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.
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Probability worksheet — GCSE Higher
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- 1.Two cards are dealt one after the other from an ordinary pack of 52 playing cards. The first card is not put back before the second is dealt. The pack contains 4 aces. Work out the probability that neither card is an ace. Give your answer as a product of two fractions.
- 2.Each turn of a game ends in a win, a draw or a loss, and no turn can end in more than one of these. The probability of a win is 1/4 and the probability of a draw is 1/3. Work out the probability of a loss.
- 3.A pupil must choose 2 different subjects at random from these five: maths, physics, biology, chemistry and history. Work out the probability that the two subjects chosen are maths and biology.
- 4.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.
- 5.On Amelia's walk to school there are two sets of traffic lights. The probability that the first set is green when she reaches it is 0.4. The probability that the second set is green when she reaches it is 0.5. The two sets work independently of each other. Work out the probability that both sets are green.
- 6.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.
- 7.A bag contains 3 red counters and 5 blue counters. Three counters are taken out at random, one after another, without being replaced. Work out the probability that all three counters taken out are red.
- 8.For two events A and B, P(A) = 0.6 and P(A and B) = 0.15. Work out P(B | A).
- 9.150 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.
- 10.Of the 70 students on the Year 10 geography trip, 42 are girls and the rest are boys. 24 of the girls and 10 of the boys brought a packed lunch. A student who did not bring a packed lunch is chosen at random. Work out the probability that this student is a girl.
- 11.A company has 400 employees. 150 of them work part time and 160 of them cycle to work. Working part time and cycling to work are independent. Work out how many of the employees you would expect both to work part time and to cycle to work.
- 12.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.
- 13.Five raffle tickets numbered 1 to 5 are in a bag. Two tickets are drawn one after the other and the first is not put back. Work out the probability that ticket 1 is drawn first and ticket 2 is drawn second.
- 14.An ordinary fair dice is rolled twice. Work out the probability of getting at least one 5.
- 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
- (b) (48/52) × (47/51) — Method: for two deals one after the other with nothing put back, multiply the probability of the first by the probability of the second worked out from the cards that are left. Working: 52 − 4 = 48 cards are not aces, so the first card is not an ace with probability 48/52. One card has now gone and it was not an ace, so 51 cards remain and 47 of them are not aces, giving 47/51. Answer: the probability is (48/52) × (47/51). The distractors: (48/52) × (48/52) comes from leaving the pack at 52 cards for the second deal, which is only true if the first card is replaced; (4/52) × (3/51) comes from working out the probability that both cards ARE aces instead of neither; (4/52) × (4/51) comes from the same misreading with the ace count left at 4 while the total is reduced, adjusting only half of the second fraction.
- (a) 5/12 — Method: a win, a draw and a loss are the only outcomes and no two can happen together, so the three probabilities form an exhaustive set of mutually exclusive events and add to 1; add the two given probabilities, then subtract from 1. Working: 1/4 + 1/3 over the common denominator 12 is 3/12 + 4/12 = 7/12, and 1 − 7/12 = 12/12 − 7/12. Answer: 5/12. The distractors: 7/12 comes from stopping at the probability of a win or a draw and never subtracting from 1; 1/12 comes from subtracting the two given probabilities from each other, 1/3 − 1/4, instead of adding them and taking the total from 1; 5/7 comes from adding 1/4 and 1/3 by adding the numerators and the denominators to get 2/7 and then subtracting that from 1.
- (b) 1/10 — Method: list every possible pair of subjects systematically, so that no pair is missed and no pair is counted twice, then compare the number of successful pairs with the size of the list. Working: pairing maths with each of the other four gives 4 pairs, physics with each subject after it gives 3, biology gives 2 and chemistry gives 1, so there are 4 + 3 + 2 + 1 = 10 pairs. Exactly one of them is maths with biology. Answer: the probability is 1/10. The distractors: 1/5 comes from counting maths-then-biology and biology-then-maths as two separate successes while still dividing by the 10 unordered pairs; 1/15 comes from a list that also pairs each subject with itself, giving 15 entries instead of 10; 1/4 comes from working out the second step alone, that 1 of the 4 subjects left after maths is biology, without allowing for the chance that maths is picked at all.
- (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.
- (a) 0.2 — Method: for independent events, the probability that both happen is the product of the two probabilities. Working: the first set is green with probability 0.4 and the second with probability 0.5, so the calculation is 0.4 × 0.5. Since 4 × 5 = 20 and the two factors carry one decimal place each, the product carries two. Answer: the probability is 0.2. The distractors: 0.9 comes from adding 0.4 and 0.5 instead of multiplying them; 0.45 comes from averaging the two probabilities; 0.1 comes from subtracting 0.4 from 0.5, treating the question as a difference.
- (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.
- (c) 1/56 — Method: for draws without replacement, multiply a chain of three fractions where both the numerator (reds remaining) and the denominator (counters remaining) fall by one after each draw. Working: P(all three red) = 3/8 × 2/7 × 1/6 = 6/336 = 1/56. Answer: 1/56. Watch out: using 3/8 for all three draws (27/512) treats the counters as if they were replaced each time. Reducing only the numerator each draw (3/8 × 2/8 × 1/8) forgets that the total number of counters left in the bag also falls. And reducing only the denominator while keeping the numerator at 3 each time (3/8 × 3/7 × 3/6) forgets that a red counter has actually left the bag.
- (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.
- (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.
- (c) 1/2 — Method: restrict to the students who did NOT bring a packed lunch, then find what fraction of that group are girls. Girls without lunch = 42 − 24 = 18. Boys = 70 − 42 = 28, so boys without lunch = 28 − 10 = 18. Total without lunch = 18 + 18 = 36. Working: P(girl | no lunch) = 18 ÷ 36 = 1/2. Answer: 1/2. Watch out: dividing 18 by 42 (the total number of girls) instead of by 36 finds P(no lunch | girl), the reverse conditional. Dividing by 70 (the whole trip) ignores that you already know the student did not bring a lunch. And using the 'brought a lunch' numbers (24 out of 34) answers the question for the wrong group entirely — you were asked about the students who did NOT bring one.
- (b) 60 — Method: independence means the proportion of part timers among the cyclists is the same as the proportion among all the employees, so find that proportion and apply it to the cyclists. Working: 150 of the 400 employees work part time, which is a proportion of 0.375. Applying it to the 160 cyclists gives 0.375 × 160 = 60 employees. Answer: you would expect 60 employees. The distractors: 310 adds 150 and 160, treating the group who do both as everyone who does one thing or the other; 10 subtracts 150 from 160, reading 'both' as the difference between the two counts; 75 halves the 150 part timers, assuming that independence means they split evenly between cyclists and non cyclists, which would need exactly half the workforce to cycle.
- (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) 1/20 — Method: two draws with nothing put back are combined by multiplying, with the second probability worked out from the tickets still in the bag. Working: the first draw takes the wanted ticket with probability 1/5. That ticket is kept out, so 4 tickets remain and only one of them is the ticket wanted second, giving 1/4. Multiplying gives 1/20. Answer: the probability is 1/20. The distractors: 1/10 comes from ignoring the order and treating the draw as a choice of two tickets from five, of which there are ten; 1/25 comes from keeping the total at 5 for the second draw, which is what happens only if the first ticket is put back; 2/5 comes from counting the two wanted tickets over the five in the bag, as though one draw decided the whole question.
- (b) 11/36 — Method: 'at least one' is the opposite of 'none at all', so work out the probability of no 5 on either roll and take it away from 1. Working: a roll that is not a 5 has probability 5/6, and the rolls are independent, so no 5 at all has probability 5/6 × 5/6 = 25/36. Taking this from 36/36 leaves 11/36. Answer: the probability is 11/36. The distractors: 25/36 is the probability of no 5 at all, written down without the final subtraction; 12/36 comes from counting the 6 pairs with a 5 on the first roll and the 6 pairs with a 5 on the second and adding them, which counts the pair (5, 5) twice; 30/36 comes from working out 1 − 1/6 as though only one roll were made.
- (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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