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.Two ordinary fair dice are rolled, one after the other. Work out the probability that the first dice shows a 6 and the second dice shows an even number.
- 2.A box contains 9 red balls and 11 green balls. Two balls are taken out at random, one after the other, without being replaced. Given that both balls taken out are the same colour, work out the probability that both balls are red.
- 3.Five balls numbered 1, 2, 3, 4 and 5 are in a bag. Two of them are taken out together at random. Work out the probability that the two numbers on them add up to more than 7.
- 4.A café offers 3 types of sandwich, cheese, ham and egg, and 4 types of drink, tea, coffee, juice and water. A customer chooses one sandwich and one drink at random. Work out the probability that the customer chooses egg and water.
- 5.Two fair five-sided dice, numbered 1 to 5, are rolled. Work out the probability that the two scores differ by at least 2.
- 6.An ordinary six-sided dice, numbered 1 to 6, is rolled 30 times and lands on a 6 seven times. Ravi says the theoretical probability of rolling a 6 and the relative frequency of rolling a 6 in this trial are the same number. Is Ravi right?
- 7.150 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.
- 8.Two fair coins are thrown together 400 times. Work out how many of the 400 throws you would expect to give one head and one tail.
- 9.A drawing pin is dropped many times and lands either point up or point down. The relative frequency of landing point up is recorded as the experiment goes on: after 50 drops it is 0.720, after 200 drops it is 0.665, and after 1000 drops it is 0.638. The pin is to be dropped a further 2000 times. Work out the best estimate of the number of times it will land point up.
- 10.The probability that Kofi passes his driving test on any attempt is 0.6, and each attempt is independent of the others. Work out the probability that he fails both his first two attempts.
- 11.A call centre finds that 40% of its calls are about billing. 35% of the calls about billing are dealt with in under five minutes. The centre takes 500 calls on Monday. Work out how many of Monday's calls you would expect to be about billing and dealt with in under five minutes.
- 12.The probability that a seed fails to germinate is 1/5. Three seeds are planted independently. Work out the probability that at least one of the three seeds germinates.
- 13.Two ordinary fair dice are rolled and the two scores are added together. Work out the probability that the total is a prime number.
- 14.For two events A and B, P(A) = 0.6 and P(A and B) = 0.15. Work out P(B | A).
- 15.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.
Answer key
- (a) 1/12 — The probability that the first dice shows a 6 is 1/6. The probability that the second dice shows an even number, 2, 4 or 6, is 3/6 = 1/2. Since the two dice are independent, multiply the probabilities: 1/6 × 1/2 = 1/12. A candidate who answers 1/6 has considered only the first dice and forgotten the condition on the second dice. A candidate who answers 1/2 has considered only the second dice and forgotten the condition on the first dice. A candidate who answers 1/36 has treated 'an even number' as a single specific value rather than three possible values, using 1/6 × 1/6.
- (d) 36/91 — Method: P(both red | same colour) = P(both red) ÷ P(same colour), where P(same colour) = P(both red) + P(both green). Working: P(both red) = 9/20 × 8/19 = 72/380 = 18/95. P(both green) = 11/20 × 10/19 = 110/380 = 11/38. P(same colour) = 18/95 + 11/38 = 36/190 + 55/190 = 91/190. P(both red | same colour) = (36/190) ÷ (91/190) = 36/91. Answer: 36/91. Watch out: stopping at 18/95 gives P(both red) itself, without dividing by the probability that the colours matched at all. Working out 55/91 finds the same-colour probability for green instead of red — check which colour's count you are putting on top. And 9/20 is just the chance the first ball drawn is red, which ignores the second draw and the without-replacement condition completely.
- (d) 1/5 — Method: list the pairs systematically, work out each total, then count the pairs that meet the condition and compare that count with the length of the list. Working: the pairs and their totals are 1 and 2 giving 3, 1 and 3 giving 4, 1 and 4 giving 5, 1 and 5 giving 6, 2 and 3 giving 5, 2 and 4 giving 6, 2 and 5 giving 7, 3 and 4 giving 7, 3 and 5 giving 8, and 4 and 5 giving 9. That is 10 pairs, of which 2 have a total of more than 7. Answer: the probability is 1/5. The distractors: 2/5 comes from counting the totals of exactly 7 as well, reading 'more than 7' as '7 or more'; 1/10 comes from finding only the pair 4 and 5 and missing that 3 and 5 also beat 7; 4/5 comes from counting the pairs on the wrong side of the condition, the 8 pairs whose total is 7 or less.
- (d) 1/12 — Method: list the full possibility space of sandwich-and-drink pairs, then divide the one matching pair by the size of the whole space. Working: there are 3 × 4 = 12 equally likely sandwich-and-drink pairs, and exactly one of them is egg and water. Answer: 1/12. Watch out: writing down 1/7 comes from adding the two counts, 3 + 4 = 7, instead of multiplying them to build the possibility space. Writing down 1/3 uses only the chance of choosing egg out of 3 sandwiches and ignores the drink altogether. And writing down 1/4 uses only the chance of choosing water out of 4 drinks and ignores the sandwich altogether.
- (b) 12/25 — List every outcome as an ordered pair (first dice, second dice) out of the 25 equally likely outcomes. The pairs with a difference of exactly 2 are (1, 3), (3, 1), (2, 4), (4, 2), (3, 5) and (5, 3), the pairs with a difference of 3 are (1, 4), (4, 1), (2, 5) and (5, 2), and the pairs with a difference of 4 are (1, 5) and (5, 1), giving 6 + 4 + 2 = 12 outcomes with a difference of at least 2, so the probability is 12/25. Choosing 13/25 comes from finding the probability that the two scores differ by LESS than 2 instead, using the remaining 13 outcomes, the opposite of what was asked. Choosing 10/25 comes from forgetting the two outcomes where the difference is 4, (1, 5) and (5, 1), and adding only the difference-2 and difference-3 outcomes, 6 + 4 = 10 out of 25. Choosing 6/25 comes from counting only the pairs with a difference of exactly 2, forgetting that differences of 3 and 4 also count as at least 2.
- (c) No — 7/30 is the relative frequency; theory stays 1/6. — The theoretical probability of rolling a 6 on an ordinary dice is fixed at 1/6, worked out from the number of equally likely outcomes, and does not change however the dice is actually rolled. The relative frequency from this trial is 7/30, found from what happened in these particular 30 rolls. Since 7/30 and 1/6 are different numbers, the correct statement is 'No — 7/30 is the relative frequency; theory stays 1/6.' Assuming the two values must always match because they describe the same event gives 'Yes — relative frequency always equals theory.' Believing that an observed result redefines the theoretical probability gives 'Yes — the theoretical probability has now become 7/30.' Refusing to work out either value at all gives 'Neither can be found — 30 rolls is too few to tell', which ignores that both numbers CAN be calculated from the information given.
- (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.
- (b) 200 — Method: list the equally likely outcomes for the two coins before writing any probability, then multiply by the number of throws. Working: the equally likely outcomes are head then head, head then tail, tail then head, and tail then tail, so there are 4 of them. Two of those 4 give one head and one tail, so the probability is 2/4, which is 1/2. Over 400 throws the expected number is 400 × 1 ÷ 2 = 200. Answer: about 200 of the throws would be expected to give one head and one tail. The distractors: 133 comes from treating two heads, two tails and one of each as three equally likely results and working out 400 ÷ 3 = 133.3, then rounding; 100 comes from counting only head then tail as a success, giving 400 × 1 ÷ 4 = 100; 300 is the expected number of throws that do not give two heads, 400 × 3 ÷ 4 = 300.
- (a) 1276 — Method: an unbiased relative frequency tends towards the theoretical probability as the number of trials increases, so use the record resting on the most trials, then multiply by the number of new trials. Working: the three records rest on 50, 200 and 1000 drops, so the most reliable is the one after 1000 drops, namely 0.638, and the run is indeed settling as the trials increase. The expected number of point up landings in 2000 further drops is 2000 × 0.638 = 1276. Answer: about 1276 times. The distractors: 1440 uses the earliest record, which rests on only 50 drops, giving 2000 × 0.720 = 1440; 1330 uses the middle record, treating 200 drops as a safe compromise when 1000 drops is better still, giving 2000 × 0.665 = 1330; 1348 comes from averaging the three records, since 0.720 + 0.665 + 0.638 = 2.023 and 2.023 ÷ 3 = 0.674, then 2000 × 0.674 = 1348, which gives the 50 drop record the same weight as the 1000 drop record.
- (a) 0.16 — The probability that Kofi fails a single attempt is 1 − 0.6 = 0.4. Since the attempts are independent, the probability he fails both is 0.4 × 0.4 = 0.16. Choosing 0.36 comes from squaring the probability of PASSING instead, 0.6 × 0.6 = 0.36, which is the probability of passing both attempts, not failing both. Choosing 0.4 comes from giving the probability of failing just one attempt, forgetting to combine two attempts. Choosing 0.24 comes from multiplying the fail probability by the pass probability, 0.4 × 0.6 = 0.24, mixing up passing and failing between the two attempts.
- (a) 70 — Method: two linked steps. Find the expected number of billing calls first, then take the 35% of those, because the 35% is quoted for billing calls only. Working: 40% of 500 is 200 billing calls. 35% of 200 is 70 calls. Answer: you would expect 70 calls. The distractors: 200 stops after the first step and gives the billing calls, forgetting that only some of them are dealt with quickly; 175 is 35% of 500, applying the quick response rate to every call the centre takes rather than to the billing calls only; 375 comes from adding 40% and 35% to get 75% and taking 75% of 500, which treats two stages of one journey as separate outcomes to be added.
- (d) 124/125 — The probability that a seed germinates is 1 − 1/5 = 4/5, so the probability that all three seeds fail to germinate is 1/5 × 1/5 × 1/5 = 1/125. The probability that at least one germinates is 1 − 1/125 = 124/125. Choosing 4/5 comes from giving the probability that a single seed germinates, forgetting to combine all three seeds. Choosing 64/125 comes from working out the probability that ALL three seeds germinate, 4/5 × 4/5 × 4/5 = 64/125, instead of at least one. Choosing 12/125 comes from working out the probability that EXACTLY one seed germinates, 3 × 4/5 × 1/5 × 1/5 = 12/125, instead of 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.
- (c) 0.25 — Method: P(B | A) = P(A and B) ÷ P(A). Working: P(B | A) = 0.15 ÷ 0.6 = 0.25. Answer: 0.25. Watch out: multiplying 0.6 by 0.15 instead of dividing gives 0.09, and subtracting 0.15 from 0.6 gives 0.45 — neither uses the conditional probability formula. Leaving the answer as 0.15 mistakes the probability of A and B happening together for the probability of B once you already know A has happened — those are different quantities.
- (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.
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