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.
Probability worksheet — GCSE Foundation
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- 1.150 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.
- 2.Two ordinary fair dice are rolled and the two scores are added together. Work out the probability that the total is a prime number.
- 3.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.
- 4.A bag contains 3 red counters, 2 blue counters and 2 green counters. One counter is taken at random from the bag. Work out the probability that the counter is red.
- 5.A fair six-sided dice is rolled 120 times. Work out how many of the rolls you would expect to give a score of 3 or more.
- 6.A frequency tree records how 90 pupils travel to school and whether they were late. The first pair of branches splits the 90 pupils into 50 who walk and 40 who do not walk. On the walking branch, 6 were late and 44 were not. On the other branch, 10 were late and 30 were not. One of the 90 pupils is chosen at random. Work out the probability that the pupil walks to school and was late.
- 7.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.
- 8.A fair spinner has 10 equal sections, numbered 1 to 10. The spinner is spun 150 times. Work out how many times you would expect it to land on a number greater than 7.
- 9.Two fair four-sided dice, numbered 1 to 4, are rolled and the two scores are added together. Work out the probability that the total is 5.
- 10.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?
- 11.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.
- 12.A quality inspector tests 145 light bulbs and finds that 33 are faulty. Work out the relative frequency of a bulb being faulty, as a percentage correct to 1 decimal place.
- 13.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.
- 14.A fair six-sided dice is rolled once. Write down the probability that the score is an even number.
- 15.A raffle sells 400 tickets at 50p each. There is one prize of £45. Aisha buys 8 tickets. Work out how much money Aisha should expect to lose from playing, giving your answer in pounds.
Answer key
- (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) 5/12 — The possible totals are 2 to 12. The prime totals are 2, 3, 5, 7 and 11, with 1, 2, 4, 6 and 2 ways to make them, giving 1 + 2 + 4 + 6 + 2 = 15 outcomes out of 36, so the probability is 15/36 = 5/12. Choosing 7/18 comes from forgetting that a total of 2 counts as prime, giving 14/36 = 7/18 instead. Choosing 19/36 comes from wrongly including a total of 9 as a prime number, adding its 4 ways to the correct 15 to get 19/36. Choosing 13/36 comes from forgetting the 2 ways of making a total of 11, giving 13/36 instead of 15/36.
- (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) 3/7 — Method: every counter is equally likely to be taken, so write the number of red counters over the total number of counters in the bag. Working: the bag holds 3 + 2 + 2 = 7 counters, of which 3 are red. Answer: 3/7, a value between 0 and 1 and just below the middle of the scale. The distractors: 4/7 comes from giving the probability that the counter is not red, counting the 2 blue and 2 green instead; 2/7 comes from counting one of the other colours by mistake and giving 2 counters over the total; 3/4 comes from writing the 3 red counters over the 4 counters that are not red instead of over all 7 counters.
- (a) 80 — Method: list the outcomes that count as a success, write the probability from them, then multiply by the number of rolls. Working: the scores of 3 or more are 3, 4, 5 and 6, which is 4 of the 6 equally likely scores, so the probability is 4/6, which cancels to 2/3. Over 120 rolls the expected number is 120 × 2 ÷ 3 = 80. Answer: about 80 of the rolls would be expected to give 3 or more. The distractors: 60 comes from reading a score of 3 or more as a score above 3 and counting only 4, 5 and 6, giving 120 × 3 ÷ 6 = 60; 40 is the expected number of rolls that are not 3 or more, 120 × 2 ÷ 6 = 40; 20 is 120 ÷ 6 and is the expected count for one single score.
- (d) 1/15 — Method: a probability read from a frequency tree is the count at the end of the branch you want, divided by the total number in the whole experiment. Working: the branch for walking followed by the branch for being late ends with 6 pupils, and the experiment covers all 90 pupils, so the probability is 6/90. Dividing the top and the bottom by 6 gives 1/15. Answer: the probability is 1/15. The distractors: 3/25 is 6/50 and comes from dividing the 6 by the 50 walkers rather than by the whole group, which answers a different question about walkers only; 8/45 is 16/90 and comes from counting every late pupil, the 6 walkers and the 10 others together, instead of only the late walkers; 1/9 is 10/90 and comes from reading the late count on the branch for pupils who do not walk.
- (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) 45 — Three of the ten numbers (8, 9 and 10) are greater than 7, so the probability is 3/10, and 150 × 3/10 = 45. Writing 105 is wrong because 150 × 7/10 = 105 uses the seven numbers that are NOT greater than 7 (1 to 7), the opposite of what is asked. Writing 60 is wrong because it counts 7, 8, 9 and 10 as four numbers greater than 7, wrongly including 7 itself: 150 × 4/10 = 60. Writing 50 is wrong because 150 ÷ 3 = 50 divides by the count of favourable numbers instead of multiplying by the correct fraction of the spinner. The expected number of spins landing on a number greater than 7 is 45.
- (b) 1/4 — There are 4 × 4 = 16 equally likely ordered outcomes for the two dice. The pairs that total 5 are: first dice 1 with second dice 4; first dice 2 with second dice 3; first dice 3 with second dice 2; and first dice 4 with second dice 1 — which is 4 outcomes, so the probability is 4/16 = 1/4. A candidate who answers 1/8 has listed only 2 of the four pairs, forgetting that first dice 1 with second dice 4 and first dice 4 with second dice 1 are separate outcomes because the dice are different. A candidate who answers 3/16 has found 3 pairs instead of 4, missing one from the list. A candidate who answers 1/16 has counted only a single pair, such as first dice 2 with second dice 3, and treated the order of the dice as not mattering.
- (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) 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.
- (b) 22.8% — Relative frequency as a percentage is the faulty count divided by the total, then multiplied by 100: 33 ÷ 145 × 100 = 22.76, which rounds to 22.8%. Giving 33.0% as the answer uses the frequency, 33, directly as a percentage without dividing by the total 145 at all. Rounding 22.76 down to 22.7% instead of up applies the wrong rounding direction at the first decimal place. Finding the relative frequency of the bulbs that were NOT faulty first: 145 − 33 = 112, and 112 ÷ 145 × 100 = 77.24, answers the opposite question and rounds to 77.2%.
- (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) 1/2 — Method: for equally likely outcomes the probability is the number of favourable outcomes divided by the total number of outcomes, written in its simplest form. Working: the dice can show 1, 2, 3, 4, 5 or 6, so there are 6 equally likely scores; the even scores are 2, 4 and 6, which is 3 of them, giving 3/6. Answer: 3/6 cancels to 1/2, which sits halfway along the 0 to 1 probability scale. The distractors: 1/3 comes from listing only 2 and 4 as the even scores and writing 2/6; 1/6 comes from giving the probability of one particular even score, such as the 2, rather than any even score; 2/3 comes from listing the even numbers as 0, 2, 4 and 6 and writing 4/6, forgetting that a dice has no 0 on it.
- (c) £3.10 — Aisha's tickets cost 8 × 50p = £4.00. Her expected winnings are (8/400) × £45 = £0.90, since she holds 8 of the 400 tickets. Her expected loss is the cost minus the expected winnings: £4.00 − £0.90 = £3.10. Writing £4.00 is wrong because it is only the cost of her tickets, with no account taken of the expected winnings she might get back. Writing £0.90 is wrong because that is her expected WINNINGS, not her loss — the cost has not been subtracted. Writing £3.89 is wrong because it uses 1 ticket instead of her actual 8 tickets when working out the expected winnings: (1/400) × £45 = £0.1125, giving £4.00 − £0.11 = £3.89. Aisha should expect to lose £3.10.
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