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.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.
- 2.A fair spinner has 8 equal sections: 5 are red and 3 are blue. Freya spins it twice, and each spin is independent of the other. Work out the probability that she gets one red and one blue in the two spins, in either order.
- 3.A = {factors of 12} and B = {factors of 18}. Work out n(A ∩ B), the number of elements in both sets.
- 4.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.
- 5.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.
- 6.A bag contains 3 yellow counters and 2 green counters. Sophie takes one counter at random and does not put it back. She then takes a second counter at random. Work out the probability that the first counter is yellow and the second is green.
- 7.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.
- 8.A pack contains 6 cards, numbered 1 to 6. A card is drawn at random from the pack, and a fair coin is flipped. Work out the probability of drawing an even-numbered card and the coin landing on tails.
- 9.In a certain town, P(rain on Monday) = 0.3. If it rains on Monday, the probability that it also rains on Tuesday is 0.6. If it does not rain on Monday, the probability that it rains on Tuesday is 0.25. Work out the probability that it rains on Tuesday.
- 10.A fair spinner is divided into 5 equal sections, 2 labelled win and 3 labelled lose. Zara spins it twice, and the two spins are independent. Work out the probability that she wins on the first spin and loses on the second spin.
- 11.A leisure centre has 150 members. 80 of the members are male and the rest are female. Every member uses either the pool or the gym, but not both. 66 members use the pool, and 35 of those pool users are male. Work out how many female members use the gym.
- 12.Spinner A has 4 equal sections, numbered 1, 2, 3 and 4. Spinner B has 5 equal sections, numbered 1, 2, 3, 4 and 5. Both spinners are spun once, and the two numbers are multiplied together. Work out the probability that the product is 12.
- 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.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.
- 15.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.
Answer key
- (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.
- (b) 15/32 — Method: there are two ways to get one of each colour, red then blue and blue then red. Work out the probability of each path by multiplying, then add the two paths. Working: red then blue is 5/8 × 3/8 = 15/64, and blue then red is 3/8 × 5/8 = 15/64. Adding the two paths gives 30/64. Answer: the probability is 15/32. The distractors: 15/64 comes from working out red then blue only and forgetting that blue then red also gives one of each; 15/28 comes from doubling correctly but reducing the total to 7 for the second spin, which is what happens to a bag when an item is kept out, not to a spinner; 39/64 comes from working from the opposite event and subtracting only the two-red case, 1 − 25/64, leaving the two-blue case inside the answer.
- (d) 4 — Method: list the elements of each set in full, then find which elements appear in both lists — that is A ∩ B. Working: factors of 12 = {1, 2, 3, 4, 6, 12}. Factors of 18 = {1, 2, 3, 6, 9, 18}. The elements in both lists are 1, 2, 3 and 6, so A ∩ B = {1, 2, 3, 6} and n(A ∩ B) = 4. Answer: 4. Watch out: writing down 6 gives n(A), the size of the factors-of-12 list on its own, not the size of the overlap. Writing down 8 comes from counting every element that appears in EITHER list, 1, 2, 3, 4, 6, 9, 12 and 18 — that is the union, a different set from the intersection. And writing down 3 misses that 1 is a factor of both 12 and 18, and so belongs in A ∩ B alongside 2, 3 and 6.
- (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.
- (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.
- (b) 3/10 — Method: on a tree diagram, follow the path that matches the description and multiply the probabilities written along it; with nothing put back, the second set of branches is worked out from the counters that are left. Working: 3 of the 5 counters are yellow, so the first branch of the path is 3/5. A yellow counter has been kept out, so 4 counters remain and both green counters are still there, making the second branch 2/4. Multiplying along the path gives 6/20. Answer: the probability is 3/10. The distractors: 6/25 comes from using 2/5 on the second branch, which is the tree for a counter that is put back; 3/20 comes from taking one off the green count as well as off the total, using 1/4 on the second branch; 3/5 comes from reading the first branch only and never multiplying along the path.
- (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.
- (b) 1/4 — There are 6 × 2 = 12 equally likely outcomes. The even-numbered cards are 2, 4 and 6, so there are 3 × 1 = 3 outcomes with an even card and tails, giving a probability of 3/12 = 1/4. Choosing 1/2 comes from working out only the probability of drawing an even card, 3/6, and forgetting to combine it with the coin landing on tails. Choosing 1/12 comes from treating only one specific outcome, such as card 6 with tails, as the only one that counts, instead of all three even cards paired with tails. Choosing 1/8 comes from doubling the coin stage when counting the total, using 6 × 2 × 2 = 24 outcomes instead of 6 × 2 = 12, and giving 3/24 = 1/8.
- (b) 0.355 — Method: use the law of total probability across the two Monday branches: P(rain Tue) = P(rain Mon) × P(rain Tue | rain Mon) + P(no rain Mon) × P(rain Tue | no rain Mon). Working: P(no rain Mon) = 1 − 0.3 = 0.7. P(rain Tue) = (0.3 × 0.6) + (0.7 × 0.25) = 0.18 + 0.175 = 0.355. Answer: 0.355. Watch out: using only the rain-Monday branch (0.3 × 0.6) or only the no-rain-Monday branch (0.7 × 0.25) accounts for just one of the two ways Tuesday can turn out rainy — both branches must be added. And swapping which weekday-probability multiplies which branch (0.7 with the rain branch, 0.3 with the no-rain branch) uses the right numbers on the wrong branches.
- (b) 6/25 — The probability of winning on any one spin is 2/5, and the probability of losing on any one spin is 3/5. Since the spins are independent, multiply the probability of winning on the first spin by the probability of losing on the second spin: 2/5 × 3/5 = 6/25. A candidate who answers 4/25 has used the winning probability for both spins, 2/5 × 2/5. A candidate who answers 9/25 has used the losing probability for both spins, 3/5 × 3/5. A candidate who answers 3/10 has treated the spins as if they were dependent, reducing the second spin's denominator to 4.
- (b) 39 — Method: put the counts into a two-way table and fill each missing cell by subtracting along a row or down a column. Working: the number of female members is 150 − 80 = 70. The pool column holds 66 members and 35 of them are male, so the number of female pool users is 66 − 35 = 31. Subtracting along the female row, 70 − 31 = 39 female members use the gym. Answer: 39 female members use the gym. The distractors: 45 comes from subtracting along the male row instead, 80 − 35 = 45, which counts male gym users; 31 is the female pool cell, written down one step before the gym cell; 84 is 150 − 66 and counts every gym user, male and female together.
- (a) 1/10 — There are 4 × 5 = 20 equally likely outcomes in total. The pairs whose product is 12 are Spinner A showing 3 with Spinner B showing 4, and Spinner A showing 4 with Spinner B showing 3, which is 2 outcomes, giving a probability of 2/20 = 1/10. Choosing 1/20 comes from finding only one of the two pairs, (3, 4), and missing (4, 3) as a separate outcome. Choosing 1/8 comes from using 16 as the total number of outcomes, 4 × 4, forgetting that Spinner B has 5 sections rather than 4. Choosing 1/5 comes from listing the factor pairs of 12 as 2 × 6 and 3 × 4 and counting each one in both orders, (2, 6), (6, 2), (3, 4) and (4, 3), giving 4 outcomes out of 20 without checking that neither spinner has a 6 on it.
- (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.
- (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.
- (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.
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