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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- (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.
- (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) 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.
- (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.
- (a) Game B, by £8 — Game A's expected total is 40 × 0.2 × £5 = £40. Game B's expected total is 40 × 0.1 × £12 = £48. Game B is higher, by £48 − £40 = £8. Writing 'Game A, by £8' is wrong because it has the right difference but the wrong game — Game A's total (£40) is actually LOWER than Game B's, not higher. Writing 'Game B, by £48' is wrong because £48 is Game B's whole expected total, not the DIFFERENCE between the two games. Writing 'Game A, by £40' is wrong in the same way, using Game A's whole total as if it were the margin, and naming the wrong game as the winner. Game B gives the higher expected total, by £8.
- (b) 8/35 — Method: the pupil picked is known to play at least one of the two sports, so first count how many pupils that is, then divide the number who play both by it. Working: 25 play football and 18 play tennis, but the 8 who play both have been counted in each figure, so the number who play at least one sport is 25 + 18 minus 8, which is 35. The pupils who play both give 8/35, which will not cancel. Answer: the probability is 8/35. The distractors: 2/15 is 8/60, dividing by the whole year group instead of by the 35 pupils who play at least one sport; 8/43 uses 25 + 18 as the denominator, forgetting that the 8 pupils who play both have been counted twice; 8/25 conditions on the footballers alone, answering the probability that a footballer also plays tennis rather than using every pupil who plays a sport.
- (d) 3/16 — Method: work out each draw's own probability first, then multiply them together since the two draws are independent. Working: there are 8 tickets in all, 3 of them blue, so P(blue) = 3/8. P(spinner number greater than 2) = 2/4 = 1/2, since 3 and 4 qualify. Multiplying gives 3/8 × 1/2, which comes to 3/16. Answer: 3/16. Watch out: writing down 3/8 stops after the first draw and never brings in the spinner at all. Writing down 1/2 does the opposite, using only the spinner and ignoring the ticket draw. And writing down 5/16 uses 5/8, the probability of a RED ticket, instead of 3/8 for blue — reading the wrong colour off the raffle.
- (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.
- (d) 1/8 — Method: a run of flips of a fair coin gives equally likely sequences of heads and tails, so count the sequences that match and divide by how many sequences there are. Working: each flip lands two ways and no flip affects another, so three flips give 2 × 2 × 2 = 8 equally likely sequences: HHH, HHT, HTH, HTT, THH, THT, TTH and TTT. Only HHH has a head at every flip, so 1 sequence of the 8 matches. Answer: the probability is 1/8. The distractors: 1/4 comes from treating 'three heads', 'two heads', 'one head' and 'no heads' as four equally likely results, which they are not, since one sequence gives three heads and three sequences give two; 1/6 comes from taking the number of sequences to be 2 + 2 + 2 = 6, adding the two ways each flip can land instead of multiplying them; 1/2 comes from reading the first flip only and giving the probability of a head on one flip, without combining it with the other two.
- (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.
- (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.
- (d) 12/17 — Method: the person picked is known to have passed, so the sample space is everyone who passed; divide the course takers who passed by that total. Working: 120 course takers and 50 others passed, so 170 people passed. The course takers who passed give 120/170, and dividing the numerator and the denominator by 10 gives 12/17. Answer: the probability is 12/17. The distractors: 4/5 is 120/150, the probability that someone passed given that they took the course, which reverses the condition and the event; 12/25 is 120/250, dividing by everyone who sat the test rather than by the 170 who passed; 17/25 is 170/250, the probability that a person picked from everyone sitting the test passed, which answers a different question altogether.
- (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%.
- (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.
- (a) 1/6 — Method: write the results of the two dice as ordered pairs, count the pairs whose scores add to the total asked for, divide by the number of ordered pairs there are, then cancel the fraction down. Working: there are 6 × 6 = 36 equally likely ordered pairs. The pairs whose scores add to 7 are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1), which is 6 pairs, so the probability is 6/36. Dividing the top and the bottom by 6 gives 1/6. Answer: the probability is 1/6. The distractors: 7/36 comes from taking the number of favourable pairs to be 7 because 7 is the total asked for, confusing the size of a total with the number of ways of making it; 1/7 comes from using the 21 different combinations of two scores as the equally likely results, finding the 3 combinations 1 and 6, 2 and 5, 3 and 4, and cancelling 3/21; 6/11 comes from counting the 6 favourable pairs correctly but dividing by the 11 possible totals from 2 to 12 rather than by the 36 pairs.
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