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.At a fair, a game costs £2 to play. The probability of winning the game is 0.15, and each win pays out £10. Amir plays the game 200 times. Work out how much money Amir should expect to lose in total.
- 2.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?
- 3.At a youth club, 65 members are asked which sports they play. 27 play football, 21 play basketball, and 9 play both. Work out the number of members who play neither sport.
- 4.A charity raffle has three types of ticket: winning, near-miss and losing, and every ticket is exactly one of these. The probability that a ticket is winning is 1/8 and the probability that it is a near-miss is 1/4. Work out the probability that a ticket is losing.
- 5.A student is estimating the probability that a spinner lands on red. In her first 85 spins it landed on red 34 times. She then spins it 165 more times, and in those it lands on red 58 times. Work out the best estimate of the probability of red from all 250 spins. Give your answer as a decimal, correct to 3 decimal places.
- 6.A biased spinner is spun 40 times and lands on red 16 times. It is then spun a further 60 times and lands on red 21 times. Work out the best estimate of the probability that the spinner lands on red, using the results of all 100 spins together.
- 7.An ordinary pack of 52 playing cards contains 13 hearts. One card is taken at random from the pack. Work out the probability that the card is not a heart. Give your answer as a percentage.
- 8.A spinner can land on red, blue, green or yellow. The probability that it lands on red is 0.24 and the probability that it lands on yellow is 0.16. The probabilities of landing on blue and on green are equal, and each is called x. Work out the value of x.
- 9.Spinner P and Spinner Q are each fair and have 5 equal sections, numbered 1 to 5. Both spinners are spun once, and every outcome is listed as an ordered pair, Spinner P then Spinner Q, such as (2, 4). Work out how many of the outcomes have the score on Spinner P less than the score on Spinner Q.
- 10.The probability that it rains on a given morning is 0.3. If it rains, the probability that Priya's bus is late is 0.8. If it does not rain, the probability that her bus is late is 0.1. Work out the probability that her bus is late.
- 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 spinner is divided into sectors of 144°, 90° and 126°, coloured purple, orange and grey in that order. The spinner is spun 300 times. Work out how many times you would expect it to land on purple.
- 13.A red fair dice and a blue fair dice are rolled together. Every outcome is listed as an ordered pair, red score then blue score. Work out the probability that the red score is less than the blue score.
- 14.Two fair spinners are each numbered 1, 2 and 3. Priya spins both spinners together and records the two numbers as a pair, listing every possible outcome systematically in a grid. Work out the probability that the two numbers are the same.
- 15.At a leisure centre, 50 members are asked which activities they do. 32 do yoga, 24 do pilates, and 11 do both. Work out how many members do yoga only, and not pilates.
Answer key
- (b) £100 — Method: find the expected number of wins, turn that into the expected pay out, then compare it with what the games cost. Working: the expected number of wins is 200 × 0.15 = 30. Each win pays £10, so the expected pay out is 30 × 10 = 300 pounds. Playing 200 times at £2 a go costs 200 × 2 = 400 pounds. The expected loss is 400 − 300 = 100 pounds. Answer: Amir should expect to be about £100 down. The distractors: £300 is the expected winnings on their own, with the cost of playing never taken off; £400 is the total cost of playing, with the winnings never taken off; £700 comes from adding the two totals, 400 + 300 = 700, instead of subtracting one from the other.
- (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.
- (c) 26 — The number who play football or basketball is 27 + 21 − 9 = 39, subtracting the 9 who play both so they are not counted twice. The number who play neither is then 65 − 39 = 26. Giving 39 stops after finding the number who play football or basketball, without taking the complement within the 65 members. Subtracting all three given numbers from 65, 65 − 27 − 21 − 9 = 8, treats the 9 who play both as a separate group rather than a correction for double-counting. Adding 27 and 21 without removing the double-counted 9, then subtracting that total from 65, gives 65 − (27 + 21) = 17.
- (a) 5/8 — Winning, near-miss and losing are exhaustive, so the three probabilities sum to 1. Writing 1/4 as 2/8 so every fraction has the same denominator, 1 − 1/8 − 2/8 = 8/8 − 1/8 − 2/8 = 5/8. Subtracting only the winning probability and forgetting the near-miss probability gives 1 − 1/8 = 7/8. Subtracting only the near-miss probability and forgetting the winning probability gives 1 − 1/4 = 3/4. Adding the two given probabilities and stopping there gives 1/8 + 2/8 = 3/8, the probability that a ticket is winning or a near-miss, not the probability that it is losing.
- (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) 0.37 — Method: pool the two runs into one combined set of results, then find the relative frequency of red across all of the spins together. Working: total reds = 16 + 21 = 37. Total spins = 40 + 60 = 100. Relative frequency = 37 ÷ 100 = 0.37. Answer: 0.37. Watch out: writing down 0.40 uses only the first run, 16 ÷ 40, and throws away the extra evidence from the second 60 spins. Writing down 0.35 uses only the second run, 21 ÷ 60, and throws away the first run instead. And writing down 0.375 averages the two runs' separate rates, (0.40 + 0.35) ÷ 2, which treats a run of 40 spins and a run of 60 spins as equally weighted, when pooling the actual counts gives the larger run its fair share of influence.
- (d) 75% — Method: count how many cards are not hearts, write that count over the total number of cards, cancel the fraction down and then turn it into a percentage. Working: 52 − 13 = 39 cards are not hearts, so the probability is 39/52; dividing the numerator and the denominator by 13 gives 3/4, and 3/4 = 0.75, so 0.75 × 100 = 75. Answer: 75%, three quarters of the way along the 0 to 1 scale. The distractors: 25% comes from giving the probability that the card is a heart, 13 out of 52, which cancels to 1/4; 50% comes from reading 'not a heart' as 'not a red card' and halving the pack; 39% comes from writing the count of 39 cards straight down as the percentage without comparing it with the 52 cards in the pack.
- (a) 0.30 — Red, blue, green and yellow are exhaustive, so all four probabilities sum to 1: 0.24 + 0.16 + x + x = 1, so 2x + 0.40 = 1, giving 2x = 0.60 and x = 0.30. Stopping at 2x = 0.60 without dividing by 2 leaves 0.60, the combined probability of both blue and green together, not the value of x on its own. Sharing the 0.60 across all four colours instead of just the two unknown ones gives 0.60 ÷ 4 = 0.15. Leaving out the 0.16 for yellow gives 2x + 0.24 = 1, so 2x = 0.76 and x = 0.38.
- (b) 10 — There are 25 equally likely ordered pairs in total. Listing the pairs where the Spinner P score is less than the Spinner Q score gives 10 outcomes, running from (1, 2) up to (4, 5). Choosing 15 comes from also including the 5 pairs where the two scores are equal, such as (1, 1) and (2, 2), which do not satisfy 'less than'. Choosing 4 comes from counting only the outcomes where the scores differ by exactly 1, such as (1, 2) and (2, 3), and missing the pairs that differ by 2, 3 or 4. Choosing 20 comes from correctly excluding the 5 outcomes where the two scores are equal, 25 − 5 = 20, but forgetting to also exclude the outcomes where Spinner P is bigger than Spinner Q, rather than smaller.
- (c) 0.31 — Method: a late bus can be reached along two paths of the tree. Multiply along each path, then add the paths that end in a late bus. Working: the rain path gives 0.3 × 0.8 = 0.24. No rain has probability 1 − 0.3 = 0.7, so the second path gives 0.7 × 0.1 = 0.07. Adding the two paths gives 0.24 + 0.07. Answer: the probability is 0.31. The distractors: 0.9 comes from adding the two branch probabilities 0.8 and 0.1 without first weighting them by how often it rains; 0.24 comes from following the rain path only and ignoring that the bus can also be late when it is dry; 0.27 comes from using 0.3 at the start of both paths, so that the no-rain path is given 0.3 × 0.1 instead of 0.7 × 0.1.
- (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) 120 — Purple's probability is its angle out of the full circle: 144° ÷ 360° = 2/5. Expected number on purple = 2/5 × 300 = 120. Assuming the three colours are equally likely, ignoring their different angles, gives 300 ÷ 3 = 100. Using orange's angle, 90°, instead of purple's 144° gives a probability of 1/4, so 300 × 1/4 = 75. Treating 144 as a percentage instead of finding its fraction of 360° gives 300 × 0.144 = 43.2, which rounds to 43.
- (b) 5/12 — There are 36 equally likely ordered pairs. Listing the pairs where the red score is smaller than the blue score gives 15 outcomes, so the probability is 15/36 = 5/12. Choosing 1/2 comes from assuming the 36 outcomes split evenly between 'red smaller' and 'red bigger', which ignores the 6 outcomes where the two dice show the same score. Choosing 7/12 comes from including the outcomes where the two scores are equal, 15 + 6 = 21, giving 21/36 = 7/12, instead of using 'less than' strictly. Choosing 5/36 comes from counting only the outcomes where the two scores differ by exactly 1, such as (1, 2) and (2, 3), which is 5 outcomes, and missing the pairs that differ by more than 1.
- (d) 1/3 — List the outcomes for the two spinners systematically in a 3 × 3 grid: 1-1, 1-2, 1-3, 2-1, 2-2, 2-3, 3-1, 3-2, 3-3, where the first number is the score on spinner A and the second is the score on spinner B — 9 equally likely outcomes in total. The pairs where the two numbers are the same are 1-1, 2-2 and 3-3, so there are 3 favourable outcomes. P(same number) = 3/9 = 1/3. 2/3 comes from working out the probability that the two numbers are different and then forgetting to take the complement the right way round, so the probability of "different" is given instead of the probability of "same". 1/2 comes from listing only the 6 unordered pairs 1-1, 2-2, 3-3, 1-2, 1-3, 2-3 instead of all 9 ordered outcomes in the grid, then taking 3 out of that 6. 1/9 comes from spotting only one of the three matching pairs, such as 1-1, and missing 2-2 and 3-3.
- (d) 21 — Method: 'yoga only' means yoga but not pilates, so subtract the number who do both from the total who do yoga. Working: 32 − 11 = 21. Answer: 21. Watch out: writing down 32, the total who do yoga, answers 'how many do yoga' rather than 'how many do yoga only' — it still includes the 11 who also do pilates. Adding the overlap instead of subtracting it, 32 + 11 = 43, moves in the wrong direction entirely. And writing down 11 gives the number who do both activities, which is the opposite of yoga only.
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