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
MathsUKwww.geekhero.co.uk
- (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.
- (b) 22.8% — Relative frequency as a percentage is the faulty count divided by the total, then multiplied by 100: 33 ÷ 145 × 100 = 22.76, which rounds to 22.8%. Giving 33.0% as the answer uses the frequency, 33, directly as a percentage without dividing by the total 145 at all. Rounding 22.76 down to 22.7% instead of up applies the wrong rounding direction at the first decimal place. Finding the relative frequency of the bulbs that were NOT faulty first: 145 − 33 = 112, and 112 ÷ 145 × 100 = 77.24, answers the opposite question and rounds to 77.2%.
- (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) 1/10 — 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.
- (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) 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.
- (b) 10 — Method: the number using at least one machine is the running total plus the weights total minus the overlap, since the overlap would otherwise be added in twice; the number using neither is the survey total minus that. Working: at least one machine = 24 + 20 − 9 = 35. Neither = 45 − 35 = 10. Answer: 10. Watch out: subtracting 24 + 20 from 45 without adding the 9 back, 45 − 24 − 20 = 1, removes the overlap twice over instead of once. Writing down 15, which is 24 − 9, gives the number who use ONLY the running machines, not the number who use neither. And writing down 9 mistakes the overlap region for the region outside both circles altogether.
- (b) 6/25 — The probability of winning on any one spin is 2/5, and the probability of losing on any one spin is 3/5. Since the spins are independent, multiply the probability of winning on the first spin by the probability of losing on the second spin: 2/5 × 3/5 = 6/25. A candidate who answers 4/25 has used the winning probability for both spins, 2/5 × 2/5. A candidate who answers 9/25 has used the losing probability for both spins, 3/5 × 3/5. A candidate who answers 3/10 has treated the spins as if they were dependent, reducing the second spin's denominator to 4.
- (c) 1/36 — Method: the two dice do not affect each other, so the probability that both results happen is the product of the two separate probabilities. Working: a 1 on the red dice has probability 1/6 and a 2 on the blue dice has probability 1/6. Multiplying gives 1 × 1 = 1 on the top and 6 × 6 = 36 on the bottom. Answer: the probability is 1/36. The distractors: 1/18 comes from allowing either order, counting a 2 on the red dice with a 1 on the blue dice as well, although the stem names which dice shows which score; 1/3 comes from adding 1/6 and 1/6 instead of multiplying; 1/6 comes from giving the probability for one dice only and treating the other as making no difference.
- (d) 7/15 — Both socks are the same colour either if both are red or if both are blue. The probability both are red is 4/6 × 3/5 = 12/30. The probability both are blue is 2/6 × 1/5 = 2/30. Adding these gives 12/30 + 2/30 = 14/30 = 7/15. Choosing 2/5 comes from only working out the 'both red' path, 12/30, and forgetting the 'both blue' path also counts. Choosing 5/9 comes from treating the draws as if the first sock were replaced, using 4/6 × 4/6 + 2/6 × 2/6 = 20/36 = 5/9, instead of reducing the totals for the second draw. Choosing 7/18 comes from reducing the number of socks removed but not the number left to choose from, using 4/6 × 3/6 + 2/6 × 1/6 = 14/36 = 7/18, instead of 5 remaining socks for the second draw.
- (c) Noah — 11/32 = 0.34375, above 34%. — Converting 11/32 to a decimal gives 11 ÷ 32 = 0.34375, which is greater than 34% (0.34), so Noah has the better save rate: 'Noah — 11/32 = 0.34375, above 34%.' Comparing the raw numbers 34 and 11 directly, without converting the fraction to the same form, gives 'Elin — 34 is bigger than 11.' Treating a larger denominator as meaning a bigger value, rather than smaller equal shares, gives 'Noah — 32 is a bigger denominator.' Rounding 34.375% to 34% to the nearest whole percent hides the difference and gives 'Equal — both round to 34% to the nearest percent.'
- (d) 3400 — Combining all three greenhouses gives 200 + 150 + 250 = 600 seeds planted in total, and 172 + 126 + 212 = 510 germinated, so the combined estimate of the germination probability is 510/600 = 0.85. Out of a new batch of 4000 seeds, the expected number to germinate is 4000 × 0.85 = 3400. Writing 3440 is wrong because it uses only Greenhouse 1's rate, 172/200 = 0.86, instead of the combined rate from all three: 4000 × 0.86 = 3440. Writing 3360 is wrong because it uses only Greenhouse 2's rate, 126/150 = 0.84: 4000 × 0.84 = 3360. Writing 510 is wrong because that is the total number that germinated in the ORIGINAL trial, not scaled up to the new batch of 4000 seeds at all. The best estimate is 3400 seeds.
- (a) £150 — Each game, the expected payout is 0.1 × £20 = £2, so the fête's expected profit per game is the £3 charged minus the £2 expected payout, £1. Over 150 games, that is 150 × £1 = £150. Writing £300 is wrong because 150 × £2 = £300 is the total expected PAYOUT, not the profit — it has not been subtracted from the entry fees. Writing £450 is wrong because 150 × £3 = £450 is the total money taken in entry fees, without accounting for what is expected to be paid out in prizes. Writing £1 is wrong because that is only the expected profit for ONE game — it has not been scaled up to all 150 games. The fête's expected profit is £150.
- (d) 9 — Method: each of the first spin's 3 outcomes can be paired with each of the second spin's 3 outcomes, so the two counts are multiplied together. Working: 3 × 3 = 9 ordered pairs. Answer: 9. Watch out: writing down 6 comes either from adding the two spins' counts, 3 + 3, instead of multiplying them, or from listing colour pairs without order, treating (red, blue) as the same entry as (blue, red) — both shortcuts miss outcomes that the ordered list contains. Writing down 3 lists only a single spin's outcomes and never pairs the two spins up at all. And writing down 27 treats the spinner as though it had been spun three times, 3 × 3 × 3, one spin too many.
- (b) 0.55 — Method: it is easier to find the probability that Ffion wins NEITHER stage, then subtract that from 1. Working: P(lose stage 1) = 1 − 0.4 = 0.6, and P(lose stage 2) = 1 − 0.25 = 0.75. P(neither) = 0.6 × 0.75 = 0.45. P(at least one) = 1 − 0.45 = 0.55. Answer: 0.55. Watch out: adding the two win probabilities, 0.4 + 0.25 = 0.65, treats winning both as impossible and overcounts — that is not how independent probabilities combine. Multiplying the two win probabilities, 0.4 × 0.25 = 0.1, gives the probability of winning BOTH stages, not at least one. And stopping at 0.45, the probability of winning neither stage, forgets the final step of subtracting from 1.
Build your own mix at the worksheet builder.