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 fair six-sided dice is rolled 300 times. Work out how many more times you would expect it to land on an even number than on a six.
- 2.A bag contains 45 sweets. 18 of the sweets are lemon flavour and the rest are orange. A sweet is taken at random, its flavour is recorded, and it is put back in the bag. This is repeated 200 times. Work out how many times you would expect an orange sweet to be taken.
- 3.A Venn diagram shows two sets, P and Q, inside a universal set. n(P) = 34, n(Q) = 27, n(P ∩ Q) = 11, and n(ξ) = 90, where ξ is the universal set. Work out n((P ∪ Q)′), the number of elements in neither P nor Q.
- 4.A fair six-sided dice is rolled 150 times. The table shows how many times each number came up: 1 came up 22 times, 2 came up 27 times, 3 came up 24 times, 4 came up 34 times, 5 came up 21 times and 6 came up 22 times. The theoretical probability of each number is 1/6. Which number is most over-represented compared with its theoretical probability?
- 5.Two games are offered at a school fair, and every one of the 40 pupils in a class plays each game once. In Game A, the probability of winning a prize is 0.2 and the prize is worth £5. In Game B, the probability of winning a prize is 0.1 and the prize is worth £12. Work out which game gives a higher expected total value of prizes for the class, and by how much.
- 6.150 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.
- 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 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.A fair six-sided dice is rolled once. Work out the probability that the score is an odd number. Give your answer as a decimal.
- 10.A bag is known to contain red, blue and green counters in equal numbers, so the theoretical probability of taking each colour is 1/3. In 90 trials with replacement, red was taken 38 times, blue was taken 26 times and green was taken 26 times. Which colour is over-represented compared with its theoretical probability?
- 11.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.
- 12.Priya spins a fair spinner with P(red) = 0.3, and separately flips a fair coin with P(heads) = 0.5. Using a tree diagram, work out the probability that she gets red AND heads.
- 13.On the probability scale from 0 (impossible) to 1 (certain), evens is exactly halfway. An event is 3/10 more likely than evens. Work out the probability of that event, giving your answer as a decimal.
- 14.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.
- 15.The probability that a plant grown from a seed packet germinates is 7/10. Work out the probability that it does not germinate.
Answer key
- (c) 100 — There are 3 even numbers on a fair dice (2, 4 and 6), so the probability of landing on an even number is 3/6 = 1/2, and 300 × 1/2 = 150. The probability of landing on a six is 1/6, so 300 × 1/6 = 50. The dice is expected to land on an even number 150 − 50 = 100 more times than on a six. Writing 50 is wrong because that is just the expected number of sixes on its own, without comparing it to the expected number of evens. Writing 150 is wrong because that is just the expected number of evens on its own, without subtracting the sixes. Writing 200 is wrong because it adds the two expected frequencies together (150 + 50 = 200) instead of finding the difference between them. The dice is expected to land on an even number 100 more times than on a six.
- (c) 120 — 27 of the 45 sweets are orange (45 − 18 = 27), so the probability of taking an orange sweet is 27/45 = 3/5, and 200 × 3/5 = 120. Writing 80 is wrong because 200 × 18/45 = 80 uses the LEMON sweets' fraction instead of orange. Writing 182 is wrong because it takes the 18 lemon sweets away from the 200 repeats (200 − 18 = 182), applying the ‘the rest are orange’ subtraction to the number of goes instead of to the 45 sweets in the bag. Writing 27 is wrong because it is simply the number of orange sweets in the bag — it has not been scaled up to account for the 200 repeats. The expected number of times an orange sweet is taken is 120.
- (b) 40 — n(P ∪ Q) = n(P) + n(Q) − n(P ∩ Q) = 34 + 27 − 11 = 50. The complement is everyone outside both sets: n((P ∪ Q)′) = 90 − 50 = 40. Adding P and Q without subtracting the overlap gives 34 + 27 = 61, so 90 − 61 = 29 double-subtracts the 11 who are in both. Reporting n(P ∪ Q) itself, 50, forgets to take the complement at all. Subtracting only n(P) from the universal set, 90 − 34 = 56, ignores set Q altogether.
- (b) 4 — With 150 rolls and probability 1/6 for each number, the expected count is 150 ÷ 6 = 25. Comparing each actual count with 25: 1 is 22 (3 below), 2 is 27 (2 above), 3 is 24 (1 below), 4 is 34 (9 above), 5 is 21 (4 below) and 6 is 22 (3 below). Number 4 is furthest above its expected count, so it is the most over-represented. Number 2 is also above its expected count, but by only 2, far less than 4's 9. Number 3's count of 24 is below the expected 25, so it is under-represented, not over. Number 6's count of 22 is also below the expected 25, so it too is under-represented.
- (a) Game B, by £8 — Game A's expected total is 40 × 0.2 × £5 = £40. Game B's expected total is 40 × 0.1 × £12 = £48. Game B is higher, by £48 − £40 = £8. Writing 'Game A, by £8' is wrong because it has the right difference but the wrong game — Game A's total (£40) is actually LOWER than Game B's, not higher. Writing 'Game B, by £48' is wrong because £48 is Game B's whole expected total, not the DIFFERENCE between the two games. Writing 'Game A, by £40' is wrong in the same way, using Game A's whole total as if it were the margin, and naming the wrong game as the winner. Game B gives the higher expected total, by £8.
- (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.
- (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) 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.
- (c) 0.5 — Method: count the favourable outcomes, write them over the total number of equally likely outcomes and then divide to turn the fraction into a decimal. Working: the odd scores are 1, 3 and 5, which is 3 of the 6 equally likely scores, so the probability is 3/6, and 3 ÷ 6 = 0.5. Answer: 0.5, the middle of the 0 to 1 probability scale. The distractors: 0.33 comes from listing only 3 and 5 as odd and working out 2 ÷ 6; 0.17 comes from giving the probability of one particular odd score, 1 ÷ 6; 0.3 comes from writing '3 out of 6' as 0.3, reading the 3 straight off as tenths instead of dividing.
- (c) Red — Theoretical probability is 1/3 ≈ 0.333 for each colour. Red's relative frequency is 38/90 ≈ 0.422, above 1/3, so red is over-represented. Blue's relative frequency is 26/90 ≈ 0.289, below 1/3, so blue is under-represented, not over. Green's relative frequency is also 26/90 ≈ 0.289, below 1/3 for the same reason. Since red's relative frequency clearly exceeds 1/3, it is not true that none of the colours are over-represented.
- (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.
- (c) 0.15 — Method: for two independent events, multiply along the branches of the tree to find the probability of both outcomes happening together. Working: P(red and heads) = P(red) × P(heads) = 0.3 × 0.5 = 0.15. Answer: 0.15. Watch out: adding the two probabilities, 0.3 + 0.5 = 0.8, does not give the probability of both — probabilities along one path of a tree are multiplied, not added. Writing down 0.5 ignores the spinner altogether and gives only the coin's probability. And writing down 0.65 is the probability of red OR heads, which is 0.3 + 0.5 − 0.15 = 0.65, a different question from the one asked here.
- (d) 0.80 — Evens is exactly halfway along the scale, at 0.5, and 3/10 written as a decimal is 0.3. A probability 3/10 higher than evens is 0.5 + 0.3 = 0.80. Giving 0.30 as the answer converts the increase but never adds it to the value of evens. Adding the increase to 1, the 'certain' end of the scale, instead of to evens gives 1 + 0.3 = 1.30, which cannot be a probability. Writing 3/10 as 0.03 instead of 0.3, a place-value slip, gives 0.5 + 0.03 = 0.53.
- (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) 3/10 — Germinating and not germinating are exhaustive, so their probabilities sum to 1: 1 − 7/10 = 3/10. Writing 7/10 again gives the probability that the plant DOES germinate, not its complement. Putting the difference 10 − 7 = 3 over the original numerator instead of the original denominator gives 3/7. Inverting the correct answer, swapping its numerator and denominator, gives 10/3, which is impossible as a probability since it is greater than 1.
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