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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- (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.
- (c) 3/16 — The probability that Priya misses a shot is 1 − 3/4 = 1/4. The two shots are independent, so multiply the probability of scoring the first shot by the probability of missing the second shot: 3/4 × 1/4 = 3/16. Choosing 9/16 comes from using the probability of scoring, 3/4, for both shots instead of switching to the miss probability for the second shot, 3/4 × 3/4 = 9/16, which is the probability that she scores with both shots. Choosing 1/2 comes from subtracting the two probabilities instead of multiplying them, 3/4 − 1/4 = 1/2. Choosing 3/4 comes from giving only the probability that she scores with the first shot, forgetting to combine it with what happens on the second shot.
- (b) 10 — The number who use the pool or the sauna (or both) is the total minus those who use neither: 70 − 12 = 58. Since pool + sauna double-counts the overlap, n(P ∩ S) = 38 + 30 − 58 = 10. Adding the pool and sauna counts without subtracting the overlap at all gives 38 + 30 = 68, more members than are in the whole gym. Subtracting the sauna count from the union, 58 − 30 = 28, actually finds the number who use ONLY the pool, not both. Reporting the 'neither' count, 12, confuses it with the 'both' region — they describe opposite corners of the diagram.
- (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) 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) 5/12 — There are 36 equally likely ordered pairs. Listing the pairs where the red score is smaller than the blue score gives 15 outcomes, so the probability is 15/36 = 5/12. Choosing 1/2 comes from assuming the 36 outcomes split evenly between 'red smaller' and 'red bigger', which ignores the 6 outcomes where the two dice show the same score. Choosing 7/12 comes from including the outcomes where the two scores are equal, 15 + 6 = 21, giving 21/36 = 7/12, instead of using 'less than' strictly. Choosing 5/36 comes from counting only the outcomes where the two scores differ by exactly 1, such as (1, 2) and (2, 3), which is 5 outcomes, and missing the pairs that differ by more than 1.
- (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.
- (d) 0.80 — Evens is exactly halfway along the scale, at 0.5, and 3/10 written as a decimal is 0.3. A probability 3/10 higher than evens is 0.5 + 0.3 = 0.80. Giving 0.30 as the answer converts the increase but never adds it to the value of evens. Adding the increase to 1, the 'certain' end of the scale, instead of to evens gives 1 + 0.3 = 1.30, which cannot be a probability. Writing 3/10 as 0.03 instead of 0.3, a place-value slip, gives 0.5 + 0.03 = 0.53.
- (d) 1/12 — Method: list the full possibility space of sandwich-and-drink pairs, then divide the one matching pair by the size of the whole space. Working: there are 3 × 4 = 12 equally likely sandwich-and-drink pairs, and exactly one of them is egg and water. Answer: 1/12. Watch out: writing down 1/7 comes from adding the two counts, 3 + 4 = 7, instead of multiplying them to build the possibility space. Writing down 1/3 uses only the chance of choosing egg out of 3 sandwiches and ignores the drink altogether. And writing down 1/4 uses only the chance of choosing water out of 4 drinks and ignores the sandwich altogether.
- (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) 3/10 — Method: on a tree diagram, follow the path that matches the description and multiply the probabilities written along it; with nothing put back, the second set of branches is worked out from the counters that are left. Working: 3 of the 5 counters are yellow, so the first branch of the path is 3/5. A yellow counter has been kept out, so 4 counters remain and both green counters are still there, making the second branch 2/4. Multiplying along the path gives 6/20. Answer: the probability is 3/10. The distractors: 6/25 comes from using 2/5 on the second branch, which is the tree for a counter that is put back; 3/20 comes from taking one off the green count as well as off the total, using 1/4 on the second branch; 3/5 comes from reading the first branch only and never multiplying along the path.
- (a) 4/9 — There are 3 × 6 = 18 equally likely outcomes. The product is a multiple of 5 whenever Spinner C shows 5, whatever Spinner D shows, which is 6 outcomes, or whenever Spinner D shows 5 and Spinner C shows 2 or 3, which is 2 more outcomes. This gives 6 + 2 = 8 outcomes, so the probability is 8/18 = 4/9. Choosing 1/3 comes from only counting the 6 outcomes where Spinner C shows 5 and forgetting the 2 outcomes where Spinner D shows 5 instead. Choosing 1/9 comes from only counting the 2 outcomes where Spinner D shows 5 and forgetting the 6 outcomes where Spinner C shows 5. Choosing 1/2 comes from counting 9 outcomes instead of 8, by listing the case where Spinner C shows 5 and Spinner D shows 5 twice over.
- (b) 15/32 — Method: there are two ways to get one of each colour, red then blue and blue then red. Work out the probability of each path by multiplying, then add the two paths. Working: red then blue is 5/8 × 3/8 = 15/64, and blue then red is 3/8 × 5/8 = 15/64. Adding the two paths gives 30/64. Answer: the probability is 15/32. The distractors: 15/64 comes from working out red then blue only and forgetting that blue then red also gives one of each; 15/28 comes from doubling correctly but reducing the total to 7 for the second spin, which is what happens to a bag when an item is kept out, not to a spinner; 39/64 comes from working from the opposite event and subtracting only the two-red case, 1 − 25/64, leaving the two-blue case inside the answer.
- (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.
- (a) 0.09 — Method: two independent events that must both happen are combined by multiplying their probabilities. Working: the same probability 0.3 applies to each day, so the calculation is 0.3 × 0.3. Written as fractions this is 3/10 × 3/10 = 9/100. Answer: the probability is 0.09. The distractors: 0.6 comes from adding 0.3 and 0.3 instead of multiplying them; 0.3 comes from quoting the single-day probability, as though the second day added no further condition; 0.9 comes from multiplying 3 by 3 correctly but keeping only one decimal place in the product instead of two.
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