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.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.
- 2.A spinner lands on red, on blue or on green, and it cannot land on more than one colour at a time. The probability that it lands on red is 0.3 and the probability that it lands on blue is 0.5. Work out the probability that it lands on red or on blue.
- 3.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.
- 4.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.
- 5.A fair four-sided dice, numbered 1 to 4, is rolled twice. Every outcome is listed as an ordered pair, first roll then second roll, such as (2, 3). Work out how many different outcomes are in the list.
- 6.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.
- 7.A student is estimating the probability that a spinner lands on red. In her first 85 spins it landed on red 34 times. She then spins it 165 more times, and in those it lands on red 58 times. Work out the best estimate of the probability of red from all 250 spins. Give your answer as a decimal, correct to 3 decimal places.
- 8.A spinner has equal sections, and the probability that it lands on blue in a single spin is 1/4. It is spun twice, and the two spins are independent. Work out the probability that it lands on blue on both spins.
- 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.A survey of 60 readers records: 21 read only fiction, 17 read only non-fiction, 14 read both, and the rest read neither. Work out the probability that a randomly chosen reader reads exactly one of the two types, giving your answer as a fraction in its simplest form.
- 11.The probability that a seed fails to germinate is 1/5. Three seeds are planted independently. Work out the probability that at least one of the three seeds germinates.
- 12.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.
- 13.Spinner E has 4 equal sections, numbered 1 to 4. Spinner F has 6 equal sections, numbered 1 to 6. Both spinners are spun once. Work out the probability of getting at least one 3.
- 14.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.
- 15.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.
Answer key
- (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.
- (a) 0.8 — Method: the spinner cannot land on red and blue at the same time, so the two events are mutually exclusive and their probabilities are added. Working: 0.3 + 0.5, lining the decimal points up. Answer: 0.8, which also tells you that the remaining colour, green, has probability 0.2 because the three must add to 1. The distractors: 0.2 comes from subtracting 0.3 from 0.5 instead of adding the two probabilities; 0.15 comes from multiplying 0.3 by 0.5 instead of adding them; 0.4 comes from finding the mean of 0.3 and 0.5 rather than their total.
- (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.
- (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) 16 — Method: keep the two rolls in order, first roll then second roll, and count every ordered pair the dice can land on, so a (2, 3) result is counted separately from a (3, 2) result. Working: the first roll can land on any of 4 numbers, and for each of those the second roll can also land on any of 4 numbers, giving 4 × 4 = 16 ordered pairs. Answer: 16. Watch out: writing down 8 comes from adding the two rolls' counts, 4 + 4, instead of multiplying them. Writing down 10 comes from listing unordered pairs, so (2, 3) and (3, 2) are counted as the same entry — that gives the 4 doubles plus 6 mixed pairs, 10 in total, instead of 16 ordered pairs. And writing down 4 comes from counting only one roll's outcomes and never pairing the two rolls together at all.
- (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.
- (a) 0.368 — Combining both samples, the spinner landed on red 34 + 58 = 92 times out of a total of 85 + 165 = 250 spins, so the best estimate of the probability is 92/250 = 0.368. Writing 0.400 is wrong because it uses only the first sample, 34/85 = 0.400, ignoring the extra 165 spins recorded afterwards. Writing 0.352 is wrong because it uses only the second sample, 58/165 = 0.352 (to 3 decimal places), ignoring the first 85 spins. Writing 0.376 is wrong because it averages the two separate estimates, (0.400 + 0.352) ÷ 2 = 0.376, instead of combining the actual numbers of reds and spins across both samples. The best estimate of the probability that the spinner lands on red, using all 250 spins, is 0.368.
- (a) 1/16 — The two spins are independent, so multiply the probability of blue on each spin: 1/4 × 1/4 = 1/16. Choosing 1/2 comes from adding the two probabilities instead of multiplying, 1/4 + 1/4 = 1/2. Choosing 1/8 comes from wrongly treating the second spin as having a 1/2 chance of blue instead of the given 1/4, giving 1/4 × 1/2 = 1/8. Choosing 1/4 comes from giving the probability for a single spin and forgetting that the spinner is spun twice.
- (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.
- (d) 19/30 — Exactly one means only fiction or only non-fiction, not both: 21 + 17 = 38 out of the 60 readers, which simplifies to 19/30. Including the 14 who read both as well gives 21 + 17 + 14 = 52, so 52/60 = 13/15 — that is at least one, not exactly one. Using only the both-count, 14, as the numerator gives 14/60 = 7/30, the probability of reading both, not exactly one. Using 52, the number who read at least one type, as the denominator instead of the full 60 readers surveyed gives 38/52 = 19/26.
- (d) 124/125 — The probability that a seed germinates is 1 − 1/5 = 4/5, so the probability that all three seeds fail to germinate is 1/5 × 1/5 × 1/5 = 1/125. The probability that at least one germinates is 1 − 1/125 = 124/125. Choosing 4/5 comes from giving the probability that a single seed germinates, forgetting to combine all three seeds. Choosing 64/125 comes from working out the probability that ALL three seeds germinate, 4/5 × 4/5 × 4/5 = 64/125, instead of at least one. Choosing 12/125 comes from working out the probability that EXACTLY one seed germinates, 3 × 4/5 × 1/5 × 1/5 = 12/125, instead of at least one.
- (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) 3/8 — There are 4 × 6 = 24 equally likely outcomes. The outcomes with no 3 at all have Spinner E showing 1, 2 or 4 and Spinner F showing 1, 2, 4, 5 or 6, giving 3 × 5 = 15 outcomes. So the outcomes with at least one 3 are 24 − 15 = 9, and the probability is 9/24 = 3/8. Choosing 5/12 comes from adding the two individual probabilities of a 3, 1/4 + 1/6, which counts the outcome where both spinners show 3 twice over. Choosing 1/4 comes from only counting the case where Spinner E shows 3, and forgetting the outcomes where Spinner F shows 3 instead. Choosing 5/8 comes from working out the probability of getting no 3 at all, 15/24 = 5/8, and giving that as the final answer instead of subtracting it from 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) 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.
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