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.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.
- 2.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.
- 3.150 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.
- 4.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?
- 5.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.
- 6.At a sports centre, 45% of the members are aged under 18. 60% of the members aged under 18 swim each week. 20% of the members aged 18 or over swim each week. Work out the percentage of all the members who swim each week.
- 7.At a school fête, a tombola stall costs £1.50 to play. The probability of winning is 0.2, and the prize is worth £6. Work out the stall's expected profit, on average, from each game played.
- 8.A machine produces bolts. In a sample of 250 bolts, 15 are faulty. Work out the relative frequency of a bolt being faulty, giving your answer as a fraction in its simplest form.
- 9.Two ordinary fair dice are rolled. Work out the probability that the difference between the two scores is 1.
- 10.A market research company found that, out of 120 shoppers surveyed, 42 preferred paper bags to plastic bags. A supermarket expects 900 shoppers this week. Using this relative frequency, work out how many of the 900 shoppers would be expected to prefer paper bags.
- 11.An ordinary pack of 52 playing cards contains 13 hearts. One card is taken at random from the pack. Work out the probability that the card is not a heart. Give your answer as a percentage.
- 12.A charity tombola has 30 tickets, 6 of which win a prize. Priya buys a ticket at random and does not return it. Her friend Tom then buys a second ticket at random from the remaining tickets. Work out the probability that both Priya and Tom win a prize.
- 13.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.
- 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 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.
Answer key
- (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.
- (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) 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) 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.'
- (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.
- (b) 38% — Method: the two swimming percentages are quoted inside different age groups, so weight each one by the size of its group and add the two results. Working: the under 18s are 45% of the members and 60% of them swim, giving 0.45 × 60 = 27% of all the members. The members aged 18 or over are 55% of the members and 20% of them swim, giving 0.55 × 20 = 11% of all the members. Adding these gives 38%. Answer: 38% of the members swim each week. The distractors: 80% comes from adding 60% and 20% straight off, treating two rates quoted inside different groups as though they could be added; 40% is the mean of 60% and 20%, which would be right only if the two age groups were the same size, and they are not; 42% comes from pairing each swimming rate with the wrong age group, working out 0.45 × 20 added to 0.55 × 60.
- (b) £0.30 profit for the stall — The stall keeps the £1.50 entry fee whatever happens, and expects to pay out prize × probability of winning = £6 × 0.2 = £1.20 on average. So its expected profit per game is £1.50 − £1.20 = £0.30. Reporting the expected pay-out of £1.20 itself as the profit forgets that the stall also keeps the entry fee. Assuming the player always wins gives an expected cost of £6 − £1.50 = £4.50, treated as a loss for the stall. Using the probability of NOT winning, 0.8, to find the expected pay-out gives £6 × 0.8 = £4.80, and £1.50 − £4.80 = −£3.30, a £3.30 loss.
- (d) 3/50 — Relative frequency is the number of faulty bolts divided by the total sample size: 15/250, which simplifies to 3/50 by dividing both the numerator and the denominator by 5. Using 235, the number of bolts that were NOT faulty, as the denominator instead of the total 250 gives 15/235, which simplifies to 3/47. Inverting the fraction, dividing the total by the number of faulty bolts instead of the other way round, gives 250/15, which simplifies to 50/3 — a value greater than 1, which cannot be a probability. Simplifying by dividing the numerator and the denominator by different numbers, 15 ÷ 15 = 1 and 250 ÷ 25 = 10, gives 1/10.
- (d) 5/18 — Method: list every ordered pair of dice scores whose difference is 1, then divide by the 36 equally likely pairs. Working: the pairs with a difference of 1 are (1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4), (5, 6) and (6, 5), which is 10 pairs out of 36, cancelling down to 5/18. Answer: 5/18. Watch out: writing down 5/36 lists only the 5 pairs going up, (1, 2), (2, 3), (3, 4), (4, 5) and (5, 6), and misses that each one has a matching pair the other way round, such as (2, 1) — the two dice are different objects, and order matters, so each of those 5 gaps counts twice. Writing down 1/6 treats the six possible differences, 0 to 5, as equally likely and picks 1 out of 6 of them, but a difference of 1 is reached by far more pairs of scores than a difference of 5 is, so the six differences are not equally likely. And writing down 1/4 comes from adding the two dice's outcome counts instead of multiplying them, 6 + 6 = 12, and grouping the six scores into just three non-overlapping pairs one apart, {1, 2}, {3, 4} and {5, 6}, giving 3 out of that wrong pool of 12.
- (d) 315 — The relative frequency from the survey is 42 ÷ 120 = 0.35, so the expected number who prefer paper bags among 900 shoppers is 0.35 × 900 = 315. Giving 42 as the answer reuses the original survey count without scaling it up to 900 shoppers at all. Finding the expected number who prefer PLASTIC bags instead of paper, using the relative frequency 78 ÷ 120 = 0.65, gives 0.65 × 900 = 585. Using 1000 shoppers instead of the 900 actually stated gives 0.35 × 1000 = 350.
- (d) 75% — Method: count how many cards are not hearts, write that count over the total number of cards, cancel the fraction down and then turn it into a percentage. Working: 52 − 13 = 39 cards are not hearts, so the probability is 39/52; dividing the numerator and the denominator by 13 gives 3/4, and 3/4 = 0.75, so 0.75 × 100 = 75. Answer: 75%, three quarters of the way along the 0 to 1 scale. The distractors: 25% comes from giving the probability that the card is a heart, 13 out of 52, which cancels to 1/4; 50% comes from reading 'not a heart' as 'not a red card' and halving the pack; 39% comes from writing the count of 39 cards straight down as the percentage without comparing it with the 52 cards in the pack.
- (a) 1/29 — The probability that Priya's ticket wins is 6/30. Since her ticket is not returned, there are now only 5 winning tickets left out of 29 tickets in total, so the probability that Tom's ticket also wins is 5/29. Multiplying these, 6/30 × 5/29 = 30/870 = 1/29. A candidate who answers 1/25 has treated Priya's ticket as returned, using 6/30 twice. A candidate who answers 1/30 has correctly reduced the winning tickets to 5 for Tom but forgotten to reduce the total number of tickets, using 5/30 instead of 5/29. A candidate who answers 11/59 has added the numerators and added the denominators, (6+5)/(30+29), instead of multiplying.
- (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.
- (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.
- (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.
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