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.A spinner has equal sections, and the probability that it lands on blue in a single spin is 1/4. It is spun twice, and the two spins are independent. Work out the probability that it lands on blue on both spins.
- 2.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.
- 3.A spinner is divided into sectors of 180°, 120° and 60°, coloured red, blue and green in that order. The spinner is spun 60 times. Work out how many times you would expect it to land on green.
- 4.A cinema sold 150 tickets for a Saturday screening. A frequency tree records the sales: 90 of the tickets were bought by adults and the rest by children. Of the adult tickets, 54 were for the 3D showing and the rest were for the standard showing. Work out how many adult tickets were for the standard showing.
- 5.A garden centre recorded 240 plant sales using a frequency tree. The first stage splits the sales into three types: shrubs, bedding plants and trees. 96 sales were shrubs, 114 were bedding plants and the rest were trees. Work out the probability that a sale chosen at random from these 240 was a tree.
- 6.A bag contains 5 red counters and 3 blue counters. A counter is taken at random, its colour is noted, and it is put back in the bag before a second counter is taken at random. Work out the probability that both counters are the same colour.
- 7.A fair coin is flipped and a fair five-sided spinner, labelled 1 to 5, is spun. All the outcomes are listed as pairs, such as (H, 3). Work out the probability that the outcome is a tail and an odd number.
- 8.Two ordinary fair dice are rolled. Work out the probability that both dice show the same score.
- 9.At a coffee shop, each customer buys tea, coffee, water or juice, and never more than one of these. The probability that a customer buys tea is 0.36 and the probability that they buy juice is 0.04. The probability that a customer buys coffee is three times the probability that they buy water. Work out the probability that a customer buys water.
- 10.The universal set is {1, 2, 3, ..., 30}. Set A is the set of multiples of 4 between 1 and 30. Work out n(A′), the number of elements not in A.
- 11.Two fair four-sided dice, numbered 1 to 4, are rolled and the two scores are added together. Work out the probability that the total is 5.
- 12.A survey of 160 employees at a company recorded whether they cycle to work, using a frequency tree. The first branch splits them into 90 who work full-time and 70 who work part-time. Of the full-time employees, 27 cycle to work. Of the part-time employees, 14 cycle to work. Work out the probability that an employee, chosen at random from the 160, cycles to work. Give your answer as a fraction in its simplest form.
- 13.The probability that a component is faulty is 1/10. Two components are tested independently. Work out the probability that at least one of the two components is faulty.
- 14.Two ordinary fair dice are rolled and the two scores are added together. Work out the probability that the total is a prime number.
- 15.One card is taken at random from an ordinary pack of 52 playing cards. 26 of the cards are red. Work out the probability that the card is red. Give your answer as a percentage.
Answer key
- (a) 1/16 — The two spins are independent, so multiply the probability of blue on each spin: 1/4 × 1/4 = 1/16. Choosing 1/2 comes from adding the two probabilities instead of multiplying, 1/4 + 1/4 = 1/2. Choosing 1/8 comes from wrongly treating the second spin as having a 1/2 chance of blue instead of the given 1/4, giving 1/4 × 1/2 = 1/8. Choosing 1/4 comes from giving the probability for a single spin and forgetting that the spinner is spun twice.
- (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.
- (b) 10 — The probability that the spinner lands on green is its angle out of the whole circle: 60° ÷ 360° = 1/6. Expected number of times on green = 1/6 × 60 = 10. Assuming the three colours are equally likely because there are three sectors, regardless of their different angles, gives 60 ÷ 3 = 20. Using red's angle of 180° instead of green's 60° gives a probability of 1/2, so 60 × 1/2 = 30. Treating green's angle, 60°, as a percentage instead of finding its fraction of 360° gives 60 × 0.60 = 36.
- (c) 36 — On the adult branch there are 90 tickets in total, and 54 of them are for the 3D showing, so the standard-showing branch is 90 − 54 = 36. Subtracting 54 from the overall total of 150 gives 96, but 150 is the total for ALL tickets, not just the adult branch, so 96 is wrong. Writing 60 is wrong because that is the number of CHILD tickets (150 − 90 = 60), not adult standard tickets. Writing 54 again is wrong because that is the number of adult 3D tickets, not the number of adult standard tickets. The adult standard-showing branch is 36.
- (b) 1/8 — Shrubs, bedding plants and trees are the three branches at the first stage of the tree, so they must total 240: tree sales = 240 − 96 − 114 = 30. So P(tree) = 30/240 = 1/8. Using the shrub count instead, 96/240 = 2/5, is the probability of a shrub sale, not a tree sale. Using the bedding-plant count instead, 114/240 = 19/40, is the probability of a bedding-plant sale. Subtracting the bedding count from the shrub count (114 − 96 = 18) instead of subtracting both from 240 gives 18/240 = 3/40, which is not the number of tree sales at all.
- (a) 17/32 — Because the counter is replaced, each pick is independent with P(red) = 5/8 and P(blue) = 3/8 every time. Both red has probability 5/8 × 5/8 = 25/64, and both blue has probability 3/8 × 3/8 = 9/64. 'Same colour' means either of these, so add them: 25/64 + 9/64 = 34/64 = 17/32. Giving 25/64 finds only the probability of both counters being red, leaving out both counters being blue, which also counts as the same colour. Giving 15/64 finds the probability of one red and one blue counter in just one of the two possible orders — the opposite of what was asked, and only half of it. Giving 13/28 works out the probabilities as if the counter had NOT been replaced, using 5/8 × 4/7 and 3/8 × 2/7, even though the question states it was put back.
- (d) 3/10 — Method: list the full possibility space as pairs of coin and spinner results, then count how many pairs satisfy both conditions and divide by the size of the whole space. Working: the coin gives 2 outcomes and the spinner gives 5, so the full space has 2 × 5 = 10 equally likely pairs. The pairs with a tail and an odd number are (T,1), (T,3) and (T,5), which is 3 out of 10. Answer: 3/10. Watch out: writing down 1/2 uses only the coin's own chance of a tail and ignores that the spinner also has to land on an odd number. Writing down 3/5 uses only the spinner's chance of landing on an odd number and ignores the coin altogether. And writing down 1/10 counts just one matching outcome, such as (T,1), instead of all three pairs that satisfy both conditions.
- (b) 1/6 — Method: list the ordered pairs where the two scores match, and divide by the 36 equally likely pairs. Working: the matching pairs are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6), which is 6 pairs out of 36, cancelling down to 1/6. Answer: 1/6. Watch out: writing down 1/36 finds the probability of one particular double, such as (6, 6), rather than any double at all. Guessing 1/2 treats 'same' and 'different' as equally likely events, when there are only 6 matching pairs against 30 non-matching ones. And writing down 1/3 comes from listing each double twice, once for each order of the two dice, giving 12 pairs out of 36 — but (1, 1) is a single outcome, and swapping the two dice over does not produce a second one.
- (a) 0.15 — Let P(water) = x, so P(coffee) = 3x. The four outcomes are exhaustive: 0.36 + 0.04 + x + 3x = 1, so 0.4 + 4x = 1, giving 4x = 0.6 and x = 0.15. So P(water) = 0.15. Reporting 3x, the coffee probability, instead of water gives 0.45. Splitting the remaining 0.6 evenly between coffee and water, ignoring the 3:1 ratio, gives 0.30. Stopping once the remaining probability 0.6 is found, without dividing by the four equal shares, gives 0.60.
- (c) 23 — Multiples of 4 from 1 to 30 are 4, 8, 12, 16, 20, 24 and 28 — 7 numbers, so n(A) = 7. The universal set has 30 elements, so n(A′) = 30 − 7 = 23. Reporting n(A) itself, 7, without subtracting it from the universal set forgets what the complement means. Estimating the count of multiples of 4 as 30 ÷ 4 = 7.5, rounded to 8, instead of listing them exactly, gives 30 − 8 = 22. Miscounting the universal set as having 31 elements instead of 30, an off-by-one slip, gives 31 − 7 = 24.
- (b) 1/4 — There are 4 × 4 = 16 equally likely ordered outcomes for the two dice. The pairs that total 5 are: first dice 1 with second dice 4; first dice 2 with second dice 3; first dice 3 with second dice 2; and first dice 4 with second dice 1 — which is 4 outcomes, so the probability is 4/16 = 1/4. A candidate who answers 1/8 has listed only 2 of the four pairs, forgetting that first dice 1 with second dice 4 and first dice 4 with second dice 1 are separate outcomes because the dice are different. A candidate who answers 3/16 has found 3 pairs instead of 4, missing one from the list. A candidate who answers 1/16 has counted only a single pair, such as first dice 2 with second dice 3, and treated the order of the dice as not mattering.
- (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.
- (d) 19/100 — It is easier to first find the probability that NEITHER component is faulty, then subtract from 1. The probability a component is not faulty is 9/10, so the probability neither is faulty is 9/10 × 9/10 = 81/100. So the probability at least one is faulty is 1 − 81/100 = 19/100. Choosing 1/5 comes from adding the two probabilities of a fault instead, 1/10 + 1/10 = 1/5, which double-counts the case where both are faulty. Choosing 1/10 comes from giving the probability for just one component being faulty. Choosing 1/100 comes from squaring the probability of a fault directly, 1/10 × 1/10 = 1/100, which is actually the probability that BOTH are faulty, not at least one.
- (d) 5/12 — The possible totals are 2 to 12. The prime totals are 2, 3, 5, 7 and 11, with 1, 2, 4, 6 and 2 ways to make them, giving 1 + 2 + 4 + 6 + 2 = 15 outcomes out of 36, so the probability is 15/36 = 5/12. Choosing 7/18 comes from forgetting that a total of 2 counts as prime, giving 14/36 = 7/18 instead. Choosing 19/36 comes from wrongly including a total of 9 as a prime number, adding its 4 ways to the correct 15 to get 19/36. Choosing 13/36 comes from forgetting the 2 ways of making a total of 11, giving 13/36 instead of 15/36.
- (d) 50% — Method: write the number of red cards over the total number of cards, then turn that fraction into a percentage by dividing and multiplying by 100. Working: the probability is 26/52; 26 ÷ 52 = 0.5 and 0.5 × 100 = 50. Answer: 50%, the exact middle of the 0 to 1 probability scale, which matches the fact that half the pack is red. The distractors: 26% comes from writing the number of red cards straight down as the percentage without dividing by the total; 25% comes from taking one of the four suits as the red ones and giving a quarter; 2% comes from dividing the wrong way round, 52 ÷ 26 = 2, and writing that as a percentage.
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