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.
Answer key: Probability worksheet — GCSE Foundation
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- (b) 41/160 — In total, 27 + 14 = 41 of the 160 employees cycle to work, so the probability is 41/160 (41 and 160 share no common factor, so this is already in its simplest form). Writing 27/160 is wrong because it only counts the full-time cyclists and leaves out the 14 part-time cyclists. Writing 41/90 is wrong because it uses the full-time total (90) as the denominator instead of the whole survey (160). Writing 1/5 is wrong because it only uses the part-time branch, simplifying 14/70 to 1/5 and ignoring the full-time cyclists completely. The probability is 41/160.
- (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.
- (a) 1/5 — Method: find the probability that the counter is black, then use the fact that an event and its complement add to 1. Working: there are 2 + 8 = 10 counters, so P(black) = 8/10 = 4/5; the complement is 1 − 4/5, and writing 1 as 5/5 gives 5/5 − 4/5. Answer: 1/5. The distractors: 4/5 comes from giving the probability that the counter is black instead of its complement; 1/2 comes from assuming the two colours are equally likely because there are only two of them; 1/4 comes from writing the 2 white counters over the 8 black counters rather than over all 10 counters.
- (c) 0.6 — Method: no two of these ways of travelling can happen on the same day, so they are mutually exclusive and the probability that one of them happens is the sum of their probabilities; they need not add to 1, because travelling by car is a fourth way. Working: 0.2 + 0.3 = 0.5, and 0.5 + 0.1 = 0.6, the decimal points lined up at each step. Answer: 0.6, which also shows that the probability of being taken by car is 0.4, since all four ways together must make 1. The distractors: 0.4 comes from carrying on past the question and giving the probability of the remaining way, by car; 0.5 comes from adding only the bus and the bike and leaving the smallest of the three probabilities out of the total; 0.006 comes from multiplying the three probabilities together instead of adding them.
- (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.
- (b) 57/100 — Pooling both trials: total heads = 24 + 33 = 57, total flips = 40 + 60 = 100, so the combined relative frequency is 57/100, which is already in its simplest form since 57 and 100 share no common factor. Averaging the two separate relative frequencies instead, (24/40 + 33/60) ÷ 2 = (0.6 + 0.55) ÷ 2 = 0.575 = 23/40, treats the two trials as equally weighted even though Ben made more flips, which is not correct. Using only Leah's data gives 24/40 = 3/5. Using only Ben's data gives 33/60 = 11/20.
- (c) 3/7 — Method: every counter is equally likely to be taken, so write the number of red counters over the total number of counters in the bag. Working: the bag holds 3 + 2 + 2 = 7 counters, of which 3 are red. Answer: 3/7, a value between 0 and 1 and just below the middle of the scale. The distractors: 4/7 comes from giving the probability that the counter is not red, counting the 2 blue and 2 green instead; 2/7 comes from counting one of the other colours by mistake and giving 2 counters over the total; 3/4 comes from writing the 3 red counters over the 4 counters that are not red instead of over all 7 counters.
- (d) 3/5 — 'Chosen from the boys' restricts the sample space to the 45 boys, of whom 27 walk. So the probability is 27/45 = 3/5. Using all 80 students as the denominator instead of just the boys gives 27/80, which is the probability that a student chosen from everyone is a boy who walks — not what was asked. Using the 18 boys who do NOT walk (45 − 27) as the numerator instead of the 27 who do gives 18/45 = 2/5. Dividing the 27 walking boys by the number of girls (80 − 45 = 35) instead of by the number of boys gives 27/35.
- (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.
- (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.
- (b) 37/75 — There are 150 − 84 = 66 men. 84 − 50 = 34 women prefer weight training, and 40 men prefer weight training, so 34 + 40 = 74 people in total prefer weight training, out of 150: 74/150 = 37/75. Writing 4/15 is wrong because it only counts the men who prefer weight training (40/150, simplified), leaving out the 34 women. Writing 17/75 is wrong because it only counts the women who prefer weight training (34/150, simplified), leaving out the 40 men. Writing 37/42 is wrong because it uses the number of women (84) as the denominator instead of the whole gym (150) — 74/84 simplifies to 37/42, but that is not a probability out of the whole group. The probability is 37/75.
- (c) 36 — On the adult branch there are 90 tickets in total, and 54 of them are for the 3D showing, so the standard-showing branch is 90 − 54 = 36. Subtracting 54 from the overall total of 150 gives 96, but 150 is the total for ALL tickets, not just the adult branch, so 96 is wrong. Writing 60 is wrong because that is the number of CHILD tickets (150 − 90 = 60), not adult standard tickets. Writing 54 again is wrong because that is the number of adult 3D tickets, not the number of adult standard tickets. The adult standard-showing branch is 36.
- (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) 136 — 15% of 160 = 0.15 × 160 = 24 patients reported side effects, so 160 − 24 = 136 did not. Stopping after finding the number who reported side effects, 24, answers the wrong question — it is not the number who did NOT report them. Misreading '15%' as a raw count of 15 patients, rather than a percentage, gives 160 − 15 = 145. Subtracting 15% of 160 twice, 160 − 24 − 24 = 112, double-counts the side-effect group.
- (d) 21 — Pink, gold and grey cover every counter, so their probabilities sum to 1: P(grey) = 1 − 0.2 − 0.45 = 0.35. Number of grey counters = 0.35 × 60 = 21. Multiplying the pink probability by 60 gives 12, the number of pink counters, not grey. Multiplying the gold probability by 60 gives 27, the number of gold counters. Adding 0.2 and 0.45 without subtracting from 1 gives 0.65, and 0.65 × 60 = 39 counts the pink and gold counters together, not the grey ones.
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