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 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.
- 2.A bag contains 5 letter tiles spelling the word MATHS, with one letter on each tile. Two tiles are drawn at random, one after the other, without the first tile being put back. Work out the probability that both tiles drawn are consonants.
- 3.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.
- 4.At a sports club, the numbers of members who play tennis and badminton are: 50 members play tennis, 40 members play badminton, and 18 members play both tennis and badminton. A member who plays tennis is chosen at random. Work out the probability that this member also plays badminton.
- 5.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.
- 6.Two events are independent. The first has probability 2/5 and the second has probability 1/3. Work out the probability that both events happen.
- 7.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.
- 8.A factory finds that the probability a randomly chosen light bulb is defective is 0.035. In a batch of 4,000 bulbs, work out how many bulbs you would expect to work correctly.
- 9.At a gym, 70 members are asked whether they use the pool or the sauna. 38 use the pool, 30 use the sauna, and 12 use neither. Work out n(P ∩ S), the number of members who use both the pool and the sauna.
- 10.A box holds 5 blue pens and 7 black pens. Two pens are taken at random, one at a time, and are not put back. The first pen taken is black. Work out the probability that the second pen taken is blue.
- 11.A two-way table records how 130 pupils travel to school. 70 of the pupils are girls and the rest are boys. 42 of the girls walk to school and the rest of the girls cycle. 33 of the boys walk to school. Work out what fraction of the girls walk to school.
- 12.Five raffle tickets numbered 1 to 5 are in a bag. Two tickets are drawn one after the other and the first is not put back. Work out the probability that ticket 1 is drawn first and ticket 2 is drawn second.
- 13.Bag A contains 3 red balls and 2 blue balls. Bag B contains 1 red ball and 3 blue balls. One ball is taken at random from each bag. Work out the probability that both balls are red.
- 14.The probability that Kofi passes his driving test on any attempt is 0.6, and each attempt is independent of the others. Work out the probability that he fails both his first two attempts.
- 15.Of the 70 students on the Year 10 geography trip, 42 are girls and the rest are boys. 24 of the girls and 10 of the boys brought a packed lunch. A student who did not bring a packed lunch is chosen at random. Work out the probability that this student is a girl.
Answer key
- (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.
- (d) 3/5 — The first tile drawn is a consonant with probability 4/5, since 4 of the 5 letters in MATHS are consonants. Since it is not replaced, 3 consonants remain out of the 4 tiles left, so the second tile is a consonant with probability 3/4. Multiplying gives 4/5 × 3/4 = 3/5. Choosing 4/5 comes from only using the probability for the first draw and forgetting the second draw changes once a tile is removed. Choosing 3/4 comes from only using the probability for the second draw on its own. Choosing 16/25 comes from treating the two draws as independent, using 4/5 × 4/5 = 16/25, which would only be correct if the first tile were replaced before the second draw.
- (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.
- (a) 9/25 — Method: P(badminton | tennis) = n(tennis and badminton) ÷ n(tennis) — restrict to the tennis-players, then find what fraction of them also play badminton. Working: n(tennis and badminton) = 18, n(tennis) = 50, so P(badminton | tennis) = 18/50 = 9/25. Answer: 9/25. Watch out: dividing by 40 (the badminton total) finds P(tennis | badminton) instead of P(badminton | tennis) — the wrong direction. Dividing by 90 (all the members named in the question) ignores that you already know the member plays tennis. And dividing by 72 (50 + 40 − 18, the number who play at least one of the two sports) answers a question about the union, not the condition you were given.
- (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.
- (c) 2/15 — Method: for independent events, the probability that both happen is the product of the two probabilities. Working: the calculation is 2/5 × 1/3. Multiplying fractions gives 2 × 1 = 2 on the top and 5 × 3 = 15 on the bottom. Answer: the probability is 2/15. The distractors: 11/15 comes from adding the two probabilities, 6/15 + 5/15, instead of multiplying them; 3/8 comes from adding the numerators and the denominators separately; 1/15 comes from subtracting one probability from the other, 6/15 − 5/15.
- (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.
- (a) 3,860 — The probability a bulb works correctly is the complement of being defective: 1 − 0.035 = 0.965. Expected number working correctly = 0.965 × 4,000 = 3,860. Using the probability of being defective instead of its complement gives 4,000 × 0.035 = 140, the expected number of DEFECTIVE bulbs, not working ones. Shifting the decimal point in the complement, using 0.0965 instead of 0.965, gives 4,000 × 0.0965 = 386. Assuming every bulb works, ignoring the 0.035 probability altogether, gives the full batch of 4,000.
- (b) 10 — The number who use the pool or the sauna (or both) is the total minus those who use neither: 70 − 12 = 58. Since pool + sauna double-counts the overlap, n(P ∩ S) = 38 + 30 − 58 = 10. Adding the pool and sauna counts without subtracting the overlap at all gives 38 + 30 = 68, more members than are in the whole gym. Subtracting the sauna count from the union, 58 − 30 = 28, actually finds the number who use ONLY the pool, not both. Reporting the 'neither' count, 12, confuses it with the 'both' region — they describe opposite corners of the diagram.
- (d) 5/11 — Method: the pen already taken was black, so update the contents of the box before working out the second probability. Working: the box held 12 pens and one black pen has gone, so 11 pens remain. None of the blue pens has been taken, so all 5 are still there, and the probability is 5/11, which will not cancel. Answer: the probability is 5/11. The distractors: 5/12 uses the box as it was at the start, which is only correct if the first pen is put back; 4/11 takes one off the blue count as well as the total, as though the pen removed had been blue; 6/11 gives the probability that the second pen is black, carrying on with the colour of the first pen instead of the colour asked for.
- (b) 3/5 — The question asks about the girls only, so use the girls' total of 70 as the denominator: 42 out of 70 girls walk, giving 42/70 = 3/5. Choosing 21/65 comes from using the whole survey of 130 pupils as the denominator instead of just the 70 girls, 42/130 = 21/65. Choosing 33/70 comes from using the boys' walking count, 33, over the girls' total of 70, mixing up the two rows of the table. Choosing 2/5 comes from using the number of girls who CYCLE, 70 − 42 = 28, instead of the number who walk, giving 28/70 = 2/5.
- (b) 1/20 — Method: two draws with nothing put back are combined by multiplying, with the second probability worked out from the tickets still in the bag. Working: the first draw takes the wanted ticket with probability 1/5. That ticket is kept out, so 4 tickets remain and only one of them is the ticket wanted second, giving 1/4. Multiplying gives 1/20. Answer: the probability is 1/20. The distractors: 1/10 comes from ignoring the order and treating the draw as a choice of two tickets from five, of which there are ten; 1/25 comes from keeping the total at 5 for the second draw, which is what happens only if the first ticket is put back; 2/5 comes from counting the two wanted tickets over the five in the bag, as though one draw decided the whole question.
- (d) 3/20 — The probability of taking a red ball from Bag A is 3/5, and the probability of taking a red ball from Bag B is 1/4. Since the two picks are independent, the probabilities are multiplied: 3/5 × 1/4 = 3/20. A candidate who answers 4/9 has added the numerators and added the denominators, (3+1)/(5+4), instead of multiplying the two fractions. A candidate who answers 17/20 has added the two fractions, 3/5 + 1/4, instead of multiplying them. A candidate who answers 3/25 has misread Bag B as also containing 5 balls, using 1/5 instead of 1/4.
- (a) 0.16 — The probability that Kofi fails a single attempt is 1 − 0.6 = 0.4. Since the attempts are independent, the probability he fails both is 0.4 × 0.4 = 0.16. Choosing 0.36 comes from squaring the probability of PASSING instead, 0.6 × 0.6 = 0.36, which is the probability of passing both attempts, not failing both. Choosing 0.4 comes from giving the probability of failing just one attempt, forgetting to combine two attempts. Choosing 0.24 comes from multiplying the fail probability by the pass probability, 0.4 × 0.6 = 0.24, mixing up passing and failing between the two attempts.
- (c) 1/2 — Method: restrict to the students who did NOT bring a packed lunch, then find what fraction of that group are girls. Girls without lunch = 42 − 24 = 18. Boys = 70 − 42 = 28, so boys without lunch = 28 − 10 = 18. Total without lunch = 18 + 18 = 36. Working: P(girl | no lunch) = 18 ÷ 36 = 1/2. Answer: 1/2. Watch out: dividing 18 by 42 (the total number of girls) instead of by 36 finds P(no lunch | girl), the reverse conditional. Dividing by 70 (the whole trip) ignores that you already know the student did not bring a lunch. And using the 'brought a lunch' numbers (24 out of 34) answers the question for the wrong group entirely — you were asked about the students who did NOT bring one.
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