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 survey found that, of 250 shoppers questioned, 68% said they had used a self-checkout in the past month. A different store expects 1400 shoppers this week. Using this relative frequency, work out how many of the 1400 shoppers would be expected to have used a self-checkout in the past month.
- 2.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.
- 3.A train company runs 25 trains a day, every day. The probability that any one train is late is 0.08. Work out how many late trains the company should expect over a period of 4 weeks.
- 4.150 people at a gym were asked whether they prefer weight training or cardio; each person chose exactly one. 84 of the people are women. 50 of the women prefer cardio. 40 of the men prefer weight training. Work out the probability that a person, chosen at random from the 150, prefers weight training. Give your answer as a fraction in its simplest form.
- 5.A charity bake sale tracked 240 items sold using a frequency tree. The first branch splits sales into 150 cake sales and 90 biscuit sales. Of the cake sales, 100 were bought by adults and the rest by children. Of the biscuit sales, 54 were bought by children. Each item sold for £1.20. Work out the total amount of money raised from sales to children.
- 6.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.
- 7.Ellie drops a bottle top and records whether it lands open end up. In her first 20 drops it lands open end up 13 times. She carries on, and after 200 drops in total it has landed open end up 84 times. Work out the best estimate of the probability that the bottle top lands open end up.
- 8.A bag contains 20 counters. 8 of the counters are red and 12 are blue. A counter is taken at random, its colour is recorded, and it is put back in the bag. This is done 150 times. Work out how many red counters you would expect to be recorded.
- 9.A biased spinner is spun 40 times and lands on red 16 times. It is then spun a further 60 times and lands on red 21 times. Work out the best estimate of the probability that the spinner lands on red, using the results of all 100 spins together.
- 10.A raffle sells 400 tickets at 50p each. There is one prize of £45. Aisha buys 8 tickets. Work out how much money Aisha should expect to lose from playing, giving your answer in pounds.
- 11.Two pupils each flip the same coin to estimate the probability of heads. Leah flips it 40 times and gets 24 heads. Ben flips it 60 times and gets 33 heads. By combining both pupils' results, work out the relative frequency of heads, giving your answer as a fraction in its simplest form.
- 12.Two goalkeepers face penalty kicks. Elin saves 34% of the penalties she faces. Noah saves 11/32 of the penalties he faces. Which statement correctly compares them?
- 13.A game has two independent stages. P(win stage 1) = 0.4, and P(win stage 2) = 0.25. Work out the probability that Ffion wins at least one of the two stages.
- 14.A library recorded loans for 250 books over a week, using a frequency tree. The first branch splits the books into 160 fiction and 90 non-fiction. Of the fiction books, 112 were returned on time and the rest were returned late. All the non-fiction books were returned on time. Work out the probability that a book, chosen at random from the 250, was returned late. Give your answer as a fraction in its simplest form.
- 15.A two-way table records 150 customers at a café. 90 of the customers bought a hot drink and the rest did not. 54 of the hot-drink customers also bought a cake. 21 of the customers who did not buy a hot drink bought a cake. Work out the percentage of all 150 customers who bought a cake.
Answer key
- (a) 952 — 68% = 0.68. The relative frequency from the survey applies to the new group of 1400 shoppers, so the expected number is 0.68 × 1400 = 952. Working out 1 − 0.68 = 0.32 and applying that instead, 0.32 × 1400 = 448, finds the number who have NOT used a self-checkout, not the number who have. Applying 68% to the original sample size of 250 instead of the new total of 1400 gives 0.68 × 250 = 170. Applying the complement percentage to the original sample size, 0.32 × 250 = 80, compounds both mistakes.
- (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.
- (d) 56 — Method: count the trials over the whole period first, then multiply the number of trials by the probability. Working: 4 weeks is 4 × 7 = 28 days, and at 25 trains a day that is 25 × 28 = 700 trains. The expected number of late trains is 700 × 0.08 = 56. Answer: about 56 late trains over the 4 weeks. The distractors: 2 is the expected number for a single day, 25 × 0.08 = 2, with the 28 days never brought in; 14 uses one week instead of four, 25 × 7 × 0.08 = 14; 644 is 700 − 56 and counts the trains expected to be on time.
- (b) 37/75 — There are 150 − 84 = 66 men. 84 − 50 = 34 women prefer weight training, and 40 men prefer weight training, so 34 + 40 = 74 people in total prefer weight training, out of 150: 74/150 = 37/75. Writing 4/15 is wrong because it only counts the men who prefer weight training (40/150, simplified), leaving out the 34 women. Writing 17/75 is wrong because it only counts the women who prefer weight training (34/150, simplified), leaving out the 40 men. Writing 37/42 is wrong because it uses the number of women (84) as the denominator instead of the whole gym (150) — 74/84 simplifies to 37/42, but that is not a probability out of the whole group. The probability is 37/75.
- (a) £124.80 — On the cake branch, 150 − 100 = 50 cakes were bought by children. Adding the 54 biscuits bought by children gives 50 + 54 = 104 items sold to children in total, and at £1.20 each that raises 104 × £1.20 = £124.80. Writing £60.00 is wrong because 50 × £1.20 = £60.00 only counts the cake sales to children and leaves out the 54 biscuits. Writing £163.20 is wrong because it uses the ADULT sales instead of children's: 100 cake adults plus 90 − 54 = 36 biscuit adults gives 136 × £1.20 = £163.20. Writing £136.80 is wrong because it finds the cake children's number by subtracting the wrong branch (150 − 90 = 60 instead of 150 − 100 = 50), giving 60 + 54 = 114 items and 114 × £1.20 = £136.80. The total raised from sales to children is £124.80.
- (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.
- (d) 0.420 — Method: use the relative frequency worked out from the larger number of trials as the best estimate of the probability, since a bigger sample tends to sit closer to the true probability. Working: 84 out of 200 drops land open end up, so the relative frequency from the total is 84 ÷ 200 = 0.420. Answer: 0.420. Watch out: writing down 0.650 comes from 13 ÷ 20, using only the first, much smaller sample instead of the total. Writing down 0.535 comes from averaging 0.650 and 0.420, treating the 20-drop run and the 200-drop run as equally reliable instead of using the larger sample on its own. And writing down 0.580 comes from 1 − 0.420, working out the probability that the bottle top lands the other way up instead of open end up.
- (a) 60 — Method: because the counter goes back each time, every draw is the same experiment, so the expected number of reds is the number of draws multiplied by P(red). Working: there are 8 red counters out of 20, so P(red) = 8 ÷ 20 = 0.4. Over 150 draws the expected number of reds is 150 × 0.4 = 60. Answer: about 60 red counters would be expected. The distractors: 90 uses P(blue) by mistake, 150 × 12 ÷ 20 = 90; 100 comes from writing the probability from the red to blue ratio of 8 to 12, giving 150 × 8 ÷ 12 = 100; 75 comes from treating red and blue as equally likely because there are two colours, which gives 150 ÷ 2 = 75.
- (c) 0.37 — Method: pool the two runs into one combined set of results, then find the relative frequency of red across all of the spins together. Working: total reds = 16 + 21 = 37. Total spins = 40 + 60 = 100. Relative frequency = 37 ÷ 100 = 0.37. Answer: 0.37. Watch out: writing down 0.40 uses only the first run, 16 ÷ 40, and throws away the extra evidence from the second 60 spins. Writing down 0.35 uses only the second run, 21 ÷ 60, and throws away the first run instead. And writing down 0.375 averages the two runs' separate rates, (0.40 + 0.35) ÷ 2, which treats a run of 40 spins and a run of 60 spins as equally weighted, when pooling the actual counts gives the larger run its fair share of influence.
- (c) £3.10 — Aisha's tickets cost 8 × 50p = £4.00. Her expected winnings are (8/400) × £45 = £0.90, since she holds 8 of the 400 tickets. Her expected loss is the cost minus the expected winnings: £4.00 − £0.90 = £3.10. Writing £4.00 is wrong because it is only the cost of her tickets, with no account taken of the expected winnings she might get back. Writing £0.90 is wrong because that is her expected WINNINGS, not her loss — the cost has not been subtracted. Writing £3.89 is wrong because it uses 1 ticket instead of her actual 8 tickets when working out the expected winnings: (1/400) × £45 = £0.1125, giving £4.00 − £0.11 = £3.89. Aisha should expect to lose £3.10.
- (b) 57/100 — Pooling both trials: total heads = 24 + 33 = 57, total flips = 40 + 60 = 100, so the combined relative frequency is 57/100, which is already in its simplest form since 57 and 100 share no common factor. Averaging the two separate relative frequencies instead, (24/40 + 33/60) ÷ 2 = (0.6 + 0.55) ÷ 2 = 0.575 = 23/40, treats the two trials as equally weighted even though Ben made more flips, which is not correct. Using only Leah's data gives 24/40 = 3/5. Using only Ben's data gives 33/60 = 11/20.
- (c) Noah — 11/32 = 0.34375, above 34%. — Converting 11/32 to a decimal gives 11 ÷ 32 = 0.34375, which is greater than 34% (0.34), so Noah has the better save rate: 'Noah — 11/32 = 0.34375, above 34%.' Comparing the raw numbers 34 and 11 directly, without converting the fraction to the same form, gives 'Elin — 34 is bigger than 11.' Treating a larger denominator as meaning a bigger value, rather than smaller equal shares, gives 'Noah — 32 is a bigger denominator.' Rounding 34.375% to 34% to the nearest whole percent hides the difference and gives 'Equal — both round to 34% to the nearest percent.'
- (b) 0.55 — Method: it is easier to find the probability that Ffion wins NEITHER stage, then subtract that from 1. Working: P(lose stage 1) = 1 − 0.4 = 0.6, and P(lose stage 2) = 1 − 0.25 = 0.75. P(neither) = 0.6 × 0.75 = 0.45. P(at least one) = 1 − 0.45 = 0.55. Answer: 0.55. Watch out: adding the two win probabilities, 0.4 + 0.25 = 0.65, treats winning both as impossible and overcounts — that is not how independent probabilities combine. Multiplying the two win probabilities, 0.4 × 0.25 = 0.1, gives the probability of winning BOTH stages, not at least one. And stopping at 0.45, the probability of winning neither stage, forgets the final step of subtracting from 1.
- (b) 24/125 — On the fiction branch, 160 − 112 = 48 books were returned late. None of the non-fiction books were late, so the total number of late books is 48, out of 250 books altogether: 48/250 = 24/125. Writing 3/10 is wrong because it divides the 48 late fiction books by the fiction total (160) instead of the whole library total (250). Writing 56/125 is wrong because 112/250 simplifies to 56/125, and 112 is the number of fiction books returned ON TIME, not late. Writing 9/25 is wrong because 90/250 simplifies to 9/25, and 90 is simply the number of non-fiction books, which has nothing to do with late returns. The probability is 24/125.
- (b) 50% — The total number of customers who bought a cake is 54 + 21 = 75, combining both hot-drink and non-hot-drink customers. As a percentage of all 150 customers, this is (75 ÷ 150) × 100 = 50%. Choosing 36% comes from only counting the hot-drink customers who bought a cake, (54 ÷ 150) × 100 = 36%, and forgetting the 21 non-hot-drink customers who also bought a cake. Choosing 14% comes from only counting the non-hot-drink customers who bought a cake, (21 ÷ 150) × 100 = 14%, and forgetting the 54 hot-drink customers who also bought a cake. Choosing 60% comes from dividing by the hot-drink total of 90 instead of the grand total of 150, (54 ÷ 90) × 100 = 60%.
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