Printable · GCSE Higher · ages 14-16
Probability worksheet — GCSE Higher
Fifteen questions across the probability statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Probability worksheet — GCSE Higher
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- 1.At a fête, a game costs £3 to play. The probability of winning is 0.1, and each win pays out £20. 150 people play the game. Work out the fête's expected profit from the game.
- 2.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.
- 3.The probability that Isla answers a quiz question correctly is 0.8. She answers three questions, and her answers are independent of each other. Work out the probability that she answers all three correctly.
- 4.A bag contains 5 letter tiles spelling the word MATHS, with one letter on each tile. Two tiles are drawn at random, one after the other, without the first tile being put back. Work out the probability that both tiles drawn are consonants.
- 5.A raffle has two independent draws. In the first, a winning ticket is picked at random from 5 red tickets and 3 blue tickets. In the second, a winning number is picked using a fair spinner with 4 equal sections, numbered 1 to 4. Work out the probability that the first winning ticket is blue and the second winning number is greater than 2.
- 6.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.
- 7.At a school car park, each car is a saloon, an estate or a hatchback, and no car is more than one of these. Among 200 cars, the probability that a car chosen at random is a saloon is 0.28 and the probability that it is an estate is 0.37. Work out how many of the 200 cars are hatchbacks.
- 8.The probability that Priya scores with a netball shot is 3/4. She takes two shots, and the shots are independent. Work out the probability that she scores with her first shot but misses her second shot.
- 9.Harry plays a game in which he flips a fair coin and rolls an ordinary fair dice. He wins a prize only if the coin shows heads and the dice shows a 6. Work out the probability that Harry wins a prize.
- 10.Two games are offered at a school fair, and every one of the 40 pupils in a class plays each game once. In Game A, the probability of winning a prize is 0.2 and the prize is worth £5. In Game B, the probability of winning a prize is 0.1 and the prize is worth £12. Work out which game gives a higher expected total value of prizes for the class, and by how much.
- 11.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.
- 12.A fair coin is flipped three times. Work out the probability of getting at least one head.
- 13.A garage tested 200 cars in one week. 70% of the cars were more than 3 years old and the rest were 3 years old or less. Of the cars more than 3 years old, 1 in 4 failed the test. 12 of the cars that were 3 years old or less failed the test. Work out how many of the 200 cars failed the test altogether.
- 14.A ferry company finds that on 20% of days the sea is rough. If the sea is rough, the probability that a crossing is delayed is 0.75. If the sea is calm, the probability that a crossing is delayed is 0.1. Given that a crossing was delayed, work out the probability that the sea was rough that day.
- 15.250 people took a theory test at one test centre. 150 of them had taken a preparation course and the rest had not. 120 of those who had taken the course passed and 50 of those who had not taken the course passed. One of the people who passed is picked at random. Work out the probability that this person had taken the preparation course.
Answer key
- (a) £150 — Each game, the expected payout is 0.1 × £20 = £2, so the fête's expected profit per game is the £3 charged minus the £2 expected payout, £1. Over 150 games, that is 150 × £1 = £150. Writing £300 is wrong because 150 × £2 = £300 is the total expected PAYOUT, not the profit — it has not been subtracted from the entry fees. Writing £450 is wrong because 150 × £3 = £450 is the total money taken in entry fees, without accounting for what is expected to be paid out in prizes. Writing £1 is wrong because that is only the expected profit for ONE game — it has not been scaled up to all 150 games. The fête's expected profit is £150.
- (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.
- (c) 0.512 — Method: independent events that must all happen are combined by multiplying their probabilities. Working: the first two questions give 0.8 × 0.8 = 0.64. Bringing in the third gives 0.64 × 0.8, and since 64 × 8 = 512 with three decimal places in the product, this is 0.512. Answer: the probability is 0.512. The distractors: 0.64 comes from multiplying only two of the three probabilities and stopping; 0.8 comes from reading 'independent' as meaning the probability never changes and writing down the single-question figure; 0.0512 comes from a place-value slip in the last multiplication, counting four decimal places instead of three.
- (d) 3/5 — The first tile drawn is a consonant with probability 4/5, since 4 of the 5 letters in MATHS are consonants. Since it is not replaced, 3 consonants remain out of the 4 tiles left, so the second tile is a consonant with probability 3/4. Multiplying gives 4/5 × 3/4 = 3/5. Choosing 4/5 comes from only using the probability for the first draw and forgetting the second draw changes once a tile is removed. Choosing 3/4 comes from only using the probability for the second draw on its own. Choosing 16/25 comes from treating the two draws as independent, using 4/5 × 4/5 = 16/25, which would only be correct if the first tile were replaced before the second draw.
- (d) 3/16 — Method: work out each draw's own probability first, then multiply them together since the two draws are independent. Working: there are 8 tickets in all, 3 of them blue, so P(blue) = 3/8. P(spinner number greater than 2) = 2/4 = 1/2, since 3 and 4 qualify. Multiplying gives 3/8 × 1/2, which comes to 3/16. Answer: 3/16. Watch out: writing down 3/8 stops after the first draw and never brings in the spinner at all. Writing down 1/2 does the opposite, using only the spinner and ignoring the ticket draw. And writing down 5/16 uses 5/8, the probability of a RED ticket, instead of 3/8 for blue — reading the wrong colour off the raffle.
- (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) 70 — Saloon, estate and hatchback are exhaustive, so their probabilities sum to 1: the probability of a hatchback is 1 − 0.28 − 0.37 = 0.35. The number of hatchbacks is 0.35 × 200 = 70. Treating the SUM of the other two probabilities, 0.28 + 0.37 = 0.65, as the probability of a hatchback instead of its complement gives 0.65 × 200 = 130. Multiplying the correct probability, 0.35, by 100 instead of the 200 cars actually surveyed gives 35. Averaging the two given probabilities, (0.28 + 0.37) ÷ 2 = 0.325, instead of subtracting them from 1, and then multiplying by 200 gives 65.
- (c) 3/16 — The probability that Priya misses a shot is 1 − 3/4 = 1/4. The two shots are independent, so multiply the probability of scoring the first shot by the probability of missing the second shot: 3/4 × 1/4 = 3/16. Choosing 9/16 comes from using the probability of scoring, 3/4, for both shots instead of switching to the miss probability for the second shot, 3/4 × 3/4 = 9/16, which is the probability that she scores with both shots. Choosing 1/2 comes from subtracting the two probabilities instead of multiplying them, 3/4 − 1/4 = 1/2. Choosing 3/4 comes from giving only the probability that she scores with the first shot, forgetting to combine it with what happens on the second shot.
- (b) 1/12 — Method: the coin does not affect the dice, so the two events are independent and the probability that both happen is the product of their probabilities. Working: heads has probability 1/2 and a 6 on an ordinary dice has probability 1/6. Multiplying gives 1 on the top and 2 × 6 = 12 on the bottom. Answer: the probability is 1/12. The distractors: 2/3 comes from adding 1/2 and 1/6 instead of multiplying them; 1/6 comes from using the dice alone and ignoring the condition on the coin; 1/8 comes from counting the possible results as 6 + 2 = 8 and treating the winning result as one of those eight.
- (a) Game B, by £8 — Game A's expected total is 40 × 0.2 × £5 = £40. Game B's expected total is 40 × 0.1 × £12 = £48. Game B is higher, by £48 − £40 = £8. Writing 'Game A, by £8' is wrong because it has the right difference but the wrong game — Game A's total (£40) is actually LOWER than Game B's, not higher. Writing 'Game B, by £48' is wrong because £48 is Game B's whole expected total, not the DIFFERENCE between the two games. Writing 'Game A, by £40' is wrong in the same way, using Game A's whole total as if it were the margin, and naming the wrong game as the winner. Game B gives the higher expected total, by £8.
- (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.
- (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) 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.
- (c) 15/23 — Method: find P(rough and delayed) and the overall P(delayed) using the tree, then divide. Working: P(rough and delayed) = 0.2 × 0.75 = 0.15. P(calm and delayed) = 0.8 × 0.1 = 0.08. P(delayed) = 0.15 + 0.08 = 0.23. P(rough | delayed) = 0.15 ÷ 0.23 = 15/23. Answer: 15/23. Watch out: leaving the answer as 0.15 (3/20) gives P(rough and delayed) itself, without dividing by the overall probability that a crossing is delayed. Giving 0.75 (3/4) is the probability you were told to start with — that a crossing is delayed GIVEN the sea is rough — which is the reverse of what's being asked. And 0.2 (1/5) is just the original probability that the sea is rough, before you take the fact that the crossing was delayed into account.
- (d) 12/17 — Method: the person picked is known to have passed, so the sample space is everyone who passed; divide the course takers who passed by that total. Working: 120 course takers and 50 others passed, so 170 people passed. The course takers who passed give 120/170, and dividing the numerator and the denominator by 10 gives 12/17. Answer: the probability is 12/17. The distractors: 4/5 is 120/150, the probability that someone passed given that they took the course, which reverses the condition and the event; 12/25 is 120/250, dividing by everyone who sat the test rather than by the 170 who passed; 17/25 is 170/250, the probability that a person picked from everyone sitting the test passed, which answers a different question altogether.
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