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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- (a) 3,860 — The probability a bulb works correctly is the complement of being defective: 1 − 0.035 = 0.965. Expected number working correctly = 0.965 × 4,000 = 3,860. Using the probability of being defective instead of its complement gives 4,000 × 0.035 = 140, the expected number of DEFECTIVE bulbs, not working ones. Shifting the decimal point in the complement, using 0.0965 instead of 0.965, gives 4,000 × 0.0965 = 386. Assuming every bulb works, ignoring the 0.035 probability altogether, gives the full batch of 4,000.
- (b) 11/36 — Method: 'at least one' is the opposite of 'none at all', so work out the probability of no 5 on either roll and take it away from 1. Working: a roll that is not a 5 has probability 5/6, and the rolls are independent, so no 5 at all has probability 5/6 × 5/6 = 25/36. Taking this from 36/36 leaves 11/36. Answer: the probability is 11/36. The distractors: 25/36 is the probability of no 5 at all, written down without the final subtraction; 12/36 comes from counting the 6 pairs with a 5 on the first roll and the 6 pairs with a 5 on the second and adding them, which counts the pair (5, 5) twice; 30/36 comes from working out 1 − 1/6 as though only one roll were made.
- (b) (48/52) × (47/51) — Method: for two deals one after the other with nothing put back, multiply the probability of the first by the probability of the second worked out from the cards that are left. Working: 52 − 4 = 48 cards are not aces, so the first card is not an ace with probability 48/52. One card has now gone and it was not an ace, so 51 cards remain and 47 of them are not aces, giving 47/51. Answer: the probability is (48/52) × (47/51). The distractors: (48/52) × (48/52) comes from leaving the pack at 52 cards for the second deal, which is only true if the first card is replaced; (4/52) × (3/51) comes from working out the probability that both cards ARE aces instead of neither; (4/52) × (4/51) comes from the same misreading with the ace count left at 4 while the total is reduced, adjusting only half of the second fraction.
- (b) 11/20 — Method: the person picked is known to own a bicycle, so work inside the 20 bicycle owners and count how many of them do not own a car. Working: 9 of the 20 bicycle owners also own a car, so 11 of them do not. The probability is 11/20, which will not cancel. Answer: the probability is 11/20. The distractors: 9/20 gives the bicycle owners who DO own a car, answering the opposite event inside the correct group; 11/50 divides by the whole group of 50, keeping the restricted numerator but the full denominator; 11/32 puts the count over the number of car owners, conditioning on the wrong group entirely.
- (c) 0.04 — Method: 'made by machine B and faulty' is the second branch of a tree followed after the first, so multiply the probability of machine B by the probability of a fault given machine B. Working: machine B makes 0.4 of the bolts, and 0.1 of those bolts are faulty, so the probability is 0.4 × 0.1 = 0.04. Answer: the probability is 0.04. The distractors: 0.5 comes from adding 0.4 and 0.1 instead of multiplying, treating two stages of one journey as two separate outcomes; 0.1 gives the fault rate for machine B on its own, as though every bolt in the factory came from machine B, so the 40% share is never used; 0.07 is 0.6 × 0.05 added to 0.4 × 0.1, the probability that a bolt is faulty whichever machine made it, which answers a question about all the production rather than about machine B.
- (c) 9/16 — There are 180 students in total and 84 are in Year 11, so Year 10 has 180 − 84 = 96 students. Of those 96, 42 travel by bus, so 96 − 42 = 54 walk. P(Year 10 student walks) = 54/96 = 9/16. Using the whole school of 180 as the denominator instead of just the 96 Year 10 students gives 54/180 = 3/10. Using the bus count, 42, as if it were the number who walk gives 42/96 = 7/16, the wrong branch of the Year 10 row. Working out the probability for Year 11 instead of Year 10 — 46 walkers out of 84 — gives 46/84 = 23/42.
- (c) No — 7/30 is the relative frequency; theory stays 1/6. — The theoretical probability of rolling a 6 on an ordinary dice is fixed at 1/6, worked out from the number of equally likely outcomes, and does not change however the dice is actually rolled. The relative frequency from this trial is 7/30, found from what happened in these particular 30 rolls. Since 7/30 and 1/6 are different numbers, the correct statement is 'No — 7/30 is the relative frequency; theory stays 1/6.' Assuming the two values must always match because they describe the same event gives 'Yes — relative frequency always equals theory.' Believing that an observed result redefines the theoretical probability gives 'Yes — the theoretical probability has now become 7/30.' Refusing to work out either value at all gives 'Neither can be found — 30 rolls is too few to tell', which ignores that both numbers CAN be calculated from the information given.
- (c) The green bag (7/24) — Method: fractions can only be ordered once they share a denominator, so rewrite all four over the lowest common denominator and compare the numerators. Working: the lowest common denominator of 12, 8, 24 and 3 is 24, and scaling gives 5/12 = 10/24, 3/8 = 9/24, 7/24 stays as it is, and 1/3 = 8/24; the numerators are then 10, 9, 7 and 8. Answer: the smallest numerator is 7, so the green bag, with 7/24, is the least likely and sits furthest to the left on the 0 to 1 scale. The distractors: the red bag (5/12) comes from finding the largest of the four probabilities instead of the smallest; the blue bag (3/8) comes from scaling 3/8 by changing only the denominator to 24, which turns it into 3/24 and makes it look the smallest; the yellow bag (1/3) comes from comparing numerators alone and assuming the fraction with the numerator 1 must be the smallest.
- (b) £100 — Over 250 games, the expected total winnings are 250 × (1/5) × £12 = £600, since a player wins on 1 of the 5 equally likely sections. The total cost of playing is 250 × £2 = £500. The players' expected profit is the winnings minus the cost: £600 − £500 = £100. Writing £500 is wrong because that is only the total cost of playing, without any winnings included. Writing £600 is wrong because that is only the total expected winnings, without subtracting what was paid to play. Writing £2,500 is wrong because it assumes a win on every single game (250 × £12 = £3,000) instead of using the 1-in-5 probability, then subtracts the cost: £3,000 − £500 = £2,500. The players' expected profit over the 250 games is £100.
- (c) 28 — Reading only is 22 − 6 = 16, and gaming only is 18 − 6 = 12, so exactly one of the two is 16 + 12 = 28. Adding 22 and 18 without removing the 6 who like both, 22 + 18 = 40, counts those 6 students twice. Giving 6 mistakes the number who like both for the number who like exactly one. Finding 22 + 18 − 6 = 34 gives the number who like at least one of reading or gaming, but stops there instead of also removing the 6 who like both to leave only those who like exactly one.
- (a) 43/70 — There are 140 − 85 = 55 Zone B households, and 55 − 21 = 34 of them recycle glass. In total, 52 + 34 = 86 households recycle glass, out of 140: 86/140 = 43/70. Writing 13/35 is wrong because 52/140 simplifies to 13/35, and 52 only counts Zone A, leaving out the 34 Zone B recyclers. Writing 17/70 is wrong because 34/140 simplifies to 17/70, and 34 only counts Zone B, leaving out the 52 Zone A recyclers. Writing 73/140 is wrong because it adds the 52 Zone A recyclers to the 21 Zone B households that do NOT recycle, mixing up two different groups instead of adding the two recycling groups. The probability is 43/70.
- (a) 50 — To find the number of shots needed for an expected 12 hits, divide the number of hits wanted by the probability of a hit: 12 ÷ 0.24 = 50. Multiplying the number of hits by the probability instead of dividing gives 12 × 0.24 = 2.88, which rounds to 3 shots. Rounding 0.24 to 0.25 before dividing gives 12 ÷ 0.25 = 48. Using the probability of missing, 1 − 0.24 = 0.76, instead of the probability of hitting, gives 12 ÷ 0.76 = 15.79, which rounds to 16.
- (c) 0.25 — Method: P(B | A) = P(A and B) ÷ P(A). Working: P(B | A) = 0.15 ÷ 0.6 = 0.25. Answer: 0.25. Watch out: multiplying 0.6 by 0.15 instead of dividing gives 0.09, and subtracting 0.15 from 0.6 gives 0.45 — neither uses the conditional probability formula. Leaving the answer as 0.15 mistakes the probability of A and B happening together for the probability of B once you already know A has happened — those are different quantities.
- (c) 26 — The number who play football or basketball is 27 + 21 − 9 = 39, subtracting the 9 who play both so they are not counted twice. The number who play neither is then 65 − 39 = 26. Giving 39 stops after finding the number who play football or basketball, without taking the complement within the 65 members. Subtracting all three given numbers from 65, 65 − 27 − 21 − 9 = 8, treats the 9 who play both as a separate group rather than a correction for double-counting. Adding 27 and 21 without removing the double-counted 9, then subtracting that total from 65, gives 65 − (27 + 21) = 17.
- (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.
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