Printable · GCSE Foundation · ages 14-16
Probability worksheet — GCSE Foundation
Fifteen questions across the probability statements at Foundation 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 Foundation
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- (c) 500 — Method: the expected number of successes is the number of trials multiplied by the probability, so to find the number of trials, divide the expected number by the probability. Working: let n be the number of seeds planted. Then n multiplied by 0.6 must come to 300, so n = 300 ÷ 0.6 = 500. Answer: the grower should plant 500 seeds. The distractors: 180 comes from multiplying instead of dividing, 300 × 0.6 = 180, which answers how many of 300 seeds would germinate; 750 comes from dividing by the probability of not germinating, 300 ÷ 0.4 = 750; 120 comes from multiplying by that same 0.4, 300 × 0.4 = 120.
- (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) 12/25 — The group holds 40 of the 250 tickets, so for any one prize the probability the group wins it is 40/250 = 4/25. There are 3 prizes and the group has the same chance at each one, so the expected number won is 3 × 4/25 = 12/25. Writing 4/25 is wrong because it is the chance of winning just ONE prize, without multiplying by the 3 prizes available. Writing 4/75 is wrong because it divides by the 3 prizes instead of multiplying (4/25 ÷ 3 = 4/75), which would mean the group did worse the more prizes were on offer. Writing 64/15625 is wrong because it multiplies the single-prize probability by itself three times, (4/25)³, as though all three prizes had to be won together, instead of adding up the expected number across the three separate prizes. The expected number of prizes won by the group is 12/25.
- (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) 30 — The numbers less than 4 are 1, 2 and 3, so the probability of that event is 3/6, and the expected count in 90 rolls is 90 × 3/6 = 45. The probability of rolling a 6 is 1/6, and the expected count is 90 × 1/6 = 15. The difference between the two expected counts is 45 − 15 = 30. A candidate who answers 45 has given the expected count for 'less than 4' only, forgetting to subtract the other expected count. A candidate who answers 15 has given the expected count for '6' only. A candidate who answers 36 has used a dice with 5 possible numbers instead of 6, giving 90 × 3/5 = 54 and 90 × 1/5 = 18, a difference of 36.
- (d) No, because 3/8 + 5/12 + 1/6 = 23/24 — Using a common denominator of 24: 3/8 = 9/24, 5/12 = 10/24 and 1/6 = 4/24. Adding these numerators gives 9 + 10 + 4 = 23, so the three probabilities sum to 23/24, which is less than 1 — Zara is not correct. Adding the original numerators (3 + 5 + 1 = 9) over a denominator of 12 instead of converting each fraction properly gives 9/12 = 3/4, still less than 1 but the wrong fraction. Converting 1/6 to 5/24 instead of 4/24 (using the wrong scaling) makes the total 9/24 + 10/24 + 5/24 = 24/24 = 1, wrongly suggesting the probabilities are valid. Judging validity from the fact that each individual fraction lies between 0 and 1 ignores that an exhaustive set must sum to exactly 1, not merely contain valid individual values.
- (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.
- (d) 136 — 15% of 160 = 0.15 × 160 = 24 patients reported side effects, so 160 − 24 = 136 did not. Stopping after finding the number who reported side effects, 24, answers the wrong question — it is not the number who did NOT report them. Misreading '15%' as a raw count of 15 patients, rather than a percentage, gives 160 − 15 = 145. Subtracting 15% of 160 twice, 160 − 24 − 24 = 112, double-counts the side-effect group.
- (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) 0.36 — Relative frequency is the number of times the event happened divided by the total number of trials: 18 ÷ 50 = 0.36. Dividing by 100 instead of the actual 50 spins gives 18 ÷ 100 = 0.18. Finding the relative frequency of NOT landing on green, using 50 − 18 = 32 spins, gives 32 ÷ 50 = 0.64. Misplacing the decimal point in the division, so that 18 ÷ 50 is carried out as 18 ÷ 500, gives 0.036 — a tenth of the correct value.
- (d) 12 — There are 120 − 70 = 50 female members. 38 of them play netball, so the rest play football: 50 − 38 = 12. Writing 45 is wrong because that is the number of MALE members who play football, not female. Writing 38 again is wrong because that is the female netball total, not the female football total — the question needs the total minus 38, not 38 itself. Writing 25 is wrong because that comes from the male branch (70 − 45 = 25 male netball players), not the female branch. 12 female members play football.
- (b) 41/160 — In total, 27 + 14 = 41 of the 160 employees cycle to work, so the probability is 41/160 (41 and 160 share no common factor, so this is already in its simplest form). Writing 27/160 is wrong because it only counts the full-time cyclists and leaves out the 14 part-time cyclists. Writing 41/90 is wrong because it uses the full-time total (90) as the denominator instead of the whole survey (160). Writing 1/5 is wrong because it only uses the part-time branch, simplifying 14/70 to 1/5 and ignoring the full-time cyclists completely. The probability is 41/160.
- (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) 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%.
- (c) Red — Theoretical probability is 1/3 ≈ 0.333 for each colour. Red's relative frequency is 38/90 ≈ 0.422, above 1/3, so red is over-represented. Blue's relative frequency is 26/90 ≈ 0.289, below 1/3, so blue is under-represented, not over. Green's relative frequency is also 26/90 ≈ 0.289, below 1/3 for the same reason. Since red's relative frequency clearly exceeds 1/3, it is not true that none of the colours are over-represented.
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