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.Spinner A has 4 equal sections, numbered 1, 2, 3 and 4. Spinner B has 5 equal sections, numbered 1, 2, 3, 4 and 5. Both spinners are spun once, and the two numbers are multiplied together. Work out the probability that the product is 12.
- 2.The probability that a pupil chosen at random has a nut allergy is 1/10. There are 30 pupils in a class. Work out how many of the 30 pupils would be expected to have a nut allergy.
- 3.The probability that the Hughes family go away during the summer holidays is 3/7. Work out the probability that they do not go away during the summer holidays.
- 4.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.
- 5.At a coffee shop, each customer buys tea, coffee, water or juice, and never more than one of these. The probability that a customer buys tea is 0.36 and the probability that they buy juice is 0.04. The probability that a customer buys coffee is three times the probability that they buy water. Work out the probability that a customer buys water.
- 6.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?
- 7.A box contains counters that are silver, bronze or copper only. The probability that a counter taken at random is silver is 1/2 and the probability that it is bronze is 1/8. The box contains 64 counters in total. Work out the number of copper counters.
- 8.A bag contains 4 red counters and 6 blue counters. A counter is taken at random, its colour noted, and it is put back in the bag before a second counter is taken at random. Work out the probability that the two counters are different colours. Give your answer as a fraction in its simplest form.
- 9.A market stall sells umbrellas. Over the last 250 days, it rained on 70 of them. Using this as an estimate of the probability of rain, work out how many rainy days would be expected in the next 365 days.
- 10.A padlock code is formed by arranging three different digits chosen from 2, 3, 4 and 5, with no digit repeated. Work out how many different three-digit codes can be made.
- 11.A garden centre recorded 240 plant sales using a frequency tree. The first stage splits the sales into three types: shrubs, bedding plants and trees. 96 sales were shrubs, 114 were bedding plants and the rest were trees. Work out the probability that a sale chosen at random from these 240 was a tree.
- 12.At a school fete, each raffle ticket wins a prize, wins a voucher or wins nothing, and no ticket can win more than one of these. The probability that a ticket wins a prize is 0.4 and the probability that it wins a voucher is 0.35. Work out the probability that a ticket wins a prize or a voucher.
- 13.An ordinary fair dice is rolled twice. Work out the probability of getting at least one 5.
- 14.A pupil must choose 2 different subjects at random from these five: maths, physics, biology, chemistry and history. Work out the probability that the two subjects chosen are maths and biology.
- 15.A spinner is divided into 8 equal sections. 3 of the sections are red and the rest are not red. The spinner is spun once. Work out the probability that it does not land on red.
Answer key
- (a) 1/10 — There are 4 × 5 = 20 equally likely outcomes in total. The pairs whose product is 12 are Spinner A showing 3 with Spinner B showing 4, and Spinner A showing 4 with Spinner B showing 3, which is 2 outcomes, giving a probability of 2/20 = 1/10. Choosing 1/20 comes from finding only one of the two pairs, (3, 4), and missing (4, 3) as a separate outcome. Choosing 1/8 comes from using 16 as the total number of outcomes, 4 × 4, forgetting that Spinner B has 5 sections rather than 4. Choosing 1/5 comes from listing the factor pairs of 12 as 2 × 6 and 3 × 4 and counting each one in both orders, (2, 6), (6, 2), (3, 4) and (4, 3), giving 4 outcomes out of 20 without checking that neither spinner has a 6 on it.
- (b) 3 pupils — Method: an expected frequency is the probability multiplied by the number of trials, so multiply the probability by the number of pupils. Working: 30 × 1/10 means finding one tenth of 30, and 30 ÷ 10 = 3. Answer: 3 pupils would be expected to have a nut allergy. The distractors: 27 pupils comes from working out how many are expected NOT to have the allergy, 30 − 3, instead of how many are; 10 pupils comes from reading the 10 in the fraction 1/10 as the number of pupils; 1 pupil comes from reading the numerator of the fraction as the expected number.
- (a) 4/7 — Method: going away and not going away are the only two possibilities, so the two probabilities form an exhaustive set and add to 1; subtract the given probability from 1. Working: writing 1 as 7/7 gives 7/7 − 3/7, and 7 − 3 = 4 sevenths. Answer: 4/7. The distractors: 3/7 comes from giving back the probability that the family do go away; 1/7 comes from taking the 1 in '1 − 3/7' as a numerator and writing it over the denominator 7; 1/2 comes from assuming that going away and not going away must be equally likely because there are only two possibilities.
- (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.
- (a) 0.15 — Let P(water) = x, so P(coffee) = 3x. The four outcomes are exhaustive: 0.36 + 0.04 + x + 3x = 1, so 0.4 + 4x = 1, giving 4x = 0.6 and x = 0.15. So P(water) = 0.15. Reporting 3x, the coffee probability, instead of water gives 0.45. Splitting the remaining 0.6 evenly between coffee and water, ignoring the 3:1 ratio, gives 0.30. Stopping once the remaining probability 0.6 is found, without dividing by the four equal shares, gives 0.60.
- (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.
- (c) 24 — Silver, bronze and copper cover every counter, so their probabilities sum to 1: P(copper) = 1 − 1/2 − 1/8 = 3/8. Number of copper counters = 3/8 × 64 = 24. Multiplying the silver probability by 64 gives 32, the number of silver counters, not copper. Multiplying the bronze probability by 64 gives 8, the number of bronze counters. Adding the silver and bronze probabilities (1/2 + 1/8 = 5/8) and multiplying by 64 gives 40, the combined number of silver and bronze counters, not the copper count.
- (a) 12/25 — P(red then blue) = 4/10 × 6/10 = 24/100. P(blue then red) is the same, 6/10 × 4/10 = 24/100. Since either order counts as different colours, add them: 24/100 + 24/100 = 48/100 = 12/25. Working out only one order, red-then-blue, and forgetting blue-then-red, gives 24/100 = 6/25. Working out the probability that the two counters are the SAME colour instead — 4/10 × 4/10 + 6/10 × 6/10 = 16/100 + 36/100 = 52/100 = 13/25 — answers a different question. Using 6/9 for the second draw, as if the first counter were not replaced, gives 2 × (4/10 × 6/9) = 48/90 = 8/15.
- (c) 102 — Method: turn the past record into a relative frequency, then use it as an estimate of the probability of rain and multiply by the number of days being predicted for. Working: relative frequency of rain = 70 ÷ 250 = 0.28. Expected rainy days in 365 days = 365 × 0.28 = 102.2, which rounds to about 102 days. Answer: about 102 days. Watch out: writing down 48 swaps which number is the sample and which is the target, working out 70 ÷ 365 × 250 instead of 70 ÷ 250 × 365. Writing down 70 just repeats the original count of rainy days without scaling it up to the new, longer period at all. And writing down 110 comes from rounding the relative frequency to 0.3 before multiplying, 365 × 0.3 = 109.5, when 70 ÷ 250 is exactly 0.28 and needs no rounding at all.
- (d) 24 — There are 4 choices for the first digit. Once that digit is used, 3 digits remain for the second position, and then 2 digits remain for the third position: 4 × 3 × 2 = 24 codes. Choosing 64 comes from allowing a digit to be reused at every position, 4 × 4 × 4 = 64, which is not allowed here since no digit repeats. Choosing 12 comes from multiplying only the first two positions, 4 × 3 = 12, and forgetting that a third digit is also chosen from the digits that remain. Choosing 6 comes from counting only the arrangements of one single set of three digits, 3 × 2 × 1 = 6, and forgetting that there are 4 different sets of three digits that can be chosen from 2, 3, 4 and 5.
- (b) 1/8 — Shrubs, bedding plants and trees are the three branches at the first stage of the tree, so they must total 240: tree sales = 240 − 96 − 114 = 30. So P(tree) = 30/240 = 1/8. Using the shrub count instead, 96/240 = 2/5, is the probability of a shrub sale, not a tree sale. Using the bedding-plant count instead, 114/240 = 19/40, is the probability of a bedding-plant sale. Subtracting the bedding count from the shrub count (114 − 96 = 18) instead of subtracting both from 240 gives 18/240 = 3/40, which is not the number of tree sales at all.
- (a) 0.75 — Method: a ticket cannot win both a prize and a voucher, so the two events are mutually exclusive and their probabilities are added. Working: 0.4 + 0.35, lining the decimal points up, gives 0.40 + 0.35. Answer: 0.75. The distractors: 0.25 comes from going one step too far and giving the probability that a ticket wins nothing, which is what is left when the total is taken from 1; 0.14 comes from multiplying 0.4 by 0.35 instead of adding them; 0.05 comes from subtracting 0.35 from 0.4 instead of adding them.
- (b) 11/36 — Method: 'at least one' is the opposite of 'none at all', so work out the probability of no 5 on either roll and take it away from 1. Working: a roll that is not a 5 has probability 5/6, and the rolls are independent, so no 5 at all has probability 5/6 × 5/6 = 25/36. Taking this from 36/36 leaves 11/36. Answer: the probability is 11/36. The distractors: 25/36 is the probability of no 5 at all, written down without the final subtraction; 12/36 comes from counting the 6 pairs with a 5 on the first roll and the 6 pairs with a 5 on the second and adding them, which counts the pair (5, 5) twice; 30/36 comes from working out 1 − 1/6 as though only one roll were made.
- (b) 1/10 — Method: list every possible pair of subjects systematically, so that no pair is missed and no pair is counted twice, then compare the number of successful pairs with the size of the list. Working: pairing maths with each of the other four gives 4 pairs, physics with each subject after it gives 3, biology gives 2 and chemistry gives 1, so there are 4 + 3 + 2 + 1 = 10 pairs. Exactly one of them is maths with biology. Answer: the probability is 1/10. The distractors: 1/5 comes from counting maths-then-biology and biology-then-maths as two separate successes while still dividing by the 10 unordered pairs; 1/15 comes from a list that also pairs each subject with itself, giving 15 entries instead of 10; 1/4 comes from working out the second step alone, that 1 of the 4 subjects left after maths is biology, without allowing for the chance that maths is picked at all.
- (c) 5/8 — Method: the sections are all the same size, so every section is equally likely; count the sections that are not red and write that count over the total number of sections. Working: 8 − 3 = 5 sections are not red, and there are 8 sections altogether. Answer: 5/8, a value between 0 and 1 and a little above the halfway point of the scale. The distractors: 3/8 comes from giving the probability that the spinner does land on red; 5/11 comes from adding the 3 red sections to the 8 sections to make a total of 11 instead of using the 8 sections that exist; 1/2 comes from assuming that 'red' and 'not red' must be equally likely because there are only two possibilities.
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