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 120 times. Work out how many of the rolls you would expect to give a score of 3 or more.
- 2.Two sets, A and B, sit inside a universal set. A region of a Venn diagram is described by the notation A ∩ B′. Work out which of these describes that region.
- 3.A market research company found that, out of 120 shoppers surveyed, 42 preferred paper bags to plastic bags. A supermarket expects 900 shoppers this week. Using this relative frequency, work out how many of the 900 shoppers would be expected to prefer paper bags.
- 4.A fair spinner has 8 equal sections. 3 of the sections are red. The spinner is spun 240 times. Work out how many times you would expect it to land on red.
- 5.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.
- 6.A spinner can land on red, blue, green or yellow, and it cannot land on more than one colour. The probability that it lands on red is 0.05 and the probability that it lands on yellow is 0.35. The probability that it lands on blue is twice the probability that it lands on green. Work out the probability that it lands on green.
- 7.Spinner P and Spinner Q are each fair and have 5 equal sections, numbered 1 to 5. Both spinners are spun once, and every outcome is listed as an ordered pair, Spinner P then Spinner Q, such as (2, 4). Work out how many of the outcomes have the score on Spinner P less than the score on Spinner Q.
- 8.Each turn of a game ends in a win, a draw or a loss, and no turn can end in more than one of these. The probability of a win is 1/4 and the probability of a draw is 1/3. Work out the probability of a loss.
- 9.The probability that a component is faulty is 1/10. Two components are tested independently. Work out the probability that at least one of the two components is faulty.
- 10.A six-sided dice is rolled 12 times. It lands on a six 4 times. Erin says that the dice must be biased. Which of these is the best comment on Erin's statement?
- 11.A school surveys 60 pupils. 32 say they walk to school, 24 say they cycle to school, and 12 say they do both, on different days. The rest travel only by car. One of the 60 pupils is chosen at random. Work out the probability that this pupil travels only by car.
- 12.The probability that it rains in Manchester tomorrow is 1/12. Work out the probability that it does not rain in Manchester tomorrow.
- 13.The probability that Priya's dentist appointment overruns is 0.35. Work out the probability that it does not overrun.
- 14.A box contains cards labelled 1, 2 and 3. A card is drawn at random and its number noted, then it is put back. A second card is then drawn at random. Every outcome is listed as an ordered pair, such as (1, 2). Work out how many of the outcomes in the full list have both numbers the same.
- 15.The probability that Isla answers a quiz question correctly is 0.8. She answers three questions, and her answers are independent of each other. Work out the probability that she answers all three correctly.
Answer key
- (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.
- (c) The elements that are in A but not in B — The symbol ∩ means 'and', so A ∩ B′ means 'in A and also in the complement of B'. B′ means 'not in B'. So A ∩ B′ describes everything that is in A but not in B. The elements in both A and B describes A ∩ B, without the dash on B. The elements in B but not in A describes B ∩ A′, with the dash on A instead of B. The elements in neither A nor B describes (A ∪ B)′, the region outside both circles entirely.
- (d) 315 — The relative frequency from the survey is 42 ÷ 120 = 0.35, so the expected number who prefer paper bags among 900 shoppers is 0.35 × 900 = 315. Giving 42 as the answer reuses the original survey count without scaling it up to 900 shoppers at all. Finding the expected number who prefer PLASTIC bags instead of paper, using the relative frequency 78 ÷ 120 = 0.65, gives 0.65 × 900 = 585. Using 1000 shoppers instead of the 900 actually stated gives 0.35 × 1000 = 350.
- (c) 90 — Method: over many future trials the expected number of successes is the number of trials multiplied by the probability of a success. Working: the sections are equal, so each is equally likely and P(red) is 3 out of 8. Over 240 spins the expected number of reds is 240 × 3 ÷ 8 = 90. Answer: about 90 of the spins would be expected to land on red. The distractors: 150 uses the 5 sections that are not red, 240 × 5 ÷ 8 = 150, which is the expected number of spins that do not land on red; 30 is 240 ÷ 8 and is the expected count for one single section; 80 comes from dividing by the number of red sections instead of by the number of sections, 240 ÷ 3 = 80.
- (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.
- (c) 0.2 — Let P(green) = x, so P(blue) = 2x. Red, blue, green and yellow are exhaustive: 0.05 + 0.35 + x + 2x = 1, so 0.4 + 3x = 1, giving 3x = 0.6 and x = 0.2. So P(green) = 0.2. Splitting the remaining 0.6 evenly between blue and green, ignoring the 2:1 ratio, gives 0.3. Working out x correctly but then reporting 2x, the probability of blue, gives 0.4. Stopping after finding that blue and green together account for 0.6, without dividing by the three equal shares of x, gives 0.6.
- (b) 10 — There are 25 equally likely ordered pairs in total. Listing the pairs where the Spinner P score is less than the Spinner Q score gives 10 outcomes, running from (1, 2) up to (4, 5). Choosing 15 comes from also including the 5 pairs where the two scores are equal, such as (1, 1) and (2, 2), which do not satisfy 'less than'. Choosing 4 comes from counting only the outcomes where the scores differ by exactly 1, such as (1, 2) and (2, 3), and missing the pairs that differ by 2, 3 or 4. Choosing 20 comes from correctly excluding the 5 outcomes where the two scores are equal, 25 − 5 = 20, but forgetting to also exclude the outcomes where Spinner P is bigger than Spinner Q, rather than smaller.
- (a) 5/12 — Method: a win, a draw and a loss are the only outcomes and no two can happen together, so the three probabilities form an exhaustive set of mutually exclusive events and add to 1; add the two given probabilities, then subtract from 1. Working: 1/4 + 1/3 over the common denominator 12 is 3/12 + 4/12 = 7/12, and 1 − 7/12 = 12/12 − 7/12. Answer: 5/12. The distractors: 7/12 comes from stopping at the probability of a win or a draw and never subtracting from 1; 1/12 comes from subtracting the two given probabilities from each other, 1/3 − 1/4, instead of adding them and taking the total from 1; 5/7 comes from adding 1/4 and 1/3 by adding the numerators and the denominators to get 2/7 and then subtracting that from 1.
- (d) 19/100 — It is easier to first find the probability that NEITHER component is faulty, then subtract from 1. The probability a component is not faulty is 9/10, so the probability neither is faulty is 9/10 × 9/10 = 81/100. So the probability at least one is faulty is 1 − 81/100 = 19/100. Choosing 1/5 comes from adding the two probabilities of a fault instead, 1/10 + 1/10 = 1/5, which double-counts the case where both are faulty. Choosing 1/10 comes from giving the probability for just one component being faulty. Choosing 1/100 comes from squaring the probability of a fault directly, 1/10 × 1/10 = 1/100, which is actually the probability that BOTH are faulty, not at least one.
- (b) She is wrong; 12 rolls is too few to judge — Method: compare the result with what is expected, then ask whether the experiment is long enough for a difference to mean anything. Working: if the dice were fair the expected number of sixes in 12 rolls is 12 × 1 ÷ 6 = 2, so 4 sixes is 2 above what was expected. But over only 12 rolls a result like this turns up often by chance: the relative frequency here is 4/12, which is 1/3, and over so few trials a relative frequency can sit well away from 1/6 with no bias at all. Answer: Erin is wrong, because 12 rolls is far too few to decide; she should roll the dice many more times and see whether the relative frequency settles near 1/6. The distractors: saying she is right because 4 beats the expected 2 uses the correct expected value but treats any difference as proof, which so short an experiment cannot give; saying a fair dice gives each score twice in 12 rolls treats an expected value as a guaranteed one; saying that 4 sixes in 12 rolls cancels down to 1 in 6 mis-cancels the fraction, because 4/12 is 1/3, which is twice 1/6, so that comment reaches the right verdict from arithmetic that is wrong.
- (a) 4/15 — Method: first find how many pupils travel only by car, then write that as a fraction of the 60 pupils surveyed. Working: pupils who walk or cycle or both = 32 + 24 − 12 = 44. Only by car = 60 − 44 = 16. P(only by car) = 16/60 = 4/15. Answer: 4/15. Watch out: writing down 1/5 takes the overlap of 12 pupils on its own, 12/60, mistaking the group who do both for the group who travel only by car. Writing down 4/5 comes from 60 − 12 = 48, subtracting only the overlap from the total instead of the whole walk-or-cycle count, so cyclists and walkers who are not in the overlap are wrongly swept into the only-car group. And writing down 7/15 comes from 60 − 32 = 28, subtracting the walkers alone and forgetting the cyclists altogether.
- (b) 11/12 — Method: raining and not raining are the only two outcomes, so their probabilities add to 1; subtract the given probability from 1. Working: writing 1 as 12/12 gives 12/12 − 1/12, and only the numerators are subtracted, 12 − 1 = 11. Answer: 11/12, close to the right-hand end of the 0 to 1 scale because rain is unlikely. The distractors: 1/12 comes from giving back the probability that it does rain; 1/11 comes from subtracting the 1 from the denominator instead of subtracting the fraction from 1; 11/11 comes from subtracting 1 from the numerator and from the denominator of 12/12 rather than from the numerator alone.
- (c) 0.65 — Overrunning and not overrunning are exhaustive: between them they cover every outcome, so their probabilities sum to 1. Work out 1 − 0.35 = 0.65. Writing 0.35 again is the probability that the appointment overruns, not its complement — the subtraction was never done. Adding instead of subtracting gives 1 + 0.35 = 1.35, which cannot be a probability at all. Subtracting each digit from 10 instead of borrowing from the 1, so 10 − 3 = 7 tenths and 10 − 5 = 5 hundredths, gives 0.75.
- (c) 3 — There are 3 outcomes for the first draw and 3 for the second, giving 3 × 3 = 9 ordered pairs in total: first draw 1 with second draw 1, 2 or 3; first draw 2 with second draw 1, 2 or 3; and first draw 3 with second draw 1, 2 or 3. Checking the list, the only pairs with both numbers the same are 1 with 1, 2 with 2, and 3 with 3, so there are 3. A candidate who answers 9 has counted every outcome instead of only the matching ones. A candidate who answers 6 has mistakenly counted pairs such as 1 with 2 and 2 with 1 as matching because they contain the same two digits. A candidate who answers 1 has stopped after finding only the first matching pair in the list.
- (c) 0.512 — Method: independent events that must all happen are combined by multiplying their probabilities. Working: the first two questions give 0.8 × 0.8 = 0.64. Bringing in the third gives 0.64 × 0.8, and since 64 × 8 = 512 with three decimal places in the product, this is 0.512. Answer: the probability is 0.512. The distractors: 0.64 comes from multiplying only two of the three probabilities and stopping; 0.8 comes from reading 'independent' as meaning the probability never changes and writing down the single-question figure; 0.0512 comes from a place-value slip in the last multiplication, counting four decimal places instead of three.
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