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.
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Answer key: Probability worksheet — GCSE Higher
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- (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.
- (c) 80 — Since 180 calls are 0.75 of all the technical support calls, the technical support total is 180 ÷ 0.75 = 240. The billing calls make up the rest of the 320 calls, so 320 − 240 = 80. Choosing 240 comes from stopping after finding the technical support total and forgetting the question asks for the billing calls, which are the rest. Choosing 185 comes from multiplying 180 × 0.75 = 135 instead of dividing, then working out 320 − 135 = 185. Choosing 140 comes from using 180 directly as the whole technical support total, ignoring the probability altogether, then working out 320 − 180 = 140.
- (c) 0.15 — Method: for two independent events, multiply along the branches of the tree to find the probability of both outcomes happening together. Working: P(red and heads) = P(red) × P(heads) = 0.3 × 0.5 = 0.15. Answer: 0.15. Watch out: adding the two probabilities, 0.3 + 0.5 = 0.8, does not give the probability of both — probabilities along one path of a tree are multiplied, not added. Writing down 0.5 ignores the spinner altogether and gives only the coin's probability. And writing down 0.65 is the probability of red OR heads, which is 0.3 + 0.5 − 0.15 = 0.65, a different question from the one asked here.
- (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) 50% — The total number of customers who bought a cake is 54 + 21 = 75, combining both hot-drink and non-hot-drink customers. As a percentage of all 150 customers, this is (75 ÷ 150) × 100 = 50%. Choosing 36% comes from only counting the hot-drink customers who bought a cake, (54 ÷ 150) × 100 = 36%, and forgetting the 21 non-hot-drink customers who also bought a cake. Choosing 14% comes from only counting the non-hot-drink customers who bought a cake, (21 ÷ 150) × 100 = 14%, and forgetting the 54 hot-drink customers who also bought a cake. Choosing 60% comes from dividing by the hot-drink total of 90 instead of the grand total of 150, (54 ÷ 90) × 100 = 60%.
- (a) 1276 — Method: an unbiased relative frequency tends towards the theoretical probability as the number of trials increases, so use the record resting on the most trials, then multiply by the number of new trials. Working: the three records rest on 50, 200 and 1000 drops, so the most reliable is the one after 1000 drops, namely 0.638, and the run is indeed settling as the trials increase. The expected number of point up landings in 2000 further drops is 2000 × 0.638 = 1276. Answer: about 1276 times. The distractors: 1440 uses the earliest record, which rests on only 50 drops, giving 2000 × 0.720 = 1440; 1330 uses the middle record, treating 200 drops as a safe compromise when 1000 drops is better still, giving 2000 × 0.665 = 1330; 1348 comes from averaging the three records, since 0.720 + 0.665 + 0.638 = 2.023 and 2.023 ÷ 3 = 0.674, then 2000 × 0.674 = 1348, which gives the 50 drop record the same weight as the 1000 drop record.
- (b) The relative frequency is settling near 0.5 — Method: turn each result into a relative frequency before comparing them, because it is the relative frequency, and not the difference between the two counts, that tends towards the theoretical probability. Working: after 10 flips the relative frequency of a head is 7 ÷ 10 = 0.7, which is a long way from 0.5. After 1000 flips it is 528 ÷ 1000 = 0.528, which is much closer to 0.5. Meanwhile the gap between the two counts has grown rather than shrunk: it was 7 − 3 = 4 after 10 flips and is 528 − 472 = 56 after 1000 flips. Answer: the relative frequency is settling near 0.5, which is what an unbiased experiment does as the sample grows. The distractors: saying the counts are levelling out is the usual form of this idea and the figures contradict it, since the gap went from 4 to 56; saying the coin is biased treats 28 extra heads in 1000 flips as proof, when 0.528 sits close to 0.5 and a fair coin gives results like this often; saying the next flip is more likely to be a tail is the gambler's fallacy, since each flip stays at 1/2 whatever came before.
- (b) 3/10 — The number who use at least one app is 90 − 20 = 70. Since 55 + 42 double-counts the overlap, n(X ∩ Y) = 55 + 42 − 70 = 27, so P(both) = 27/90 = 3/10. Forgetting to subtract the 20 who use neither, and using the full 90 as the union, gives 55 + 42 − 90 = 7, so 7/90. Reporting the probability of using X or Y (or both), 70/90 = 7/9, answers a different question about the union, not the overlap. Reporting the probability of using neither app, 20/90 = 2/9, is the complement of the union, not the intersection.
- (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.
- (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.
- (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.
- (b) 0.225 — The relative frequency of rain is the number of rainy days out of all days recorded: 9 ÷ 40 = 0.225, which is noticeably less than the forecaster's claimed 0.3. Using the number of dry days, 40 − 9 = 31, as the denominator instead of the total of 40 gives 9 ÷ 31 = 0.29 (2 d.p.). Simply reporting the forecaster's claimed value, 0.3, without calculating anything from the data at all, ignores the recorded results completely. Misplacing the decimal point, treating 9 out of 40 as 9%, gives 0.09 instead of 0.225.
- (b) 38% — Method: the two swimming percentages are quoted inside different age groups, so weight each one by the size of its group and add the two results. Working: the under 18s are 45% of the members and 60% of them swim, giving 0.45 × 60 = 27% of all the members. The members aged 18 or over are 55% of the members and 20% of them swim, giving 0.55 × 20 = 11% of all the members. Adding these gives 38%. Answer: 38% of the members swim each week. The distractors: 80% comes from adding 60% and 20% straight off, treating two rates quoted inside different groups as though they could be added; 40% is the mean of 60% and 20%, which would be right only if the two age groups were the same size, and they are not; 42% comes from pairing each swimming rate with the wrong age group, working out 0.45 × 20 added to 0.55 × 60.
- (b) 80 — To find the number of attempts needed, divide the target number of successes by the probability of success: 60 ÷ 0.75 = 80. Writing 45 is wrong because 60 × 0.75 = 45 multiplies instead of dividing — that is the number of successes expected from 60 attempts, not the number of attempts needed for 60 successes. Writing 240 is wrong because 60 ÷ 0.25 = 240 uses 0.25, the probability of MISSING, instead of 0.75, the probability of scoring. Writing 90 is wrong because it comes from misremembering 0.75 as 2/3 and dividing by that instead: 60 ÷ (2/3) = 90. She needs to attempt 80 free throws.
- (b) 500 — Method: when a dice is known to be fair, the theoretical probability is the best thing to work from, and the more trials there are the closer the results tend to it. Working: for a fair dice the probability of a six is 1/6, so the expected number of sixes in 3000 rolls is 3000 × 1 ÷ 6 = 500. The class experiment gave a relative frequency of 14/60, but 60 trials is far too few to overturn a known theoretical value, and the school's 3000 rolls will tend towards 1/6 in any case. Answer: about 500 sixes. The distractors: 700 comes from using the class relative frequency instead of the theory, 3000 × 14 ÷ 60 = 700; 600 comes from splitting the difference between the two, since 1/6 is about 0.167 and 14/60 is about 0.233, whose mean is 0.2, and 3000 × 0.2 = 600; 2500 uses 5/6 instead of 1/6 and counts the rolls expected not to be a six.
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