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.
Answer key: Probability worksheet — GCSE Foundation
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- (a) 0.16 — The probability that Kofi fails a single attempt is 1 − 0.6 = 0.4. Since the attempts are independent, the probability he fails both is 0.4 × 0.4 = 0.16. Choosing 0.36 comes from squaring the probability of PASSING instead, 0.6 × 0.6 = 0.36, which is the probability of passing both attempts, not failing both. Choosing 0.4 comes from giving the probability of failing just one attempt, forgetting to combine two attempts. Choosing 0.24 comes from multiplying the fail probability by the pass probability, 0.4 × 0.6 = 0.24, mixing up passing and failing between the two attempts.
- (b) 3 — Method: build the list of possible results systematically, taking the first flip as a head and then as a tail, and keeping the two flips in order so that a head then a tail is a different result from a tail then a head; then count the results that match the description. Working: with a head at the first flip the results are HH and HT, and with a tail at the first flip they are TH and TT, so the list is HH, HT, TH, TT — four results in all. The results containing at least one tail are HT, TH and TT. Answer: 3 of the results in the list contain at least one tail. The distractors: 2 comes from treating a head and a tail as one result however they are ordered, which shortens the list to HH, HT, TT so that only two entries hold a tail; 1 comes from reading 'at least one tail' as a tail at both flips, which is the single result TT; 4 comes from counting the tails written across the list — one in HT, one in TH and two in TT — instead of counting the results that contain a tail.
- (c) 2/15 — Method: for independent events, the probability that both happen is the product of the two probabilities. Working: the calculation is 2/5 × 1/3. Multiplying fractions gives 2 × 1 = 2 on the top and 5 × 3 = 15 on the bottom. Answer: the probability is 2/15. The distractors: 11/15 comes from adding the two probabilities, 6/15 + 5/15, instead of multiplying them; 3/8 comes from adding the numerators and the denominators separately; 1/15 comes from subtracting one probability from the other, 6/15 − 5/15.
- (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.
- (d) 5/18 — Method: list every ordered pair of dice scores whose difference is 1, then divide by the 36 equally likely pairs. Working: the pairs with a difference of 1 are (1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4), (5, 6) and (6, 5), which is 10 pairs out of 36, cancelling down to 5/18. Answer: 5/18. Watch out: writing down 5/36 lists only the 5 pairs going up, (1, 2), (2, 3), (3, 4), (4, 5) and (5, 6), and misses that each one has a matching pair the other way round, such as (2, 1) — the two dice are different objects, and order matters, so each of those 5 gaps counts twice. Writing down 1/6 treats the six possible differences, 0 to 5, as equally likely and picks 1 out of 6 of them, but a difference of 1 is reached by far more pairs of scores than a difference of 5 is, so the six differences are not equally likely. And writing down 1/4 comes from adding the two dice's outcome counts instead of multiplying them, 6 + 6 = 12, and grouping the six scores into just three non-overlapping pairs one apart, {1, 2}, {3, 4} and {5, 6}, giving 3 out of that wrong pool of 12.
- (b) 50 — There are 180 − 100 = 80 south-plot gardeners. 30 of them do not grow organically, so the rest do: 80 − 30 = 50. Writing 64 is wrong because that is the number of NORTH-plot gardeners who grow organically, not south. Writing 30 again is wrong because that is the number of south-plot gardeners who do NOT grow organically — the question asks for those who do. Writing 80 is wrong because that is the whole south-plot total, without subtracting the 30 who do not grow organically. The answer is 50.
- (d) 56 — Method: count the trials over the whole period first, then multiply the number of trials by the probability. Working: 4 weeks is 4 × 7 = 28 days, and at 25 trains a day that is 25 × 28 = 700 trains. The expected number of late trains is 700 × 0.08 = 56. Answer: about 56 late trains over the 4 weeks. The distractors: 2 is the expected number for a single day, 25 × 0.08 = 2, with the 28 days never brought in; 14 uses one week instead of four, 25 × 7 × 0.08 = 14; 644 is 700 − 56 and counts the trains expected to be on time.
- (a) 1/6 — Method: write the results of the two dice as ordered pairs, count the pairs whose scores add to the total asked for, divide by the number of ordered pairs there are, then cancel the fraction down. Working: there are 6 × 6 = 36 equally likely ordered pairs. The pairs whose scores add to 7 are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1), which is 6 pairs, so the probability is 6/36. Dividing the top and the bottom by 6 gives 1/6. Answer: the probability is 1/6. The distractors: 7/36 comes from taking the number of favourable pairs to be 7 because 7 is the total asked for, confusing the size of a total with the number of ways of making it; 1/7 comes from using the 21 different combinations of two scores as the equally likely results, finding the 3 combinations 1 and 6, 2 and 5, 3 and 4, and cancelling 3/21; 6/11 comes from counting the 6 favourable pairs correctly but dividing by the 11 possible totals from 2 to 12 rather than by the 36 pairs.
- (d) 15 — Method: each of Spinner A's outcomes can be paired with each of Spinner B's outcomes, so the two counts are combined by multiplying, not adding. Working: Spinner A has 3 outcomes and Spinner B has 5, so there are 3 × 5 = 15 equally likely outcomes in total. Answer: 15. Watch out: adding the two counts, 3 + 5 = 8, undercounts drastically — every one of Spinner A's 3 outcomes pairs with all 5 of Spinner B's outcomes, not just one each. Writing down 3 counts Spinner A alone and leaves Spinner B out completely. And writing down 25 comes from squaring Spinner B's own outcome count, 5 × 5, and forgetting Spinner A altogether.
- (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.
- (b) 10 — Method: A′ means everything in the universal set that is NOT in A, so n(A′) = n(universal set) − n(A). Working: the universal set has 15 elements. A = {3, 6, 9, 12, 15}, so n(A) = 5. n(A′) = 15 − 5 = 10. Answer: 10. Watch out: writing down 5 gives n(A) itself, the size of the multiples-of-3 set, which is the opposite of its complement. Writing down 11 comes from missing 15 off the list of multiples of 3, treating A as only {3, 6, 9, 12}, so A is undercounted as 4 and A′ is overstated as 15 − 4. And writing down 12 comes from only listing the multiples of 3 up to 9 — 3, 6 and 9 — and missing that 12 and 15 also belong to A, undercounting A as 3 rather than 5.
- (c) 39.5% — Win, draw and lose are mutually exclusive and exhaustive, so their probabilities sum to 100%: 42% + 18.5% = 60.5% is the percentage that wins or draws. 100% − 60.5% = 39.5% is the percentage that neither wins nor draws. Adding 42% and 18.5% and stopping there, 60.5%, is the probability of winning or drawing, not of neither. Subtracting only the 42% from 100% gives 58.0%, ignoring the draw percentage. Subtracting only the 18.5% from 100% gives 81.5%, ignoring the win percentage.
- (b) 40% — 32 out of the 80 people chose Coffee, so the percentage is (32 ÷ 80) × 100 = 40%. Choosing 32% comes from treating 32 as if it were already a percentage out of 100, instead of dividing by the actual total of 80 people surveyed. Choosing 25% comes from using the Tea row instead of the Coffee row, (20 ÷ 80) × 100 = 25%. Choosing 20% comes from using the Juice row instead of the Coffee row, (16 ÷ 80) × 100 = 20%.
- (c) 24 — Silver, bronze and copper cover every counter, so their probabilities sum to 1: P(copper) = 1 − 1/2 − 1/8 = 3/8. Number of copper counters = 3/8 × 64 = 24. Multiplying the silver probability by 64 gives 32, the number of silver counters, not copper. Multiplying the bronze probability by 64 gives 8, the number of bronze counters. Adding the silver and bronze probabilities (1/2 + 1/8 = 5/8) and multiplying by 64 gives 40, the combined number of silver and bronze counters, not the copper count.
- (a) 195 — The relative frequency from the trial is 52 ÷ 80 = 0.65, and the expected number of point-up landings in 300 drops is 0.65 × 300 = 195. Giving 52 as the answer reuses the original count from the 80-drop trial without scaling it up to 300 drops at all. Misreading 52 out of 80 as 52% and finding 52% of 300 gives 156. Finding the expected number of point-DOWN landings instead of point-up, using the relative frequency 28 ÷ 80 = 0.35, gives 0.35 × 300 = 105.
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