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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Answer key: Probability worksheet — GCSE Foundation
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- (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.
- (c) No — 7/30 is the relative frequency; theory stays 1/6. — The theoretical probability of rolling a 6 on an ordinary dice is fixed at 1/6, worked out from the number of equally likely outcomes, and does not change however the dice is actually rolled. The relative frequency from this trial is 7/30, found from what happened in these particular 30 rolls. Since 7/30 and 1/6 are different numbers, the correct statement is 'No — 7/30 is the relative frequency; theory stays 1/6.' Assuming the two values must always match because they describe the same event gives 'Yes — relative frequency always equals theory.' Believing that an observed result redefines the theoretical probability gives 'Yes — the theoretical probability has now become 7/30.' Refusing to work out either value at all gives 'Neither can be found — 30 rolls is too few to tell', which ignores that both numbers CAN be calculated from the information given.
- (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.
- (b) 10 — The probability that the spinner lands on green is its angle out of the whole circle: 60° ÷ 360° = 1/6. Expected number of times on green = 1/6 × 60 = 10. Assuming the three colours are equally likely because there are three sectors, regardless of their different angles, gives 60 ÷ 3 = 20. Using red's angle of 180° instead of green's 60° gives a probability of 1/2, so 60 × 1/2 = 30. Treating green's angle, 60°, as a percentage instead of finding its fraction of 360° gives 60 × 0.60 = 36.
- (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) 0.150 — The six scores are exhaustive, so their probabilities sum to 1. The probability of not landing on 6 is 1 − 0.25 = 0.75, and this is shared equally between the other five scores, so each has probability 0.75 ÷ 5 = 0.150. Giving 0.750 as the answer stops after finding the probability of not landing on 6, without sharing it out between the five remaining scores. Dividing 0.75 by 6 instead of by 5 gives 0.125, wrongly including the score of 6 among the equally likely scores. Ignoring the bias completely and dividing 1 by all six scores gives 1 ÷ 6 = 0.167.
- (c) 2 — Set A, the multiples of 3, is {3, 6, 9}. Removing the numbers that are also in set B, the even numbers {2, 4, 6, 8, 10}, leaves {3, 9} — 2 numbers are in A but not B. Giving 3, the whole size of set A, forgets to remove the number 6, which is also even. Giving 7 counts every number in A or B combined, {2, 3, 4, 6, 8, 9, 10}, rather than only those in A but not B. Giving 4 counts the numbers in set B but not set A, {2, 4, 8, 10}, the wrong way round.
- (a) Ben, because a larger sample is closer to the theory — Method: a relative frequency is an estimate of a probability, and for an unbiased experiment that estimate tends towards the theoretical value as the sample grows. Working: Priya's estimate rests on 50 results, so a few unexpected heads move it a long way; one extra head shifts her relative frequency by 1 ÷ 50 = 0.02. Ben's estimate rests on 500 results, where one extra head shifts his relative frequency by only 1 ÷ 500 = 0.002. The larger sample therefore swings far less around the true value. Answer: Ben's relative frequency is the one more likely to be close, because a larger unbiased sample tends closer to the theoretical probability. The distractors: saying a small sample is less affected by luck reverses the result, since it is the small sample that swings most; saying every flip is a separate random event is true of the flips themselves but says nothing about the estimates, and is often used to argue wrongly that the number of trials does not matter; saying 500 flips must give exactly 250 heads confuses an expected value with a guaranteed one, and 500 flips very rarely give exactly 250 heads.
- (d) 19/30 — Exactly one means only fiction or only non-fiction, not both: 21 + 17 = 38 out of the 60 readers, which simplifies to 19/30. Including the 14 who read both as well gives 21 + 17 + 14 = 52, so 52/60 = 13/15 — that is at least one, not exactly one. Using only the both-count, 14, as the numerator gives 14/60 = 7/30, the probability of reading both, not exactly one. Using 52, the number who read at least one type, as the denominator instead of the full 60 readers surveyed gives 38/52 = 19/26.
- (d) 15 — Method: each of Spinner A's outcomes can be paired with each of Spinner B's outcomes, so the two counts are combined by multiplying, not adding. Working: Spinner A has 3 outcomes and Spinner B has 5, so there are 3 × 5 = 15 equally likely outcomes in total. Answer: 15. Watch out: adding the two counts, 3 + 5 = 8, undercounts drastically — every one of Spinner A's 3 outcomes pairs with all 5 of Spinner B's outcomes, not just one each. Writing down 3 counts Spinner A alone and leaves Spinner B out completely. And writing down 25 comes from squaring Spinner B's own outcome count, 5 × 5, and forgetting Spinner A altogether.
- (b) 0.49 — Adding all four pupils' flips gives 40 + 25 + 20 + 15 = 100 flips in total, and adding their heads gives 24 + 10 + 9 + 6 = 49 heads in total, so the combined estimate is 49/100 = 0.49. Writing 0.46 is wrong because it averages the four pupils' individual rates (0.60, 0.40, 0.45 and 0.40) as if they all came from the same number of flips, which they do not — this ignores that Sam and Priti flipped far more times than Leo and Fatima. Writing 0.60 is wrong because it only uses Sam's own result (24/40 = 0.60), ignoring the other three pupils. Writing 0.40 is wrong because it only combines Priti and Fatima's results (16/40 = 0.40), leaving out Sam and Leo entirely. The combined estimate from all four pupils is 0.49.
- (c) 0.5 — Method: count the favourable outcomes, write them over the total number of equally likely outcomes and then divide to turn the fraction into a decimal. Working: the odd scores are 1, 3 and 5, which is 3 of the 6 equally likely scores, so the probability is 3/6, and 3 ÷ 6 = 0.5. Answer: 0.5, the middle of the 0 to 1 probability scale. The distractors: 0.33 comes from listing only 3 and 5 as odd and working out 2 ÷ 6; 0.17 comes from giving the probability of one particular odd score, 1 ÷ 6; 0.3 comes from writing '3 out of 6' as 0.3, reading the 3 straight off as tenths instead of dividing.
- (d) 5/18 — Method: list every ordered pair of dice scores whose difference is 1, then divide by the 36 equally likely pairs. Working: the pairs with a difference of 1 are (1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4), (5, 6) and (6, 5), which is 10 pairs out of 36, cancelling down to 5/18. Answer: 5/18. Watch out: writing down 5/36 lists only the 5 pairs going up, (1, 2), (2, 3), (3, 4), (4, 5) and (5, 6), and misses that each one has a matching pair the other way round, such as (2, 1) — the two dice are different objects, and order matters, so each of those 5 gaps counts twice. Writing down 1/6 treats the six possible differences, 0 to 5, as equally likely and picks 1 out of 6 of them, but a difference of 1 is reached by far more pairs of scores than a difference of 5 is, so the six differences are not equally likely. And writing down 1/4 comes from adding the two dice's outcome counts instead of multiplying them, 6 + 6 = 12, and grouping the six scores into just three non-overlapping pairs one apart, {1, 2}, {3, 4} and {5, 6}, giving 3 out of that wrong pool of 12.
- (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.
- (b) 3 — Method: build the list of possible results systematically, taking the first flip as a head and then as a tail, and keeping the two flips in order so that a head then a tail is a different result from a tail then a head; then count the results that match the description. Working: with a head at the first flip the results are HH and HT, and with a tail at the first flip they are TH and TT, so the list is HH, HT, TH, TT — four results in all. The results containing at least one tail are HT, TH and TT. Answer: 3 of the results in the list contain at least one tail. The distractors: 2 comes from treating a head and a tail as one result however they are ordered, which shortens the list to HH, HT, TT so that only two entries hold a tail; 1 comes from reading 'at least one tail' as a tail at both flips, which is the single result TT; 4 comes from counting the tails written across the list — one in HT, one in TH and two in TT — instead of counting the results that contain a tail.
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