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 clinic recorded 300 booked appointments using a frequency tree. The first branch splits them into 210 adult appointments and the rest child appointments. Of the adult appointments, 189 were attended and the rest were missed. Of the child appointments, 81 were attended. Work out the probability that a booked appointment, chosen at random from the 300, was missed. Give your answer as a fraction in its simplest form.
- 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.A frequency tree records the results of 160 patients who took a new medicine. It splits them into those who reported side effects and those who did not. 15% of the patients reported side effects. Work out how many of the 160 patients did not report side effects.
- 4.A survey of 160 employees at a company recorded whether they cycle to work, using a frequency tree. The first branch splits them into 90 who work full-time and 70 who work part-time. Of the full-time employees, 27 cycle to work. Of the part-time employees, 14 cycle to work. Work out the probability that an employee, chosen at random from the 160, cycles to work. Give your answer as a fraction in its simplest form.
- 5.An archer hits the bullseye with probability 0.24 on any one shot. She wants to know how many shots she must take to expect to hit the bullseye 12 times. Work out how many shots this is.
- 6.A basketball player takes two free throws, and the throws are independent. The probability of scoring on each throw is 0.6, and the probability of missing is 0.4. Work out the probability that she scores exactly one of the two throws.
- 7.At a fun run, each runner finishes, retires or is disqualified, and cannot do more than one of these. The probability that a runner finishes is 68.5% and the probability that a runner retires is 24.75%. Work out the probability, as a percentage, that a runner is disqualified.
- 8.A test for a medical condition is given to 1000 people. 50 of the people have the condition and 950 do not. The test is positive for 45 of the 50 people who have the condition, and it is also positive for 95 of the 950 people who do not have the condition. One of the people whose test is positive is picked at random. Work out the probability that this person has the condition.
- 9.A biased spinner is spun 200 times. It lands on red 70 times, on blue 50 times, and on green 80 times. Using these results, work out the expected number of times the spinner does NOT land on red, in 500 spins of the same spinner.
- 10.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.
- 11.A game uses a fair spinner with 5 equal sections numbered 1 to 5. A player wins £12 if the spinner lands on 5, and wins nothing otherwise. It costs £2 to play. The game is played 250 times. Work out the expected profit for the players, in total, over the 250 games.
- 12.A basketball player has a free-throw success probability of 0.75. She wants to expect to score 60 successful free throws. Work out how many free throws she needs to attempt.
- 13.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.
- 14.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?
- 15.A council surveyed 140 households about recycling. 85 of the households are in Zone A. 52 of the Zone A households recycle glass. 21 of the Zone B households do not recycle glass. Work out the probability that a household, chosen at random from the 140, recycles glass. Give your answer as a fraction in its simplest form.
Answer key
- (c) 1/10 — On the adult branch, 210 − 189 = 21 appointments were missed. There are 300 − 210 = 90 child appointments, and 90 − 81 = 9 of those were missed. In total, 21 + 9 = 30 appointments were missed, out of 300: 30/300 = 1/10. Writing 7/100 is wrong because 21/300 simplifies to 7/100, and 21 only counts the adult branch, leaving out the 9 missed child appointments. Writing 3/100 is wrong because 9/300 simplifies to 3/100, and 9 only counts the child branch, leaving out the 21 missed adult appointments. Writing 1/9 is wrong because it divides the 30 missed appointments by the 270 that were attended (300 − 30) instead of by the whole 300 booked. The probability is 1/10.
- (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.
- (d) 136 — 15% of 160 = 0.15 × 160 = 24 patients reported side effects, so 160 − 24 = 136 did not. Stopping after finding the number who reported side effects, 24, answers the wrong question — it is not the number who did NOT report them. Misreading '15%' as a raw count of 15 patients, rather than a percentage, gives 160 − 15 = 145. Subtracting 15% of 160 twice, 160 − 24 − 24 = 112, double-counts the side-effect group.
- (b) 41/160 — In total, 27 + 14 = 41 of the 160 employees cycle to work, so the probability is 41/160 (41 and 160 share no common factor, so this is already in its simplest form). Writing 27/160 is wrong because it only counts the full-time cyclists and leaves out the 14 part-time cyclists. Writing 41/90 is wrong because it uses the full-time total (90) as the denominator instead of the whole survey (160). Writing 1/5 is wrong because it only uses the part-time branch, simplifying 14/70 to 1/5 and ignoring the full-time cyclists completely. The probability is 41/160.
- (a) 50 — To find the number of shots needed for an expected 12 hits, divide the number of hits wanted by the probability of a hit: 12 ÷ 0.24 = 50. Multiplying the number of hits by the probability instead of dividing gives 12 × 0.24 = 2.88, which rounds to 3 shots. Rounding 0.24 to 0.25 before dividing gives 12 ÷ 0.25 = 48. Using the probability of missing, 1 − 0.24 = 0.76, instead of the probability of hitting, gives 12 ÷ 0.76 = 15.79, which rounds to 16.
- (a) 0.48 — There are two ways to score exactly one throw: scoring on the first and missing the second, 0.6 × 0.4 = 0.24, or missing the first and scoring the second, 0.4 × 0.6 = 0.24. Adding these gives 0.24 + 0.24 = 0.48. Choosing 0.24 comes from working out only one of the two paths and forgetting the other one also gives exactly one score. Choosing 0.36 comes from working out the probability of scoring BOTH throws, 0.6 × 0.6 = 0.36, instead of exactly one. Choosing 0.84 comes from working out the probability of scoring AT LEAST one throw, 1 − 0.4 × 0.4 = 0.84, instead of exactly one.
- (a) 6.75% — Finishing, retiring and being disqualified are exhaustive, so the three percentages sum to 100%: 100% − 68.5% − 24.75% = 6.75%. Adding the two given percentages instead of subtracting them from 100% gives 68.5% + 24.75% = 93.25%, the combined probability of finishing or retiring, not of being disqualified. Subtracting only the retiring percentage from 100% and forgetting the finishing percentage gives 100% − 24.75% = 75.25%. Subtracting only the finishing percentage and forgetting the retiring percentage gives 100% − 68.5% = 31.50%.
- (c) 9/28 — Method: two linked steps. Total everyone whose test is positive, since the person picked is known to be one of them, then divide the positive tests that belong to people with the condition by that total. Working: 45 positive tests come from people who have the condition and 95 come from people who do not, so 140 tests are positive. The people with the condition give 45/140, and dividing the numerator and the denominator by 5 gives 9/28. Answer: the probability is 9/28. The distractors: 9/10 is 45/50, the probability of a positive test given that the person has the condition, which is the condition and the event the wrong way round and is the figure a candidate quotes when the two are confused; 9/200 is 45/1000, dividing by everyone tested rather than by the 140 who tested positive; 1/20 is 50/1000, the probability that a person has the condition before the test result is used at all.
- (c) 325 — Method: first find the relative frequency of NOT landing on red from the 200 spins, then scale that up to 500 spins. Working: non-red results = 50 + 80 = 130, out of 200 spins, so P(not red) = 130 ÷ 200 = 0.65. Expected non-red results in 500 spins = 500 × 0.65 = 325. Answer: 325. Watch out: writing down 175 finds the expected number of RED results instead, 70 ÷ 200 × 500 = 175, answering the opposite of what was asked. Writing down 250 assumes landing red or not landing red must be a fair 50-50 split, but the spinner is biased and the actual results do not split evenly. And writing down 130 stops after finding how many of the 200 spins were non-red and forgets to scale that figure up to the 500 spins asked for.
- (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.
- (b) £100 — Over 250 games, the expected total winnings are 250 × (1/5) × £12 = £600, since a player wins on 1 of the 5 equally likely sections. The total cost of playing is 250 × £2 = £500. The players' expected profit is the winnings minus the cost: £600 − £500 = £100. Writing £500 is wrong because that is only the total cost of playing, without any winnings included. Writing £600 is wrong because that is only the total expected winnings, without subtracting what was paid to play. Writing £2,500 is wrong because it assumes a win on every single game (250 × £12 = £3,000) instead of using the 1-in-5 probability, then subtracts the cost: £3,000 − £500 = £2,500. The players' expected profit over the 250 games is £100.
- (b) 80 — To find the number of attempts needed, divide the target number of successes by the probability of success: 60 ÷ 0.75 = 80. Writing 45 is wrong because 60 × 0.75 = 45 multiplies instead of dividing — that is the number of successes expected from 60 attempts, not the number of attempts needed for 60 successes. Writing 240 is wrong because 60 ÷ 0.25 = 240 uses 0.25, the probability of MISSING, instead of 0.75, the probability of scoring. Writing 90 is wrong because it comes from misremembering 0.75 as 2/3 and dividing by that instead: 60 ÷ (2/3) = 90. She needs to attempt 80 free throws.
- (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.
- (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) 43/70 — There are 140 − 85 = 55 Zone B households, and 55 − 21 = 34 of them recycle glass. In total, 52 + 34 = 86 households recycle glass, out of 140: 86/140 = 43/70. Writing 13/35 is wrong because 52/140 simplifies to 13/35, and 52 only counts Zone A, leaving out the 34 Zone B recyclers. Writing 17/70 is wrong because 34/140 simplifies to 17/70, and 34 only counts Zone B, leaving out the 52 Zone A recyclers. Writing 73/140 is wrong because it adds the 52 Zone A recyclers to the 21 Zone B households that do NOT recycle, mixing up two different groups instead of adding the two recycling groups. The probability is 43/70.
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