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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- (d) 3/5 — The first tile drawn is a consonant with probability 4/5, since 4 of the 5 letters in MATHS are consonants. Since it is not replaced, 3 consonants remain out of the 4 tiles left, so the second tile is a consonant with probability 3/4. Multiplying gives 4/5 × 3/4 = 3/5. Choosing 4/5 comes from only using the probability for the first draw and forgetting the second draw changes once a tile is removed. Choosing 3/4 comes from only using the probability for the second draw on its own. Choosing 16/25 comes from treating the two draws as independent, using 4/5 × 4/5 = 16/25, which would only be correct if the first tile were replaced before the second draw.
- (c) 0.5 — Method: count the favourable outcomes, write them over the total number of equally likely outcomes and then divide to turn the fraction into a decimal. Working: the odd scores are 1, 3 and 5, which is 3 of the 6 equally likely scores, so the probability is 3/6, and 3 ÷ 6 = 0.5. Answer: 0.5, the middle of the 0 to 1 probability scale. The distractors: 0.33 comes from listing only 3 and 5 as odd and working out 2 ÷ 6; 0.17 comes from giving the probability of one particular odd score, 1 ÷ 6; 0.3 comes from writing '3 out of 6' as 0.3, reading the 3 straight off as tenths instead of dividing.
- (a) 43/70 — There are 140 − 85 = 55 Zone B households, and 55 − 21 = 34 of them recycle glass. In total, 52 + 34 = 86 households recycle glass, out of 140: 86/140 = 43/70. Writing 13/35 is wrong because 52/140 simplifies to 13/35, and 52 only counts Zone A, leaving out the 34 Zone B recyclers. Writing 17/70 is wrong because 34/140 simplifies to 17/70, and 34 only counts Zone B, leaving out the 52 Zone A recyclers. Writing 73/140 is wrong because it adds the 52 Zone A recyclers to the 21 Zone B households that do NOT recycle, mixing up two different groups instead of adding the two recycling groups. The probability is 43/70.
- (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.
- (b) 1/20 — Method: two draws with nothing put back are combined by multiplying, with the second probability worked out from the tickets still in the bag. Working: the first draw takes the wanted ticket with probability 1/5. That ticket is kept out, so 4 tickets remain and only one of them is the ticket wanted second, giving 1/4. Multiplying gives 1/20. Answer: the probability is 1/20. The distractors: 1/10 comes from ignoring the order and treating the draw as a choice of two tickets from five, of which there are ten; 1/25 comes from keeping the total at 5 for the second draw, which is what happens only if the first ticket is put back; 2/5 comes from counting the two wanted tickets over the five in the bag, as though one draw decided the whole question.
- (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.
- (d) 0.420 — Method: use the relative frequency worked out from the larger number of trials as the best estimate of the probability, since a bigger sample tends to sit closer to the true probability. Working: 84 out of 200 drops land open end up, so the relative frequency from the total is 84 ÷ 200 = 0.420. Answer: 0.420. Watch out: writing down 0.650 comes from 13 ÷ 20, using only the first, much smaller sample instead of the total. Writing down 0.535 comes from averaging 0.650 and 0.420, treating the 20-drop run and the 200-drop run as equally reliable instead of using the larger sample on its own. And writing down 0.580 comes from 1 − 0.420, working out the probability that the bottle top lands the other way up instead of open end up.
- (b) 0.225 — The relative frequency of rain is the number of rainy days out of all days recorded: 9 ÷ 40 = 0.225, which is noticeably less than the forecaster's claimed 0.3. Using the number of dry days, 40 − 9 = 31, as the denominator instead of the total of 40 gives 9 ÷ 31 = 0.29 (2 d.p.). Simply reporting the forecaster's claimed value, 0.3, without calculating anything from the data at all, ignores the recorded results completely. Misplacing the decimal point, treating 9 out of 40 as 9%, gives 0.09 instead of 0.225.
- (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.
- (b) 6/25 — The probability of winning on any one spin is 2/5, and the probability of losing on any one spin is 3/5. Since the spins are independent, multiply the probability of winning on the first spin by the probability of losing on the second spin: 2/5 × 3/5 = 6/25. A candidate who answers 4/25 has used the winning probability for both spins, 2/5 × 2/5. A candidate who answers 9/25 has used the losing probability for both spins, 3/5 × 3/5. A candidate who answers 3/10 has treated the spins as if they were dependent, reducing the second spin's denominator to 4.
- (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.
- (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) Priya — The larger the number of trials, the closer a relative frequency tends to be to the true probability. Priya made 200 drops, more than Freya's 20, Malik's 50 or Tom's 80, so her relative frequency gives the best estimate. Freya's estimate is based on only 20 drops, the smallest sample, so it is the least reliable of the four. Malik's 50 drops and Tom's 80 drops are both larger than Freya's but still well short of Priya's 200.
- (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.
- (d) 1/8 — The multiples of 3 from 1 to 8 are 3 and 6, so the probability of that event is 2/8, which simplifies to 1/4. The probability of the coin landing on heads is 1/2. Since the spin and the toss are independent, multiply the two probabilities: 1/4 × 1/2 = 1/8. A candidate who answers 1/4 has considered only the spinner and forgotten to combine it with the coin toss. A candidate who answers 1/2 has considered only the coin and forgotten the spinner condition entirely. A candidate who answers 1/16 has counted only one number, 6, as a multiple of 3 instead of two, giving 1/8 × 1/2.
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