Printable · GCSE Higher · ages 14-16
Probability worksheet — GCSE Higher
Fifteen questions across the probability statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Answer key: Probability worksheet — GCSE Higher
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- (b) 2/11 — Method: restrict the 36 equally likely outcomes to those where at least one die shows a 5, then find what fraction of THOSE give a total of 8. Working: outcomes with at least one 5: (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5) — 11 outcomes. Among these, the total is 8 for (5, 3) and (3, 5) — 2 outcomes. P(total 8 | at least one 5) = 2/11. Answer: 2/11. Watch out: 5/36 is P(total 8) over the full 36 outcomes — it ignores that you already know one die shows a 5. Treating the condition as 'the first die shows a 5' instead of 'at least one die shows a 5' restricts you to only 6 outcomes and misses the (3, 5) case, giving 1/6. And counting only (5, 3) but not its reverse (3, 5) inside the correct 11-outcome list gives 1/11 instead of 2/11.
- (b) 100 — The sample shows a proportion of 8/60 = 2/15 faulty. Apply that proportion to the new batch of 750: 750 × 2/15 = 100. Flipping the ratio, calculating 8/750 × 60 instead of 8/60 × 750, gives 0.64, which rounds to about 1. Assuming the same number of faulty items applies to the new batch, without scaling for its larger size, just repeats the sample's count of 8. Rounding the proportion 8/60 = 0.1333... down to 0.1 before multiplying gives 750 × 0.1 = 75.
- (a) 0.16 — The probability that Kofi fails a single attempt is 1 − 0.6 = 0.4. Since the attempts are independent, the probability he fails both is 0.4 × 0.4 = 0.16. Choosing 0.36 comes from squaring the probability of PASSING instead, 0.6 × 0.6 = 0.36, which is the probability of passing both attempts, not failing both. Choosing 0.4 comes from giving the probability of failing just one attempt, forgetting to combine two attempts. Choosing 0.24 comes from multiplying the fail probability by the pass probability, 0.4 × 0.6 = 0.24, mixing up passing and failing between the two attempts.
- (a) 70 — Method: two linked steps. Find the expected number of billing calls first, then take the 35% of those, because the 35% is quoted for billing calls only. Working: 40% of 500 is 200 billing calls. 35% of 200 is 70 calls. Answer: you would expect 70 calls. The distractors: 200 stops after the first step and gives the billing calls, forgetting that only some of them are dealt with quickly; 175 is 35% of 500, applying the quick response rate to every call the centre takes rather than to the billing calls only; 375 comes from adding 40% and 35% to get 75% and taking 75% of 500, which treats two stages of one journey as separate outcomes to be added.
- (a) 1/6 — Method: write the results of the two dice as ordered pairs, count the pairs whose scores add to the total asked for, divide by the number of ordered pairs there are, then cancel the fraction down. Working: there are 6 × 6 = 36 equally likely ordered pairs. The pairs whose scores add to 7 are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1), which is 6 pairs, so the probability is 6/36. Dividing the top and the bottom by 6 gives 1/6. Answer: the probability is 1/6. The distractors: 7/36 comes from taking the number of favourable pairs to be 7 because 7 is the total asked for, confusing the size of a total with the number of ways of making it; 1/7 comes from using the 21 different combinations of two scores as the equally likely results, finding the 3 combinations 1 and 6, 2 and 5, 3 and 4, and cancelling 3/21; 6/11 comes from counting the 6 favourable pairs correctly but dividing by the 11 possible totals from 2 to 12 rather than by the 36 pairs.
- (a) 5/8 — Winning, near-miss and losing are exhaustive, so the three probabilities sum to 1. Writing 1/4 as 2/8 so every fraction has the same denominator, 1 − 1/8 − 2/8 = 8/8 − 1/8 − 2/8 = 5/8. Subtracting only the winning probability and forgetting the near-miss probability gives 1 − 1/8 = 7/8. Subtracting only the near-miss probability and forgetting the winning probability gives 1 − 1/4 = 3/4. Adding the two given probabilities and stopping there gives 1/8 + 2/8 = 3/8, the probability that a ticket is winning or a near-miss, not the probability that it is losing.
- (b) 1/4 — Method: the person picked is known to be aged 30 or over, so the sample space is those 140 people; divide the number of them who had been to the cinema by 140. Working: 35 of the 140 people aged 30 or over had been to the cinema, giving 35/140. Dividing the numerator and the denominator by 35 gives 1/4. Answer: the probability is 1/4. The distractors: 7/20 is 35/100, taking the count from the older group but the total from the under 30s, which is reading across the wrong row; 7/48 is 35/240, dividing by everyone surveyed instead of by the age group named; 3/4 is 105/140, the probability that someone aged 30 or over had NOT been to the cinema, the opposite event inside the correct group.
- (a) 20 — Dice A is fair, so its expected number of sixes is 150 × 1/6 = 25. Dice B has P(6) = 0.3, so its expected number of sixes is 150 × 0.3 = 45. The difference is 45 − 25 = 20. Adding the two expected values instead of subtracting them gives 25 + 45 = 70. Reporting Dice B's expected sixes on their own, without comparing to Dice A, gives 45. Using the fair probability 1/6 for Dice B as well as Dice A ignores the bias altogether, giving 150 × 1/6 = 25 for both dice and a difference of 0.
- (b) 0.0309 — Method: P(both defective | at least one defective) = P(both defective) ÷ P(at least one defective). Find each using independence: P(both) = 0.06², P(at least one) = 1 − P(neither) = 1 − 0.94². Working: P(both) = 0.06² = 0.0036. P(neither) = 0.94² = 0.8836, so P(at least one) = 1 − 0.8836 = 0.1164. P(both | at least one) = 0.0036 ÷ 0.1164 = 0.0309 (3 s.f.). Answer: 0.0309. Watch out: leaving the answer as 0.0036 gives P(both defective) itself, not the probability once you already know at least one is defective — you still need to divide by P(at least one defective). Giving 0.0600 answers with the single-component defect rate, ignoring the condition altogether. And 0.5000 assumes that 'at least one' makes the outcomes 'exactly one defective' and 'both defective' equally likely, which is not how these probabilities combine.
- (c) 102 — Method: turn the past record into a relative frequency, then use it as an estimate of the probability of rain and multiply by the number of days being predicted for. Working: relative frequency of rain = 70 ÷ 250 = 0.28. Expected rainy days in 365 days = 365 × 0.28 = 102.2, which rounds to about 102 days. Answer: about 102 days. Watch out: writing down 48 swaps which number is the sample and which is the target, working out 70 ÷ 365 × 250 instead of 70 ÷ 250 × 365. Writing down 70 just repeats the original count of rainy days without scaling it up to the new, longer period at all. And writing down 110 comes from rounding the relative frequency to 0.3 before multiplying, 365 × 0.3 = 109.5, when 70 ÷ 250 is exactly 0.28 and needs no rounding at all.
- (a) 0.45, different from 0.4 for all the households — Method: work out the probability inside the restricted group of garden owners, then work out the probability across the whole survey, and compare the two. Working: 54 of the 120 households with a garden own a dog, so the conditional probability is 54 divided by 120, which is 0.45. Across the whole survey 80 of the 200 households own a dog, which is 0.4. Since 0.45 is not 0.4, having a garden changes the chance of owning a dog and the two events are not independent. Answer: 0.45, different from 0.4 for all the households. The distractors: 0.27 is 54/200, dividing the households with both by the whole survey instead of by the 120 with a garden; 0.675 is 54/80, the probability that a household has a garden given that it owns a dog, which is the condition and the event the wrong way round; 0.4 is 80/200, the probability of owning a dog with the garden information never used, which is why that route also reports no difference.
- (d) 3/20 — The probability of taking a red ball from Bag A is 3/5, and the probability of taking a red ball from Bag B is 1/4. Since the two picks are independent, the probabilities are multiplied: 3/5 × 1/4 = 3/20. A candidate who answers 4/9 has added the numerators and added the denominators, (3+1)/(5+4), instead of multiplying the two fractions. A candidate who answers 17/20 has added the two fractions, 3/5 + 1/4, instead of multiplying them. A candidate who answers 3/25 has misread Bag B as also containing 5 balls, using 1/5 instead of 1/4.
- (c) 9/25 — The frequency tree already shows the swim-and-cycle branch directly: 54 out of 150, which simplifies to 9/25. Adding both cycling branches together, 54 + 18 = 72, gives the total number of cyclists, so 72/150 = 12/25, not just those who also swim. Using the non-swimmers' cycling figure, 18, gives 18/150 = 3/25, the probability of cycling WITHOUT swimming. Using the swimmers' total of 90, before splitting by cycling, gives 90/150 = 3/5, the probability of swimming on its own.
- (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.
- (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.
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