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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Probability worksheet — GCSE Foundation
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- 1.A fair coin is flipped three times. Work out the probability of getting at least one head.
- 2.150 pupils are asked about swimming and cycling, and the results are recorded on a frequency tree. The first pair of branches splits the 150 pupils into 90 who swim and 60 who do not swim. Of the 90 who swim, 54 also cycle. Of the 60 who do not swim, 18 cycle. Work out P(S ∩ C), the probability that a pupil, chosen at random, swims and cycles. Give your answer as a fraction in its simplest form.
- 3.Priya and Ben each want to estimate the probability that a coin lands on heads. Priya flips the coin 50 times and works out her relative frequency. Ben flips the same coin 500 times and works out his relative frequency. Whose relative frequency is more likely to be close to the true probability? Give a reason for your answer.
- 4.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.
- 5.In a game, Zara says the probability of scoring is 3/8, the probability of missing is 5/12 and the probability of a rebound is 1/6, and that these are the only three outcomes. Is Zara correct that her three probabilities are valid?
- 6.A survey of 60 readers records: 21 read only fiction, 17 read only non-fiction, 14 read both, and the rest read neither. Work out the probability that a randomly chosen reader reads exactly one of the two types, giving your answer as a fraction in its simplest form.
- 7.A school trip took 200 pupils to a museum, travelling either by coach or by minibus. A frequency tree records the trip: 150 pupils travelled by coach, split into 90 in Year 8 and the rest in Year 9. All the pupils who travelled by minibus were in Year 8. Work out the total number of Year 8 pupils on the trip.
- 8.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.
- 9.A card is taken at random from an ordinary pack of 52 playing cards. The probability that the card is a diamond is 1/4. Work out the probability that the card is not a diamond.
- 10.A leisure centre has 150 members. 80 of the members are male and the rest are female. Every member uses either the pool or the gym, but not both. 66 members use the pool, and 35 of those pool users are male. Work out how many female members use the gym.
- 11.An ordinary fair dice is rolled once. Set A is the set of even scores and set B is the set of scores greater than 4. Work out the probability that the score rolled is in set A or in set B or in both.
- 12.A fair spinner is divided into 5 equal sections, 2 labelled win and 3 labelled lose. Zara spins it twice, and the two spins are independent. Work out the probability that she wins on the first spin and loses on the second spin.
- 13.A biased spinner is spun 200 times. It lands on red 70 times, on blue 50 times, and on green 80 times. Using these results, work out the expected number of times the spinner does NOT land on red, in 500 spins of the same spinner.
- 14.A biased six-sided dice is rolled once. The probability that it lands on 6 is 0.25. The other five scores are all equally likely. Work out the probability that it lands on 3.
- 15.A fair coin is flipped and an ordinary fair dice is rolled. Every possible result is listed in a table. Work out how many different results the table contains altogether.
Answer key
- (c) 7/8 — Method: 'at least one head' is the opposite of 'no heads at all', so work out the probability of three tails and take it away from 1. Working: a flip that is not a head has probability 1/2, and the flips are independent, so three tails in a row has probability 1/2 × 1/2 × 1/2 = 1/8. Taking this from 8/8 leaves 7/8. Answer: the probability is 7/8. The distractors: 1/8 is the probability of three tails, written down without the final subtraction; 3/8 is the probability of exactly one head, which comes from reading 'at least one' as 'exactly one'; 1/2 comes from giving the probability of a head on a single flip and ignoring that three flips are made.
- (c) 9/25 — The frequency tree already shows the swim-and-cycle branch directly: 54 out of 150, which simplifies to 9/25. Adding both cycling branches together, 54 + 18 = 72, gives the total number of cyclists, so 72/150 = 12/25, not just those who also swim. Using the non-swimmers' cycling figure, 18, gives 18/150 = 3/25, the probability of cycling WITHOUT swimming. Using the swimmers' total of 90, before splitting by cycling, gives 90/150 = 3/5, the probability of swimming on its own.
- (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) 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.
- (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) 19/30 — Exactly one means only fiction or only non-fiction, not both: 21 + 17 = 38 out of the 60 readers, which simplifies to 19/30. Including the 14 who read both as well gives 21 + 17 + 14 = 52, so 52/60 = 13/15 — that is at least one, not exactly one. Using only the both-count, 14, as the numerator gives 14/60 = 7/30, the probability of reading both, not exactly one. Using 52, the number who read at least one type, as the denominator instead of the full 60 readers surveyed gives 38/52 = 19/26.
- (d) 140 — 90 pupils were in Year 8 on the coach branch. The minibus branch has 200 − 150 = 50 pupils, and all of them are Year 8 too, so the total number of Year 8 pupils is 90 + 50 = 140. Writing 90 alone is wrong because it only counts the coach's Year 8 pupils and misses the minibus ones. Writing 60 is wrong because that is the number of Year 9 pupils on the coach (150 − 90 = 60), not Year 8 at all. Writing 50 alone is wrong because it only counts the minibus pupils and misses the coach's Year 8 pupils. The total is 140.
- (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.
- (b) 3/4 — Method: a card either is a diamond or is not a diamond, so those two outcomes form an exhaustive set and their probabilities add to 1; subtract the given probability from 1. Working: P(diamond) = 1/4, so P(not a diamond) = 1 − 1/4; writing 1 as 4/4 gives 4/4 − 1/4. Answer: 3/4. The distractors: 1/4 comes from giving back the probability that the card is a diamond instead of its complement; 1/2 comes from reading 'not a diamond' as 'not a red card' and halving the pack; 3/52 comes from doing the subtraction 4 − 1 = 3 on the suits but then writing that 3 over the 52 cards in the pack instead of over the 4 suits.
- (b) 39 — Method: put the counts into a two-way table and fill each missing cell by subtracting along a row or down a column. Working: the number of female members is 150 − 80 = 70. The pool column holds 66 members and 35 of them are male, so the number of female pool users is 66 − 35 = 31. Subtracting along the female row, 70 − 31 = 39 female members use the gym. Answer: 39 female members use the gym. The distractors: 45 comes from subtracting along the male row instead, 80 − 35 = 45, which counts male gym users; 31 is the female pool cell, written down one step before the gym cell; 84 is 150 − 66 and counts every gym user, male and female together.
- (a) 2/3 — Method: 'in A or in B' means every score that belongs to at least one of the two sets; a score that belongs to both is still only one outcome, so it is listed once. Working: the even scores are 2, 4 and 6; the scores greater than 4 are 5 and 6. Listing the scores that appear in either set gives 2, 4, 5 and 6, with 6 written once. That is 4 of the 6 faces, or 4/6. Answer: the probability is 2/3. The distractors: 5/6 comes from adding the sizes of the two sets, 3 + 2, so that the score 6 is counted in both and appears twice; 1/2 comes from using the even scores alone; 1/6 comes from giving the probability that the score is in both sets, which is the single score 6, rather than in either of them.
- (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.
- (c) 325 — Method: first find the relative frequency of NOT landing on red from the 200 spins, then scale that up to 500 spins. Working: non-red results = 50 + 80 = 130, out of 200 spins, so P(not red) = 130 ÷ 200 = 0.65. Expected non-red results in 500 spins = 500 × 0.65 = 325. Answer: 325. Watch out: writing down 175 finds the expected number of RED results instead, 70 ÷ 200 × 500 = 175, answering the opposite of what was asked. Writing down 250 assumes landing red or not landing red must be a fair 50-50 split, but the spinner is biased and the actual results do not split evenly. And writing down 130 stops after finding how many of the 200 spins were non-red and forgets to scale that figure up to the 500 spins asked for.
- (b) 0.150 — The six scores are exhaustive, so their probabilities sum to 1. The probability of not landing on 6 is 1 − 0.25 = 0.75, and this is shared equally between the other five scores, so each has probability 0.75 ÷ 5 = 0.150. Giving 0.750 as the answer stops after finding the probability of not landing on 6, without sharing it out between the five remaining scores. Dividing 0.75 by 6 instead of by 5 gives 0.125, wrongly including the score of 6 among the equally likely scores. Ignoring the bias completely and dividing 1 by all six scores gives 1 ÷ 6 = 0.167.
- (d) 12 — Method: each result of the first experiment can be paired with every result of the second, so the two counts are combined by multiplying. Working: the coin lands in 2 ways and the dice lands in 6 ways. Each of the 6 dice scores can appear with a head or with a tail, so the table has 6 × 2 rows. Answer: there are 12 different results. The distractors: 8 comes from adding the two counts, 6 + 2, instead of multiplying them; 6 comes from listing the dice scores only and treating the coin as making no difference to the table; 36 comes from working out 6 × 6, counting the coin as though it too had six equally likely results.
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