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) 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.
- (d) 5 — Being red and not being red are exhaustive, so their probabilities sum to 1: the probability of red is 1 − 0.8 = 0.2. The number of red counters is 0.2 × 25 = 5. Using 0.8 directly as the probability of red, without taking the complement, gives 0.8 × 25 = 20 — the number of counters that are NOT red. Sharing the 25 counters equally between the three colours, ignoring the given probability altogether, gives 25 ÷ 3 ≈ 8. Misreading the total as 20 counters instead of 25 gives 0.2 × 20 = 4.
- (c) 0.31 — Method: a late bus can be reached along two paths of the tree. Multiply along each path, then add the paths that end in a late bus. Working: the rain path gives 0.3 × 0.8 = 0.24. No rain has probability 1 − 0.3 = 0.7, so the second path gives 0.7 × 0.1 = 0.07. Adding the two paths gives 0.24 + 0.07. Answer: the probability is 0.31. The distractors: 0.9 comes from adding the two branch probabilities 0.8 and 0.1 without first weighting them by how often it rains; 0.24 comes from following the rain path only and ignoring that the bus can also be late when it is dry; 0.27 comes from using 0.3 at the start of both paths, so that the no-rain path is given 0.3 × 0.1 instead of 0.7 × 0.1.
- (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.
- (a) 5/36 — Method: list every result of the two dice as an ordered pair, first score then second score, count the pairs that give the total asked for and divide by how many pairs the list holds. Working: each dice can show 6 scores, so there are 6 × 6 = 36 equally likely ordered pairs. The pairs whose scores add to 6 are (1, 5), (2, 4), (3, 3), (4, 2) and (5, 1), which is 5 pairs out of the 36. Answer: the probability is 5/36. The distractors: 4/36 comes from listing (1, 5), (5, 1), (2, 4) and (4, 2) and leaving (3, 3) out, because a double does not look like a pair that can be turned round; 5/12 comes from finding the 5 pairs but taking the number of possible results to be 6 + 6 = 12, adding the two dice instead of multiplying them; 1/11 comes from treating the eleven possible totals 2, 3, 4 and so on up to 12 as equally likely, so that a total of 6 is one result out of eleven.
- (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.
- (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.
- (b) 1/20 — Method: two draws with nothing put back are combined by multiplying, with the second probability worked out from the tickets still in the bag. Working: the first draw takes the wanted ticket with probability 1/5. That ticket is kept out, so 4 tickets remain and only one of them is the ticket wanted second, giving 1/4. Multiplying gives 1/20. Answer: the probability is 1/20. The distractors: 1/10 comes from ignoring the order and treating the draw as a choice of two tickets from five, of which there are ten; 1/25 comes from keeping the total at 5 for the second draw, which is what happens only if the first ticket is put back; 2/5 comes from counting the two wanted tickets over the five in the bag, as though one draw decided the whole question.
- (a) 0.15 — Let P(water) = x, so P(coffee) = 3x. The four outcomes are exhaustive: 0.36 + 0.04 + x + 3x = 1, so 0.4 + 4x = 1, giving 4x = 0.6 and x = 0.15. So P(water) = 0.15. Reporting 3x, the coffee probability, instead of water gives 0.45. Splitting the remaining 0.6 evenly between coffee and water, ignoring the 3:1 ratio, gives 0.30. Stopping once the remaining probability 0.6 is found, without dividing by the four equal shares, gives 0.60.
- (b) 1/3 — Method: the first sweet has already been taken and it was red, so work out the second probability from what is actually left in the bag. Working: one red sweet has gone, so 3 red sweets remain out of 9 sweets altogether, giving 3/9. Dividing the numerator and the denominator by 3 gives 1/3. Answer: the probability is 1/3. The distractors: 2/5 is 4/10, the probability for the first draw used again, which is only right if the first sweet is put back; 3/10 takes one off the red count but leaves the total at 10, updating half of the fraction; 4/9 takes one off the total but leaves the red count at 4, updating the other half of the fraction.
- (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) 50 — There are 180 − 100 = 80 south-plot gardeners. 30 of them do not grow organically, so the rest do: 80 − 30 = 50. Writing 64 is wrong because that is the number of NORTH-plot gardeners who grow organically, not south. Writing 30 again is wrong because that is the number of south-plot gardeners who do NOT grow organically — the question asks for those who do. Writing 80 is wrong because that is the whole south-plot total, without subtracting the 30 who do not grow organically. The answer is 50.
- (a) 17/32 — Because the counter is replaced, each pick is independent with P(red) = 5/8 and P(blue) = 3/8 every time. Both red has probability 5/8 × 5/8 = 25/64, and both blue has probability 3/8 × 3/8 = 9/64. 'Same colour' means either of these, so add them: 25/64 + 9/64 = 34/64 = 17/32. Giving 25/64 finds only the probability of both counters being red, leaving out both counters being blue, which also counts as the same colour. Giving 15/64 finds the probability of one red and one blue counter in just one of the two possible orders — the opposite of what was asked, and only half of it. Giving 13/28 works out the probabilities as if the counter had NOT been replaced, using 5/8 × 4/7 and 3/8 × 2/7, even though the question states it was put back.
- (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.
- (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.
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