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) 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.
- (d) 36/91 — Method: P(both red | same colour) = P(both red) ÷ P(same colour), where P(same colour) = P(both red) + P(both green). Working: P(both red) = 9/20 × 8/19 = 72/380 = 18/95. P(both green) = 11/20 × 10/19 = 110/380 = 11/38. P(same colour) = 18/95 + 11/38 = 36/190 + 55/190 = 91/190. P(both red | same colour) = (36/190) ÷ (91/190) = 36/91. Answer: 36/91. Watch out: stopping at 18/95 gives P(both red) itself, without dividing by the probability that the colours matched at all. Working out 55/91 finds the same-colour probability for green instead of red — check which colour's count you are putting on top. And 9/20 is just the chance the first ball drawn is red, which ignores the second draw and the without-replacement condition completely.
- (c) 7/8 — Method: 'at least one head' is the opposite of 'no heads at all', so work out the probability of three tails and take it away from 1. Working: a flip that is not a head has probability 1/2, and the flips are independent, so three tails in a row has probability 1/2 × 1/2 × 1/2 = 1/8. Taking this from 8/8 leaves 7/8. Answer: the probability is 7/8. The distractors: 1/8 is the probability of three tails, written down without the final subtraction; 3/8 is the probability of exactly one head, which comes from reading 'at least one' as 'exactly one'; 1/2 comes from giving the probability of a head on a single flip and ignoring that three flips are made.
- (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) 0.2 — Method: for independent events, the probability that both happen is the product of the two probabilities. Working: the first set is green with probability 0.4 and the second with probability 0.5, so the calculation is 0.4 × 0.5. Since 4 × 5 = 20 and the two factors carry one decimal place each, the product carries two. Answer: the probability is 0.2. The distractors: 0.9 comes from adding 0.4 and 0.5 instead of multiplying them; 0.45 comes from averaging the two probabilities; 0.1 comes from subtracting 0.4 from 0.5, treating the question as a difference.
- (d) 124/125 — The probability that a seed germinates is 1 − 1/5 = 4/5, so the probability that all three seeds fail to germinate is 1/5 × 1/5 × 1/5 = 1/125. The probability that at least one germinates is 1 − 1/125 = 124/125. Choosing 4/5 comes from giving the probability that a single seed germinates, forgetting to combine all three seeds. Choosing 64/125 comes from working out the probability that ALL three seeds germinate, 4/5 × 4/5 × 4/5 = 64/125, instead of at least one. Choosing 12/125 comes from working out the probability that EXACTLY one seed germinates, 3 × 4/5 × 1/5 × 1/5 = 12/125, instead of at least one.
- (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) 195 — The relative frequency from the trial is 52 ÷ 80 = 0.65, and the expected number of point-up landings in 300 drops is 0.65 × 300 = 195. Giving 52 as the answer reuses the original count from the 80-drop trial without scaling it up to 300 drops at all. Misreading 52 out of 80 as 52% and finding 52% of 300 gives 156. Finding the expected number of point-DOWN landings instead of point-up, using the relative frequency 28 ÷ 80 = 0.35, gives 0.35 × 300 = 105.
- (d) 19/30 — Exactly one means only fiction or only non-fiction, not both: 21 + 17 = 38 out of the 60 readers, which simplifies to 19/30. Including the 14 who read both as well gives 21 + 17 + 14 = 52, so 52/60 = 13/15 — that is at least one, not exactly one. Using only the both-count, 14, as the numerator gives 14/60 = 7/30, the probability of reading both, not exactly one. Using 52, the number who read at least one type, as the denominator instead of the full 60 readers surveyed gives 38/52 = 19/26.
- (d) 4 — Method: list the elements of each set in full, then find which elements appear in both lists — that is A ∩ B. Working: factors of 12 = {1, 2, 3, 4, 6, 12}. Factors of 18 = {1, 2, 3, 6, 9, 18}. The elements in both lists are 1, 2, 3 and 6, so A ∩ B = {1, 2, 3, 6} and n(A ∩ B) = 4. Answer: 4. Watch out: writing down 6 gives n(A), the size of the factors-of-12 list on its own, not the size of the overlap. Writing down 8 comes from counting every element that appears in EITHER list, 1, 2, 3, 4, 6, 9, 12 and 18 — that is the union, a different set from the intersection. And writing down 3 misses that 1 is a factor of both 12 and 18, and so belongs in A ∩ B alongside 2, 3 and 6.
- (c) 15/23 — Method: find P(rough and delayed) and the overall P(delayed) using the tree, then divide. Working: P(rough and delayed) = 0.2 × 0.75 = 0.15. P(calm and delayed) = 0.8 × 0.1 = 0.08. P(delayed) = 0.15 + 0.08 = 0.23. P(rough | delayed) = 0.15 ÷ 0.23 = 15/23. Answer: 15/23. Watch out: leaving the answer as 0.15 (3/20) gives P(rough and delayed) itself, without dividing by the overall probability that a crossing is delayed. Giving 0.75 (3/4) is the probability you were told to start with — that a crossing is delayed GIVEN the sea is rough — which is the reverse of what's being asked. And 0.2 (1/5) is just the original probability that the sea is rough, before you take the fact that the crossing was delayed into account.
- (b) 1/6 — Method: list the ordered pairs where the two scores match, and divide by the 36 equally likely pairs. Working: the matching pairs are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6), which is 6 pairs out of 36, cancelling down to 1/6. Answer: 1/6. Watch out: writing down 1/36 finds the probability of one particular double, such as (6, 6), rather than any double at all. Guessing 1/2 treats 'same' and 'different' as equally likely events, when there are only 6 matching pairs against 30 non-matching ones. And writing down 1/3 comes from listing each double twice, once for each order of the two dice, giving 12 pairs out of 36 — but (1, 1) is a single outcome, and swapping the two dice over does not produce a second one.
- (c) 30 — The numbers less than 4 are 1, 2 and 3, so the probability of that event is 3/6, and the expected count in 90 rolls is 90 × 3/6 = 45. The probability of rolling a 6 is 1/6, and the expected count is 90 × 1/6 = 15. The difference between the two expected counts is 45 − 15 = 30. A candidate who answers 45 has given the expected count for 'less than 4' only, forgetting to subtract the other expected count. A candidate who answers 15 has given the expected count for '6' only. A candidate who answers 36 has used a dice with 5 possible numbers instead of 6, giving 90 × 3/5 = 54 and 90 × 1/5 = 18, a difference of 36.
- (d) 7/18 — Method: the pupil picked is known to study French, so the sample space shrinks to the 18 French students; divide the number who study both languages by 18. Working: 7 of the pupils study both French and German, and all 7 of them are among the 18 French students, so the probability is 7/18, which will not cancel. Answer: the probability is 7/18. The distractors: 7/30 divides by the whole class, keeping the restricted numerator but the full denominator; 1/2 is 7/14, which conditions on the German students instead, answering the probability that a German student also studies French; 7/25 uses 18 + 14 minus 7 = 25, the number who study at least one language, which is a larger group than the one the question restricts you to.
- (c) 24 — Silver, bronze and copper cover every counter, so their probabilities sum to 1: P(copper) = 1 − 1/2 − 1/8 = 3/8. Number of copper counters = 3/8 × 64 = 24. Multiplying the silver probability by 64 gives 32, the number of silver counters, not copper. Multiplying the bronze probability by 64 gives 8, the number of bronze counters. Adding the silver and bronze probabilities (1/2 + 1/8 = 5/8) and multiplying by 64 gives 40, the combined number of silver and bronze counters, not the copper count.
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