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) 40% — Method: a fraction becomes a percentage by scaling it so that its denominator is 100, or equivalently by dividing the numerator by the denominator and multiplying the decimal by 100. Working: 5 × 20 = 100, so the numerator is scaled by 20 as well, 2 × 20 = 40, giving 40/100; the same figure comes from 2 ÷ 5 = 0.4 and 0.4 × 100 = 40. Answer: 40%. The distractors: 20% comes from dividing 100 by the denominator alone, 100 ÷ 5 = 20, which is the percentage for 1/5 and never uses the numerator 2; 4% comes from finding the decimal 0.4 correctly and then writing the 4 down as the percentage instead of multiplying by 100; 2% comes from scaling only the denominator up to 100 and leaving the numerator as it was, giving 2/100.
- (d) 1/20 — Method: add the counts on the faulty end branches, divide by the total number of items in the experiment, then cancel. Working: the faulty items number 15 + 5 = 20, and 400 items were checked, so the probability is 20/400. Dividing the top and the bottom by 20 gives 1/20. Answer: the probability is 1/20. The distractors: 3/50 is 15/250 and comes from dividing machine A's faults by machine A's output, which is that machine's own fault rate rather than the probability for the whole batch; 1/30 is 5/150 and does the same on machine B's branch; 19/20 is 380/400 and gives the probability that the item picked is not faulty.
- (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) 26 — Method: n(G ∪ H) = n(G) + n(H) − n(G ∩ H), taking off the overlap once so the pupils who study both subjects are not counted twice. Working: 19 + 15 − 8 = 26. Answer: 26. Watch out: adding 19 and 15 without taking off the overlap gives 34, which counts the 8 pupils who study both subjects twice. Taking the 8 off both totals before adding, 19 − 8 + 15 − 8 = 18, counts only the pupils who study exactly one of the two subjects and leaves out the 8 who study both. And writing down 11, which is 19 − 8, gives the number who study geography only, not the number who study geography or history or both.
- (c) 0.37 — Method: pool the two runs into one combined set of results, then find the relative frequency of red across all of the spins together. Working: total reds = 16 + 21 = 37. Total spins = 40 + 60 = 100. Relative frequency = 37 ÷ 100 = 0.37. Answer: 0.37. Watch out: writing down 0.40 uses only the first run, 16 ÷ 40, and throws away the extra evidence from the second 60 spins. Writing down 0.35 uses only the second run, 21 ÷ 60, and throws away the first run instead. And writing down 0.375 averages the two runs' separate rates, (0.40 + 0.35) ÷ 2, which treats a run of 40 spins and a run of 60 spins as equally weighted, when pooling the actual counts gives the larger run its fair share of influence.
- (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.
- (a) 0.48 — There are two ways to score exactly one throw: scoring on the first and missing the second, 0.6 × 0.4 = 0.24, or missing the first and scoring the second, 0.4 × 0.6 = 0.24. Adding these gives 0.24 + 0.24 = 0.48. Choosing 0.24 comes from working out only one of the two paths and forgetting the other one also gives exactly one score. Choosing 0.36 comes from working out the probability of scoring BOTH throws, 0.6 × 0.6 = 0.36, instead of exactly one. Choosing 0.84 comes from working out the probability of scoring AT LEAST one throw, 1 − 0.4 × 0.4 = 0.84, instead of exactly one.
- (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.
- (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.
- (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.
- (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.
- (a) 0.54 — The four colours are exhaustive, so all four probabilities sum to 1: the probability of damson is 1 − 0.18 − 0.22 − 0.24 = 0.36. Amber and damson cannot both happen on one spin, so the probability of amber or damson is 0.18 + 0.36 = 0.54. Stopping after finding the probability of damson alone, without adding the probability of amber, gives 0.36. Adding the three given probabilities together, 0.18 + 0.22 + 0.24 = 0.64, and treating that total as the answer never finds the probability of damson at all. Subtracting only the probability of amber from 1, 1 − 0.18 = 0.82, ignores black, cyan and damson completely.
- (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.
- (a) 160 — Expected number = probability × number of trials = 0.08 × 2,000 = 160. Moving the decimal point one place too far, using 0.008 instead of 0.08, gives 2,000 × 0.008 = 16. Working out the expected number of customers who do NOT buy a bag, using the complement 1 − 0.08 = 0.92, gives 2,000 × 0.92 = 1,840. Rounding 0.08 up to 0.1 before multiplying gives 2,000 × 0.1 = 200.
- (b) 39 — Method: put the counts into a two-way table and fill each missing cell by subtracting along a row or down a column. Working: the number of female members is 150 − 80 = 70. The pool column holds 66 members and 35 of them are male, so the number of female pool users is 66 − 35 = 31. Subtracting along the female row, 70 − 31 = 39 female members use the gym. Answer: 39 female members use the gym. The distractors: 45 comes from subtracting along the male row instead, 80 − 35 = 45, which counts male gym users; 31 is the female pool cell, written down one step before the gym cell; 84 is 150 − 66 and counts every gym user, male and female together.
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