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.
Probability worksheet — GCSE Foundation
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- 1.The numbers 1 to 10 are each written on a card. Set A is the set of multiples of 3 from 1 to 10 and set B is the set of even numbers from 1 to 10. Work out how many of the numbers from 1 to 10 are in set A but not in set B.
- 2.A survey asks 40 people whether they own a cat, a dog, both, or neither. 12 people own only a cat, 15 own only a dog, 5 own both, and 8 own neither. Work out the probability that a randomly chosen person owns a cat or a dog (or both). Give your answer as a fraction in its simplest form.
- 3.A = {factors of 12} and B = {factors of 18}. Work out n(A ∩ B), the number of elements in both sets.
- 4.80 people are asked whether they drink tea and whether they drink coffee, and the results are recorded on a frequency tree. The first pair of branches splits the 80 people into 50 tea drinkers and 30 non-tea-drinkers. Of the 50 tea drinkers, 32 also drink coffee. Of the 30 non-tea-drinkers, 6 drink coffee. Work out n(T ∩ C), the number of people who drink both tea and coffee.
- 5.A fair six-sided dice is rolled 300 times. Work out how many more times you would expect it to land on an even number than on a six.
- 6.A factory checked 400 items. A frequency tree splits them into 250 items made by machine A and 150 items made by machine B. On machine A's branch, 15 of the items were faulty. On machine B's branch, 5 of the items were faulty. One of the 400 items is picked at random. Work out the probability that it is faulty. Give your answer as a fraction in its simplest form.
- 7.A seed company tests germination using results from three greenhouses. Greenhouse 1 plants 200 seeds and 172 germinate. Greenhouse 2 plants 150 seeds and 126 germinate. Greenhouse 3 plants 250 seeds and 212 germinate. Using the combined results from all three greenhouses, work out the best estimate of the number of seeds, out of a new batch of 4000 seeds, that would be expected to germinate.
- 8.A fair coin is flipped and a fair three-sided spinner, labelled X, Y and Z, is spun. Every outcome is listed as a pair, such as (H, X). Work out the probability that the outcome is a head and Z.
- 9.Spinner A has 4 equal sections, numbered 1, 2, 3 and 4. Spinner B has 3 equal sections, coloured red, red and blue. Both spinners are spun once. Work out the probability of getting an even number on Spinner A and red on Spinner B.
- 10.Four pupils each flip the same coin a number of times and record the results. Sam flips it 40 times and gets 24 heads. Priti flips it 25 times and gets 10 heads. Leo flips it 20 times and gets 9 heads. Fatima flips it 15 times and gets 6 heads. Combine all four pupils' results to estimate the probability that the coin lands on heads.
- 11.A train company runs 25 trains a day, every day. The probability that any one train is late is 0.08. Work out how many late trains the company should expect over a period of 4 weeks.
- 12.At a school fete, each raffle ticket wins a prize, wins a voucher or wins nothing, and no ticket can win more than one of these. The probability that a ticket wins a prize is 0.4 and the probability that it wins a voucher is 0.35. Work out the probability that a ticket wins a prize or a voucher.
- 13.A survey found that, of 250 shoppers questioned, 68% said they had used a self-checkout in the past month. A different store expects 1400 shoppers this week. Using this relative frequency, work out how many of the 1400 shoppers would be expected to have used a self-checkout in the past month.
- 14.Ellie drops a bottle top and records whether it lands open end up. In her first 20 drops it lands open end up 13 times. She carries on, and after 200 drops in total it has landed open end up 84 times. Work out the best estimate of the probability that the bottle top lands open end up.
- 15.At a fun run, each runner finishes, retires or is disqualified, and cannot do more than one of these. The probability that a runner finishes is 68.5% and the probability that a runner retires is 24.75%. Work out the probability, as a percentage, that a runner is disqualified.
Answer key
- (c) 2 — Set A, the multiples of 3, is {3, 6, 9}. Removing the numbers that are also in set B, the even numbers {2, 4, 6, 8, 10}, leaves {3, 9} — 2 numbers are in A but not B. Giving 3, the whole size of set A, forgets to remove the number 6, which is also even. Giving 7 counts every number in A or B combined, {2, 3, 4, 6, 8, 9, 10}, rather than only those in A but not B. Giving 4 counts the numbers in set B but not set A, {2, 4, 8, 10}, the wrong way round.
- (c) 4/5 — Add the counts of everyone who owns a cat, a dog, or both: 12 + 15 + 5 = 32 out of 40 people, which simplifies to 4/5. Leaving out the 5 people who own both, and adding only the two only-groups, gives 27/40. Using the 8 people who own neither, instead of everyone who owns at least one pet, gives 8/40 = 1/5. Counting the 5 people who own both twice, once alongside each only-group as well as on their own, gives 12 + 15 + 5 + 5 = 37 out of 40, or 37/40.
- (d) 4 — Method: list the elements of each set in full, then find which elements appear in both lists — that is A ∩ B. Working: factors of 12 = {1, 2, 3, 4, 6, 12}. Factors of 18 = {1, 2, 3, 6, 9, 18}. The elements in both lists are 1, 2, 3 and 6, so A ∩ B = {1, 2, 3, 6} and n(A ∩ B) = 4. Answer: 4. Watch out: writing down 6 gives n(A), the size of the factors-of-12 list on its own, not the size of the overlap. Writing down 8 comes from counting every element that appears in EITHER list, 1, 2, 3, 4, 6, 9, 12 and 18 — that is the union, a different set from the intersection. And writing down 3 misses that 1 is a factor of both 12 and 18, and so belongs in A ∩ B alongside 2, 3 and 6.
- (d) 32 — The frequency tree already shows the tea-and-coffee branch directly: of the 50 tea drinkers, 32 also drink coffee, so n(T ∩ C) = 32. Adding both coffee branches together, 32 + 6 = 38, gives n(C), the total number of coffee drinkers, not just those who also drink tea. Using the non-tea branch's figure, 6, describes people who drink coffee but NOT tea. Subtracting to get 50 − 32 = 18 finds the tea drinkers who do NOT drink coffee, the opposite region to the one asked for.
- (c) 100 — There are 3 even numbers on a fair dice (2, 4 and 6), so the probability of landing on an even number is 3/6 = 1/2, and 300 × 1/2 = 150. The probability of landing on a six is 1/6, so 300 × 1/6 = 50. The dice is expected to land on an even number 150 − 50 = 100 more times than on a six. Writing 50 is wrong because that is just the expected number of sixes on its own, without comparing it to the expected number of evens. Writing 150 is wrong because that is just the expected number of evens on its own, without subtracting the sixes. Writing 200 is wrong because it adds the two expected frequencies together (150 + 50 = 200) instead of finding the difference between them. The dice is expected to land on an even number 100 more times than on a six.
- (d) 1/20 — Method: add the counts on the faulty end branches, divide by the total number of items in the experiment, then cancel. Working: the faulty items number 15 + 5 = 20, and 400 items were checked, so the probability is 20/400. Dividing the top and the bottom by 20 gives 1/20. Answer: the probability is 1/20. The distractors: 3/50 is 15/250 and comes from dividing machine A's faults by machine A's output, which is that machine's own fault rate rather than the probability for the whole batch; 1/30 is 5/150 and does the same on machine B's branch; 19/20 is 380/400 and gives the probability that the item picked is not faulty.
- (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.
- (b) 1/6 — There are 2 outcomes for the coin and 3 outcomes for the spinner, giving 2 × 3 = 6 equally likely pairs: (H,X), (H,Y), (H,Z), (T,X), (T,Y), (T,Z). Only one of these, (H,Z), matches both conditions, so the probability is 1/6. A candidate who answers 1/3 has listed only the spinner's 3 outcomes and ignored the coin. A candidate who answers 1/2 has considered only the coin and ignored the spinner. A candidate who answers 1/5 has missed one pair when listing the possibility space, treating it as 5 outcomes instead of 6.
- (a) 1/3 — Method: for two independent spinners, multiply the probability of each separate outcome, but first work out each spinner's own probability correctly, using how many of its equal sections actually carry that result. Working: Spinner A has 2 even numbers, 2 and 4, out of 4 sections, so P(even) = 2/4 = 1/2. Spinner B has 2 red sections out of 3, so P(red) = 2/3. Multiplying gives 1/2 × 2/3, which cancels down to 1/3. Answer: 1/3. Watch out: writing down 1/4 treats Spinner B's two colours as equally likely and uses 1/2 for red, when in fact 2 of its 3 sections are red — the sections are not split evenly between the two colours. Writing down 1/6 undercounts Spinner A's even numbers as just one out of four instead of two. And writing down 5/6 applies the 'at least one' formula, P(A) + P(B) − P(A)×P(B), which answers a different question about EITHER spinner landing the right way, not both together.
- (b) 0.49 — Adding all four pupils' flips gives 40 + 25 + 20 + 15 = 100 flips in total, and adding their heads gives 24 + 10 + 9 + 6 = 49 heads in total, so the combined estimate is 49/100 = 0.49. Writing 0.46 is wrong because it averages the four pupils' individual rates (0.60, 0.40, 0.45 and 0.40) as if they all came from the same number of flips, which they do not — this ignores that Sam and Priti flipped far more times than Leo and Fatima. Writing 0.60 is wrong because it only uses Sam's own result (24/40 = 0.60), ignoring the other three pupils. Writing 0.40 is wrong because it only combines Priti and Fatima's results (16/40 = 0.40), leaving out Sam and Leo entirely. The combined estimate from all four pupils is 0.49.
- (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) 0.75 — Method: a ticket cannot win both a prize and a voucher, so the two events are mutually exclusive and their probabilities are added. Working: 0.4 + 0.35, lining the decimal points up, gives 0.40 + 0.35. Answer: 0.75. The distractors: 0.25 comes from going one step too far and giving the probability that a ticket wins nothing, which is what is left when the total is taken from 1; 0.14 comes from multiplying 0.4 by 0.35 instead of adding them; 0.05 comes from subtracting 0.35 from 0.4 instead of adding them.
- (a) 952 — 68% = 0.68. The relative frequency from the survey applies to the new group of 1400 shoppers, so the expected number is 0.68 × 1400 = 952. Working out 1 − 0.68 = 0.32 and applying that instead, 0.32 × 1400 = 448, finds the number who have NOT used a self-checkout, not the number who have. Applying 68% to the original sample size of 250 instead of the new total of 1400 gives 0.68 × 250 = 170. Applying the complement percentage to the original sample size, 0.32 × 250 = 80, compounds both mistakes.
- (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.
- (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%.
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