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) 47 — Method: fill the first pair of branches of the frequency tree, then the failures on each branch, then add only the failing end branches. Working: 70% of 200 is 140, so 140 cars were more than 3 years old and 200 − 140 = 60 cars were 3 years old or less. One quarter of the older cars failed: 140 ÷ 4 = 35. The newer branch gives 12 failures. Adding the two failing branches gives 35 + 12 = 47. Answer: 47 of the cars failed the test. The distractors: 35 is the older branch on its own, with the 12 newer failures never added; 62 comes from taking 1 in 4 of all the cars, 200 ÷ 4 = 50, and then adding the 12; 153 is 200 − 47 and counts the cars that passed.
- (a) 12/25 — P(red then blue) = 4/10 × 6/10 = 24/100. P(blue then red) is the same, 6/10 × 4/10 = 24/100. Since either order counts as different colours, add them: 24/100 + 24/100 = 48/100 = 12/25. Working out only one order, red-then-blue, and forgetting blue-then-red, gives 24/100 = 6/25. Working out the probability that the two counters are the SAME colour instead — 4/10 × 4/10 + 6/10 × 6/10 = 16/100 + 36/100 = 52/100 = 13/25 — answers a different question. Using 6/9 for the second draw, as if the first counter were not replaced, gives 2 × (4/10 × 6/9) = 48/90 = 8/15.
- (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.
- (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.
- (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.
- (c) 10 — There are 5 possible cards and 2 possible coin results, so listing every pair gives 5 × 2 = 10 equally likely outcomes. Choosing 5 comes from listing only the card outcomes and forgetting the coin flip adds a second stage to each one. Choosing 7 comes from adding the two stages instead of combining them, 5 + 2 = 7, rather than pairing every card with every coin result. Choosing 20 comes from counting each coin result twice for every card, 5 × 2 × 2 = 20, effectively pairing every card with the coin twice over.
- (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%.
- (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) 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.
- (a) Ben, because a larger sample is closer to the theory — Method: a relative frequency is an estimate of a probability, and for an unbiased experiment that estimate tends towards the theoretical value as the sample grows. Working: Priya's estimate rests on 50 results, so a few unexpected heads move it a long way; one extra head shifts her relative frequency by 1 ÷ 50 = 0.02. Ben's estimate rests on 500 results, where one extra head shifts his relative frequency by only 1 ÷ 500 = 0.002. The larger sample therefore swings far less around the true value. Answer: Ben's relative frequency is the one more likely to be close, because a larger unbiased sample tends closer to the theoretical probability. The distractors: saying a small sample is less affected by luck reverses the result, since it is the small sample that swings most; saying every flip is a separate random event is true of the flips themselves but says nothing about the estimates, and is often used to argue wrongly that the number of trials does not matter; saying 500 flips must give exactly 250 heads confuses an expected value with a guaranteed one, and 500 flips very rarely give exactly 250 heads.
- (c) 0.6 — Method: no two of these ways of travelling can happen on the same day, so they are mutually exclusive and the probability that one of them happens is the sum of their probabilities; they need not add to 1, because travelling by car is a fourth way. Working: 0.2 + 0.3 = 0.5, and 0.5 + 0.1 = 0.6, the decimal points lined up at each step. Answer: 0.6, which also shows that the probability of being taken by car is 0.4, since all four ways together must make 1. The distractors: 0.4 comes from carrying on past the question and giving the probability of the remaining way, by car; 0.5 comes from adding only the bus and the bike and leaving the smallest of the three probabilities out of the total; 0.006 comes from multiplying the three probabilities together instead of adding them.
- (a) 90% — Method: turn the fraction into a decimal by dividing the numerator by the denominator, then multiply the decimal by 100 to turn it into a percentage. Working: 9 ÷ 10 = 0.9, and 0.9 × 100 = 90. Answer: 90%, which is the same probability written in a third form, and all three forms sit at the same point on the 0 to 1 scale. The distractors: 9% comes from reading the numerator of the fraction straight off as the percentage; 10% comes from reading the denominator straight off as the percentage; 0.9% comes from correctly finding the decimal 0.9 and then writing a percentage sign after it without multiplying by 100.
- (a) unlikely — 3/8 = 3 ÷ 8 = 0.375, which is between 0 and 0.5, so the probability is best described as unlikely. Rounding 3/8 loosely to 'about a half' without converting it properly leads to evens, which is wrong since 0.375 is noticeably below 0.5. Reading 3/8 as the complement and using 1 − 3/8 = 5/8 = 0.625 instead gives likely, but the question asks about the probability of 3/8 itself, not its complement. Flipping the fraction to 8/3 ≈ 2.67, which is impossible for a probability, and rounding it towards the top of the scale gives certain.
- (c) 170 — 68% of the 250 packets germinated: 250 × 0.68 = 170. Using the complement, the 32% that did NOT germinate, gives 250 × 0.32 = 80. Shifting the decimal point, using 0.068 instead of 0.68, gives 250 × 0.068 = 17. Rounding 68% up to 70% before multiplying gives 250 × 0.70 = 175.
- (a) 60 — Method: because the counter goes back each time, every draw is the same experiment, so the expected number of reds is the number of draws multiplied by P(red). Working: there are 8 red counters out of 20, so P(red) = 8 ÷ 20 = 0.4. Over 150 draws the expected number of reds is 150 × 0.4 = 60. Answer: about 60 red counters would be expected. The distractors: 90 uses P(blue) by mistake, 150 × 12 ÷ 20 = 90; 100 comes from writing the probability from the red to blue ratio of 8 to 12, giving 150 × 8 ÷ 12 = 100; 75 comes from treating red and blue as equally likely because there are two colours, which gives 150 ÷ 2 = 75.
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