 9, 7 and 8. Answer: the smallest numerator is 7, so the green bag, with 7/24, is the least likely and sits furthest to the left on the 0 to 1 scale. The distractors: the red bag (5/12) comes from finding the largest of the four probabilities instead of the smallest; the blue bag (3/8) comes from scaling 3/8 by changing only the denominator to 24, which turns it into 3/24 and makes it look the smallest; the yellow bag (1/3) comes from comparing numerators alone and assuming the fraction with the numerator 1 must be the smallest.
- (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.
- (a) 2/3 — Method: 'in A or in B' means every score that belongs to at least one of the two sets; a score that belongs to both is still only one outcome, so it is listed once. Working: the even scores are 2, 4 and 6; the scores greater than 4 are 5 and 6. Listing the scores that appear in either set gives 2, 4, 5 and 6, with 6 written once. That is 4 of the 6 faces, or 4/6. Answer: the probability is 2/3. The distractors: 5/6 comes from adding the sizes of the two sets, 3 + 2, so that the score 6 is counted in both and appears twice; 1/2 comes from using the even scores alone; 1/6 comes from giving the probability that the score is in both sets, which is the single score 6, rather than in either of them.
- (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) 15/32 — Method: there are two ways to get one of each colour, red then blue and blue then red. Work out the probability of each path by multiplying, then add the two paths. Working: red then blue is 5/8 × 3/8 = 15/64, and blue then red is 3/8 × 5/8 = 15/64. Adding the two paths gives 30/64. Answer: the probability is 15/32. The distractors: 15/64 comes from working out red then blue only and forgetting that blue then red also gives one of each; 15/28 comes from doubling correctly but reducing the total to 7 for the second spin, which is what happens to a bag when an item is kept out, not to a spinner; 39/64 comes from working from the opposite event and subtracting only the two-red case, 1 − 25/64, leaving the two-blue case inside the answer.
- (d) 3/36 — Method: list the results as ordered pairs, decide which totals satisfy the condition, count the pairs that give those totals and divide by the number of pairs there are. Working: there are 6 × 6 = 36 equally likely ordered pairs. Greater than 10 means a total of 11 or a total of 12. A total of 11 comes from (5, 6) and (6, 5); a total of 12 comes from (6, 6) alone, because both dice must show a 6. That is 2 + 1 = 3 pairs out of the 36. Answer: the probability is 3/36. The distractors: 2/36 comes from counting the two ways of making 11 and forgetting that 12 is greater than 10 as well; 4/36 comes from writing (6, 6) down twice, applying the rule that a pair can be turned round to a double that can only happen one way; 33/36 comes from reading the condition the wrong way round and giving the probability that the total is 10 or less.
- (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.
- (d) 315 — The relative frequency from the survey is 42 ÷ 120 = 0.35, so the expected number who prefer paper bags among 900 shoppers is 0.35 × 900 = 315. Giving 42 as the answer reuses the original survey count without scaling it up to 900 shoppers at all. Finding the expected number who prefer PLASTIC bags instead of paper, using the relative frequency 78 ÷ 120 = 0.65, gives 0.65 × 900 = 585. Using 1000 shoppers instead of the 900 actually stated gives 0.35 × 1000 = 350.
- (b) 1/4 — Method: the person picked is known to be aged 30 or over, so the sample space is those 140 people; divide the number of them who had been to the cinema by 140. Working: 35 of the 140 people aged 30 or over had been to the cinema, giving 35/140. Dividing the numerator and the denominator by 35 gives 1/4. Answer: the probability is 1/4. The distractors: 7/20 is 35/100, taking the count from the older group but the total from the under 30s, which is reading across the wrong row; 7/48 is 35/240, dividing by everyone surveyed instead of by the age group named; 3/4 is 105/140, the probability that someone aged 30 or over had NOT been to the cinema, the opposite event inside the correct group.
- (d) 5 — Being red and not being red are exhaustive, so their probabilities sum to 1: the probability of red is 1 − 0.8 = 0.2. The number of red counters is 0.2 × 25 = 5. Using 0.8 directly as the probability of red, without taking the complement, gives 0.8 × 25 = 20 — the number of counters that are NOT red. Sharing the 25 counters equally between the three colours, ignoring the given probability altogether, gives 25 ÷ 3 ≈ 8. Misreading the total as 20 counters instead of 25 gives 0.2 × 20 = 4.
- (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) The relative frequency is settling near 0.5 — Method: turn each result into a relative frequency before comparing them, because it is the relative frequency, and not the difference between the two counts, that tends towards the theoretical probability. Working: after 10 flips the relative frequency of a head is 7 ÷ 10 = 0.7, which is a long way from 0.5. After 1000 flips it is 528 ÷ 1000 = 0.528, which is much closer to 0.5. Meanwhile the gap between the two counts has grown rather than shrunk: it was 7 − 3 = 4 after 10 flips and is 528 − 472 = 56 after 1000 flips. Answer: the relative frequency is settling near 0.5, which is what an unbiased experiment does as the sample grows. The distractors: saying the counts are levelling out is the usual form of this idea and the figures contradict it, since the gap went from 4 to 56; saying the coin is biased treats 28 extra heads in 1000 flips as proof, when 0.528 sits close to 0.5 and a fair coin gives results like this often; saying the next flip is more likely to be a tail is the gambler's fallacy, since each flip stays at 1/2 whatever came before.
- (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%.
- (a) 4/9 — There are 3 × 6 = 18 equally likely outcomes. The product is a multiple of 5 whenever Spinner C shows 5, whatever Spinner D shows, which is 6 outcomes, or whenever Spinner D shows 5 and Spinner C shows 2 or 3, which is 2 more outcomes. This gives 6 + 2 = 8 outcomes, so the probability is 8/18 = 4/9. Choosing 1/3 comes from only counting the 6 outcomes where Spinner C shows 5 and forgetting the 2 outcomes where Spinner D shows 5 instead. Choosing 1/9 comes from only counting the 2 outcomes where Spinner D shows 5 and forgetting the 6 outcomes where Spinner C shows 5. Choosing 1/2 comes from counting 9 outcomes instead of 8, by listing the case where Spinner C shows 5 and Spinner D shows 5 twice over.
- (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.
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