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.
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Probability worksheet — GCSE Foundation
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- 1.A fair coin is flipped again and again. After the first 10 flips there have been 7 heads. After 1000 flips there have been 528 heads. Which statement best describes what these results show?
- 2.A two-way table records 200 members of a gym. 80 of the members are women and the rest are men. 60 of the women attend yoga classes. Work out the percentage of the women who attend yoga classes.
- 3.A spinner has three equal sections, coloured red, blue and green. It is spun twice, and each outcome is listed as an ordered pair of colours, such as (red, blue). Work out how many outcomes are in the full list.
- 4.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.
- 5.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.
- 6.At a gym, 45 members are surveyed. 24 use the running machines, 20 use the weights machines, and 9 use both. Work out how many of the 45 members use neither machine.
- 7.A fair spinner has 10 equal sections, numbered 1 to 10. The spinner is spun 150 times. Work out how many times you would expect it to land on a number greater than 7.
- 8.120 people took a driving theory test. A frequency tree records the results. The first pair of branches splits the 120 people into 70 men and 50 women. On the men's branch, 45 passed and the rest failed. On the women's branch, 38 passed and the rest failed. Work out how many of the 120 people failed the test.
- 9.In a survey of 45 students, 22 said they like reading, 18 said they like gaming, and 6 said they like both. Work out how many students like exactly one of reading or gaming.
- 10.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.
- 11.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.
- 12.A bag contains 5 red counters and 5 green counters. Three counters are taken out one at a time and are not put back. Work out the probability that all three counters are red.
- 13.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.
- 14.A frequency tree records how 90 pupils travel to school and whether they were late. The first pair of branches splits the 90 pupils into 50 who walk and 40 who do not walk. On the walking branch, 6 were late and 44 were not. On the other branch, 10 were late and 30 were not. One of the 90 pupils is chosen at random. Work out the probability that the pupil walks to school and was late.
- 15.Oliver flips three fair coins at the same time. Work out the probability that exactly two of the three coins land on heads.
Answer key
- (b) The relative frequency is settling near 0.5 — Method: turn each result into a relative frequency before comparing them, because it is the relative frequency, and not the difference between the two counts, that tends towards the theoretical probability. Working: after 10 flips the relative frequency of a head is 7 ÷ 10 = 0.7, which is a long way from 0.5. After 1000 flips it is 528 ÷ 1000 = 0.528, which is much closer to 0.5. Meanwhile the gap between the two counts has grown rather than shrunk: it was 7 − 3 = 4 after 10 flips and is 528 − 472 = 56 after 1000 flips. Answer: the relative frequency is settling near 0.5, which is what an unbiased experiment does as the sample grows. The distractors: saying the counts are levelling out is the usual form of this idea and the figures contradict it, since the gap went from 4 to 56; saying the coin is biased treats 28 extra heads in 1000 flips as proof, when 0.528 sits close to 0.5 and a fair coin gives results like this often; saying the next flip is more likely to be a tail is the gambler's fallacy, since each flip stays at 1/2 whatever came before.
- (a) 75% — 'Percentage of the women' restricts the group to the 80 women, of whom 60 attend yoga: 60/80 = 0.75 = 75%. Dividing by the number of men (200 − 80 = 120) instead of the number of women gives 60/120 = 0.5 = 50%. Dividing by all 200 members instead of just the 80 women gives 60/200 = 0.3 = 30%. Using the 20 women who do NOT attend yoga (80 − 60) as the numerator instead of the 60 who do gives 20/80 = 0.25 = 25%.
- (d) 9 — Method: each of the first spin's 3 outcomes can be paired with each of the second spin's 3 outcomes, so the two counts are multiplied together. Working: 3 × 3 = 9 ordered pairs. Answer: 9. Watch out: writing down 6 comes either from adding the two spins' counts, 3 + 3, instead of multiplying them, or from listing colour pairs without order, treating (red, blue) as the same entry as (blue, red) — both shortcuts miss outcomes that the ordered list contains. Writing down 3 lists only a single spin's outcomes and never pairs the two spins up at all. And writing down 27 treats the spinner as though it had been spun three times, 3 × 3 × 3, one spin too many.
- (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.
- (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) 10 — Method: the number using at least one machine is the running total plus the weights total minus the overlap, since the overlap would otherwise be added in twice; the number using neither is the survey total minus that. Working: at least one machine = 24 + 20 − 9 = 35. Neither = 45 − 35 = 10. Answer: 10. Watch out: subtracting 24 + 20 from 45 without adding the 9 back, 45 − 24 − 20 = 1, removes the overlap twice over instead of once. Writing down 15, which is 24 − 9, gives the number who use ONLY the running machines, not the number who use neither. And writing down 9 mistakes the overlap region for the region outside both circles altogether.
- (c) 45 — Three of the ten numbers (8, 9 and 10) are greater than 7, so the probability is 3/10, and 150 × 3/10 = 45. Writing 105 is wrong because 150 × 7/10 = 105 uses the seven numbers that are NOT greater than 7 (1 to 7), the opposite of what is asked. Writing 60 is wrong because it counts 7, 8, 9 and 10 as four numbers greater than 7, wrongly including 7 itself: 150 × 4/10 = 60. Writing 50 is wrong because 150 ÷ 3 = 50 divides by the count of favourable numbers instead of multiplying by the correct fraction of the spinner. The expected number of spins landing on a number greater than 7 is 45.
- (b) 37 — Method: on a frequency tree each pair of branches adds back up to the number it came from, so work out the failures on each branch and then add the two end branches. Working: on the men's branch 70 − 45 = 25 men failed. On the women's branch 50 − 38 = 12 women failed. Adding the two failing end branches gives 25 + 12 = 37. Answer: 37 of the 120 people failed the test. The distractors: 83 comes from adding the two passing branches, 45 + 38 = 83, and so reads the tree for the wrong outcome; 25 is the men's failing branch on its own, with the women never added; 12 is the women's failing branch on its own, with the men never added.
- (c) 28 — Reading only is 22 − 6 = 16, and gaming only is 18 − 6 = 12, so exactly one of the two is 16 + 12 = 28. Adding 22 and 18 without removing the 6 who like both, 22 + 18 = 40, counts those 6 students twice. Giving 6 mistakes the number who like both for the number who like exactly one. Finding 22 + 18 − 6 = 34 gives the number who like at least one of reading or gaming, but stops there instead of also removing the 6 who like both to leave only those who like exactly one.
- (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.
- (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.
- (a) 1/12 — Method: for draws with nothing put back, multiply the probabilities of the three draws, reducing both the number of red counters and the total each time a red counter is removed. Working: the first counter is red with probability 5/10. One red counter has gone, so the second is red with probability 4/9, and then the third is red with probability 3/8. Multiplying gives 60/720. Answer: the probability is 1/12. The distractors: 1/8 comes from using 5/10 three times, which is what happens only if each counter is put back; 2/9 comes from stopping after two draws and giving 5/10 × 4/9; 3/50 comes from taking one off the red count each time but leaving the total at 10, giving 5/10 × 4/10 × 3/10.
- (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) 1/15 — Method: a probability read from a frequency tree is the count at the end of the branch you want, divided by the total number in the whole experiment. Working: the branch for walking followed by the branch for being late ends with 6 pupils, and the experiment covers all 90 pupils, so the probability is 6/90. Dividing the top and the bottom by 6 gives 1/15. Answer: the probability is 1/15. The distractors: 3/25 is 6/50 and comes from dividing the 6 by the 50 walkers rather than by the whole group, which answers a different question about walkers only; 8/45 is 16/90 and comes from counting every late pupil, the 6 walkers and the 10 others together, instead of only the late walkers; 1/9 is 10/90 and comes from reading the late count on the branch for pupils who do not walk.
- (c) 3/8 — Method: write out every result of the three coins as a string of three letters, H for heads and T for tails, count the results that match the description and divide by how many results the list holds. Working: each coin lands two ways and no coin affects another, so the list holds 2 × 2 × 2 = 8 equally likely results. Exactly two heads means one coin lands on tails and the other two on heads, so the results are HHT, HTH and THH — 3 of the 8. Answer: the probability is 3/8. The distractors: 4/8 comes from reading 'exactly two heads' as 'at least two heads' and counting HHH as well; 2/8 comes from a list made without a system, in which HHT and THH are written down and HTH, the result with the tail between the two heads, is missed; 6/8 comes from counting 3 × 2 = 6 ways of picking which two of the three coins show heads, which counts every pair of coins twice, once in each order.
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