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.A game has two independent stages. P(win stage 1) = 0.4, and P(win stage 2) = 0.25. Work out the probability that Ffion wins at least one of the two stages.
- 2.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.
- 3.At a fair, a game costs £2 to play. The probability of winning the game is 0.15, and each win pays out £10. Amir plays the game 200 times. Work out how much money Amir should expect to lose in total.
- 4.A frequency tree records 320 calls to a helpline. The calls first split into calls about billing and calls about technical support. Of the technical support calls, the probability that a call was resolved on the first contact is 0.75. If 180 technical support calls were resolved on the first contact, work out how many calls were about billing.
- 5.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.
- 6.A bag contains 3 yellow counters and 2 green counters. Sophie takes one counter at random and does not put it back. She then takes a second counter at random. Work out the probability that the first counter is yellow and the second is green.
- 7.At a quiz night, each team's result is a win, a draw or a loss, and no team can have more than one of these results. The probability that a team wins is 42% and the probability that a team draws is 18.5%. Work out the probability, as a percentage, that a team does not win and does not draw.
- 8.The probability that Amelia's bus is late on a school morning is 9/10. Write this probability as a percentage.
- 9.150 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.
- 10.A fair six-sided dice is rolled once. Write down the probability that the score is an even number.
- 11.A raffle at a school fair sells 500 tickets. Mia buys 25 of the tickets. One ticket is drawn at random to win the prize. Work out the probability that Mia wins the raffle. Give your answer as a fraction in its simplest form.
- 12.A bag contains 5 yellow marbles and 3 blue marbles. Noah takes a marble at random, notes its colour and puts it back in the bag. He then takes a second marble at random. Work out the probability that the first marble is yellow and the second marble is blue.
- 13.One card is taken at random from an ordinary pack of 52 playing cards. 26 of the cards are red. Work out the probability that the card is red. Give your answer as a percentage.
- 14.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.
- 15.An ordinary fair dice is rolled once. Set A is the set of even scores and set B is the set of scores greater than 4. Work out the probability that the score rolled is in set A or in set B or in both.
Answer key
- (b) 0.55 — Method: it is easier to find the probability that Ffion wins NEITHER stage, then subtract that from 1. Working: P(lose stage 1) = 1 − 0.4 = 0.6, and P(lose stage 2) = 1 − 0.25 = 0.75. P(neither) = 0.6 × 0.75 = 0.45. P(at least one) = 1 − 0.45 = 0.55. Answer: 0.55. Watch out: adding the two win probabilities, 0.4 + 0.25 = 0.65, treats winning both as impossible and overcounts — that is not how independent probabilities combine. Multiplying the two win probabilities, 0.4 × 0.25 = 0.1, gives the probability of winning BOTH stages, not at least one. And stopping at 0.45, the probability of winning neither stage, forgets the final step of subtracting from 1.
- (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) £100 — Method: find the expected number of wins, turn that into the expected pay out, then compare it with what the games cost. Working: the expected number of wins is 200 × 0.15 = 30. Each win pays £10, so the expected pay out is 30 × 10 = 300 pounds. Playing 200 times at £2 a go costs 200 × 2 = 400 pounds. The expected loss is 400 − 300 = 100 pounds. Answer: Amir should expect to be about £100 down. The distractors: £300 is the expected winnings on their own, with the cost of playing never taken off; £400 is the total cost of playing, with the winnings never taken off; £700 comes from adding the two totals, 400 + 300 = 700, instead of subtracting one from the other.
- (c) 80 — Since 180 calls are 0.75 of all the technical support calls, the technical support total is 180 ÷ 0.75 = 240. The billing calls make up the rest of the 320 calls, so 320 − 240 = 80. Choosing 240 comes from stopping after finding the technical support total and forgetting the question asks for the billing calls, which are the rest. Choosing 185 comes from multiplying 180 × 0.75 = 135 instead of dividing, then working out 320 − 135 = 185. Choosing 140 comes from using 180 directly as the whole technical support total, ignoring the probability altogether, then working out 320 − 180 = 140.
- (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) 3/10 — Method: on a tree diagram, follow the path that matches the description and multiply the probabilities written along it; with nothing put back, the second set of branches is worked out from the counters that are left. Working: 3 of the 5 counters are yellow, so the first branch of the path is 3/5. A yellow counter has been kept out, so 4 counters remain and both green counters are still there, making the second branch 2/4. Multiplying along the path gives 6/20. Answer: the probability is 3/10. The distractors: 6/25 comes from using 2/5 on the second branch, which is the tree for a counter that is put back; 3/20 comes from taking one off the green count as well as off the total, using 1/4 on the second branch; 3/5 comes from reading the first branch only and never multiplying along the path.
- (c) 39.5% — Win, draw and lose are mutually exclusive and exhaustive, so their probabilities sum to 100%: 42% + 18.5% = 60.5% is the percentage that wins or draws. 100% − 60.5% = 39.5% is the percentage that neither wins nor draws. Adding 42% and 18.5% and stopping there, 60.5%, is the probability of winning or drawing, not of neither. Subtracting only the 42% from 100% gives 58.0%, ignoring the draw percentage. Subtracting only the 18.5% from 100% gives 81.5%, ignoring the win percentage.
- (a) 90% — Method: turn the fraction into a decimal by dividing the numerator by the denominator, then multiply the decimal by 100 to turn it into a percentage. Working: 9 ÷ 10 = 0.9, and 0.9 × 100 = 90. Answer: 90%, which is the same probability written in a third form, and all three forms sit at the same point on the 0 to 1 scale. The distractors: 9% comes from reading the numerator of the fraction straight off as the percentage; 10% comes from reading the denominator straight off as the percentage; 0.9% comes from correctly finding the decimal 0.9 and then writing a percentage sign after it without multiplying by 100.
- (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.
- (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.
- (d) 1/20 — Mia has 25 of the 500 tickets, so the probability she wins is 25/500 = 1/20, dividing the top and the bottom by 25. Writing 19/20 is wrong because that is the probability she does NOT win (1 − 1/20 = 19/20), the opposite of what is asked. Writing 1/25 is wrong because it puts 1 over the number of tickets Mia holds, as though her 25 tickets were the whole raffle — the denominator has to be the 500 tickets sold, not her own share. Writing 1/10 is wrong because it treats the raffle as having only 250 tickets instead of the actual 500: 25/250 = 1/10. The probability that Mia wins is 1/20.
- (d) 15/64 — Method: because the first marble is put back, the second draw is made from the same 8 marbles, so the two draws are independent and their probabilities are multiplied. Working: the bag holds 8 marbles both times, so yellow first has probability 5/8 and blue second has probability 3/8. Multiplying fractions gives 5 × 3 = 15 on the top and 8 × 8 = 64 on the bottom. Answer: the probability is 15/64. The distractors: 9/64 comes from working out the probability that BOTH marbles are blue, 3/8 × 3/8, instead of one of each colour in the stated order; 15/16 comes from multiplying the numerators but adding the denominators; 5/8 comes from writing down the first draw only and never combining it with the second.
- (d) 50% — Method: write the number of red cards over the total number of cards, then turn that fraction into a percentage by dividing and multiplying by 100. Working: the probability is 26/52; 26 ÷ 52 = 0.5 and 0.5 × 100 = 50. Answer: 50%, the exact middle of the 0 to 1 probability scale, which matches the fact that half the pack is red. The distractors: 26% comes from writing the number of red cards straight down as the percentage without dividing by the total; 25% comes from taking one of the four suits as the red ones and giving a quarter; 2% comes from dividing the wrong way round, 52 ÷ 26 = 2, and writing that as a percentage.
- (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.
- (a) 2/3 — Method: 'in A or in B' means every score that belongs to at least one of the two sets; a score that belongs to both is still only one outcome, so it is listed once. Working: the even scores are 2, 4 and 6; the scores greater than 4 are 5 and 6. Listing the scores that appear in either set gives 2, 4, 5 and 6, with 6 written once. That is 4 of the 6 faces, or 4/6. Answer: the probability is 2/3. The distractors: 5/6 comes from adding the sizes of the two sets, 3 + 2, so that the score 6 is counted in both and appears twice; 1/2 comes from using the even scores alone; 1/6 comes from giving the probability that the score is in both sets, which is the single score 6, rather than in either of them.
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