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 red fair dice and a blue fair dice are rolled together. Every outcome is listed as an ordered pair, red score then blue score. Work out the probability that the red score is less than the blue score.
- 2.Oliver flips three fair coins at the same time. Work out the probability that exactly two of the three coins land on heads.
- 3.In a game, Freddie says the probability of winning is 0.45, the probability of drawing is 0.3 and the probability of losing is 0.35, and that all three of his probabilities are correct. Which statement about Freddie's probabilities is correct?
- 4.Two ordinary fair dice are rolled and the two scores are multiplied together. Work out the probability that the product is odd.
- 5.A two-way table records 200 members of a gym. 80 of the members are women and the rest are men. 60 of the women attend yoga classes. Work out the percentage of the women who attend yoga classes.
- 6.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.
- 7.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.
- 8.A small pack contains 5 cards, numbered 1 to 5. One card is drawn at random from the pack, and a fair coin is flipped. Every outcome is listed as a pair, such as (3, H). Work out how many equally likely outcomes there are in total.
- 9.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.
- 10.A bag contains counters that are red, blue or green only. The probability that a counter taken at random is not red is 0.8. The bag contains 25 counters in total. Work out how many of the counters are red.
- 11.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.
- 12.A raffle has two independent draws. In the first, a winning ticket is picked at random from 5 red tickets and 3 blue tickets. In the second, a winning number is picked using a fair spinner with 4 equal sections, numbered 1 to 4. Work out the probability that the first winning ticket is blue and the second winning number is greater than 2.
- 13.On a probability scale from 0 to 1, the points impossible (0), unlikely (0.25), evens (0.5) and certain (1) are marked. An event has a probability of 0.3. Between which two neighbouring marked points does 0.3 lie?
- 14.Two cards are dealt one after the other from an ordinary pack of 52 playing cards. The first card is not put back before the second is dealt. The pack contains 4 aces. Work out the probability that neither card is an ace. Give your answer as a product of two fractions.
- 15.A library recorded loans for 250 books over a week, using a frequency tree. The first branch splits the books into 160 fiction and 90 non-fiction. Of the fiction books, 112 were returned on time and the rest were returned late. All the non-fiction books were returned on time. Work out the probability that a book, chosen at random from the 250, was returned late. Give your answer as a fraction in its simplest form.
Answer key
- (b) 5/12 — There are 36 equally likely ordered pairs. Listing the pairs where the red score is smaller than the blue score gives 15 outcomes, so the probability is 15/36 = 5/12. Choosing 1/2 comes from assuming the 36 outcomes split evenly between 'red smaller' and 'red bigger', which ignores the 6 outcomes where the two dice show the same score. Choosing 7/12 comes from including the outcomes where the two scores are equal, 15 + 6 = 21, giving 21/36 = 7/12, instead of using 'less than' strictly. Choosing 5/36 comes from counting only the outcomes where the two scores differ by exactly 1, such as (1, 2) and (2, 3), which is 5 outcomes, and missing the pairs that differ by more than 1.
- (c) 3/8 — Method: write out every result of the three coins as a string of three letters, H for heads and T for tails, count the results that match the description and divide by how many results the list holds. Working: each coin lands two ways and no coin affects another, so the list holds 2 × 2 × 2 = 8 equally likely results. Exactly two heads means one coin lands on tails and the other two on heads, so the results are HHT, HTH and THH — 3 of the 8. Answer: the probability is 3/8. The distractors: 4/8 comes from reading 'exactly two heads' as 'at least two heads' and counting HHH as well; 2/8 comes from a list made without a system, in which HHT and THH are written down and HTH, the result with the tail between the two heads, is missed; 6/8 comes from counting 3 × 2 = 6 ways of picking which two of the three coins show heads, which counts every pair of coins twice, once in each order.
- (a) No — the three probabilities sum to 1.10, over 1. — Winning, drawing and losing are exhaustive and mutually exclusive, so their probabilities must sum to exactly 1. Adding Freddie's three values gives 0.45 + 0.3 + 0.35 = 1.10, which is more than 1, so his probabilities cannot all be correct: 'No — the three probabilities sum to 1.10, over 1.' Checking only that each value lies between 0 and 1 accepts them as 'Yes — each probability lies between 0 and 1' without ever adding the three together. Judging by which outcome sounds most likely leads to 'Yes — winning has the highest single probability', which never checks the total either. Noting that a runner cannot win, draw and lose at once, and treating that alone as enough, gives 'Yes — the three outcomes are mutually exclusive' — but mutually exclusive outcomes that are also exhaustive must still sum to 1, and 1.10 does not.
- (d) 1/4 — Method: a product is odd only when BOTH factors are odd, so list the ordered pairs where both scores are odd and divide by 36. Working: the odd scores on a dice are 1, 3 and 5, so there are 3 × 3 = 9 ordered pairs where both scores are odd, out of the 36 equally likely pairs, cancelling down to 1/4. Answer: 1/4. Watch out: writing down 3/4 finds the probability that AT LEAST ONE score is odd, 1 minus the probability both are even, which is a different, easier condition to meet than both being odd. Considering only the first dice's score and ignoring the second gives 1/2, since 3 of the first dice's 6 scores are odd — but the product also depends on what the second dice shows. And writing down 1/12 comes from counting only the pairs where the SAME odd number appears twice, (1, 1), (3, 3) and (5, 5), missing pairs like (1, 3) and (5, 1) where the two odd scores differ.
- (a) 75% — 'Percentage of the women' restricts the group to the 80 women, of whom 60 attend yoga: 60/80 = 0.75 = 75%. Dividing by the number of men (200 − 80 = 120) instead of the number of women gives 60/120 = 0.5 = 50%. Dividing by all 200 members instead of just the 80 women gives 60/200 = 0.3 = 30%. Using the 20 women who do NOT attend yoga (80 − 60) as the numerator instead of the 60 who do gives 20/80 = 0.25 = 25%.
- (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.
- (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) 10 — There are 5 possible cards and 2 possible coin results, so listing every pair gives 5 × 2 = 10 equally likely outcomes. Choosing 5 comes from listing only the card outcomes and forgetting the coin flip adds a second stage to each one. Choosing 7 comes from adding the two stages instead of combining them, 5 + 2 = 7, rather than pairing every card with every coin result. Choosing 20 comes from counting each coin result twice for every card, 5 × 2 × 2 = 20, effectively pairing every card with the coin twice over.
- (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.
- (d) 5 — Being red and not being red are exhaustive, so their probabilities sum to 1: the probability of red is 1 − 0.8 = 0.2. The number of red counters is 0.2 × 25 = 5. Using 0.8 directly as the probability of red, without taking the complement, gives 0.8 × 25 = 20 — the number of counters that are NOT red. Sharing the 25 counters equally between the three colours, ignoring the given probability altogether, gives 25 ÷ 3 ≈ 8. Misreading the total as 20 counters instead of 25 gives 0.2 × 20 = 4.
- (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) 3/16 — Method: work out each draw's own probability first, then multiply them together since the two draws are independent. Working: there are 8 tickets in all, 3 of them blue, so P(blue) = 3/8. P(spinner number greater than 2) = 2/4 = 1/2, since 3 and 4 qualify. Multiplying gives 3/8 × 1/2, which comes to 3/16. Answer: 3/16. Watch out: writing down 3/8 stops after the first draw and never brings in the spinner at all. Writing down 1/2 does the opposite, using only the spinner and ignoring the ticket draw. And writing down 5/16 uses 5/8, the probability of a RED ticket, instead of 3/8 for blue — reading the wrong colour off the raffle.
- (d) unlikely and evens — The marked points are impossible (0), unlikely (0.25), evens (0.5) and certain (1). Since 0.25 < 0.3 < 0.5, the probability 0.3 lies between unlikely and evens. Choosing 'impossible and unlikely' treats 0.3 as below 0.25, which it is not. Choosing 'evens and certain' treats 0.3 as above 0.5, which it is not. Choosing 'impossible and evens' skips over the unlikely marker altogether, even though 0.3 is above it.
- (b) (48/52) × (47/51) — Method: for two deals one after the other with nothing put back, multiply the probability of the first by the probability of the second worked out from the cards that are left. Working: 52 − 4 = 48 cards are not aces, so the first card is not an ace with probability 48/52. One card has now gone and it was not an ace, so 51 cards remain and 47 of them are not aces, giving 47/51. Answer: the probability is (48/52) × (47/51). The distractors: (48/52) × (48/52) comes from leaving the pack at 52 cards for the second deal, which is only true if the first card is replaced; (4/52) × (3/51) comes from working out the probability that both cards ARE aces instead of neither; (4/52) × (4/51) comes from the same misreading with the ace count left at 4 while the total is reduced, adjusting only half of the second fraction.
- (b) 24/125 — On the fiction branch, 160 − 112 = 48 books were returned late. None of the non-fiction books were late, so the total number of late books is 48, out of 250 books altogether: 48/250 = 24/125. Writing 3/10 is wrong because it divides the 48 late fiction books by the fiction total (160) instead of the whole library total (250). Writing 56/125 is wrong because 112/250 simplifies to 56/125, and 112 is the number of fiction books returned ON TIME, not late. Writing 9/25 is wrong because 90/250 simplifies to 9/25, and 90 is simply the number of non-fiction books, which has nothing to do with late returns. The probability is 24/125.
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