Printable · GCSE Foundation · ages 14-16
Sets, Venn diagrams and tree diagrams worksheet — GCSE Foundation
Fifteen questions on "sets, venn diagrams and tree diagrams" — DfE statement P6. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Sets, Venn diagrams and tree diagrams worksheet — GCSE Foundation
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- 1.Priya spins a fair spinner with P(red) = 0.3, and separately flips a fair coin with P(heads) = 0.5. Using a tree diagram, work out the probability that she gets red AND heads.
- 2.At a gym, 45 members are surveyed. 24 use the running machines, 20 use the weights machines, and 9 use both. Work out how many of the 45 members use neither machine.
- 3.A survey of 60 readers records: 21 read only fiction, 17 read only non-fiction, 14 read both, and the rest read neither. Work out the probability that a randomly chosen reader reads exactly one of the two types, giving your answer as a fraction in its simplest form.
- 4.A = {factors of 12} and B = {factors of 18}. Work out n(A ∩ B), the number of elements in both sets.
- 5.A bag contains 5 red counters and 3 blue counters. A counter is taken at random, its colour is noted, and it is put back in the bag before a second counter is taken at random. Work out the probability that both counters are the same colour.
- 6.An ordinary fair dice is rolled once. Set A is the set of even scores and set B is the set of scores greater than 4. Work out the probability that the score rolled is in set A or in set B or in both.
- 7.A survey asks 40 people whether they own a cat, a dog, both, or neither. 12 people own only a cat, 15 own only a dog, 5 own both, and 8 own neither. Work out the probability that a randomly chosen person owns a cat or a dog (or both). Give your answer as a fraction in its simplest form.
- 8.In a class of 32 pupils, 19 study geography and 15 study history. 8 study both subjects. Work out n(G ∪ H), the number of pupils who study geography or history or both.
- 9.A bag contains 4 red counters and 6 blue counters. A counter is taken at random, its colour noted, and it is put back in the bag before a second counter is taken at random. Work out the probability that the two counters are different colours. Give your answer as a fraction in its simplest form.
- 10.Two fair spinners are each numbered 1, 2 and 3. Priya spins both spinners together and records the two numbers as a pair, listing every possible outcome systematically in a grid. Work out the probability that the two numbers are the same.
- 11.At a leisure centre, 50 members are asked which activities they do. 32 do yoga, 24 do pilates, and 11 do both. Work out how many members do yoga only, and not pilates.
- 12.150 pupils are asked about swimming and cycling, and the results are recorded on a frequency tree. The first pair of branches splits the 150 pupils into 90 who swim and 60 who do not swim. Of the 90 who swim, 54 also cycle. Of the 60 who do not swim, 18 cycle. Work out P(S ∩ C), the probability that a pupil, chosen at random, swims and cycles. Give your answer as a fraction in its simplest form.
- 13.Five balls numbered 1, 2, 3, 4 and 5 are in a bag. Two of them are taken out together at random. Work out the probability that the two numbers on them add up to more than 7.
- 14.80 people are asked whether they drink tea and whether they drink coffee, and the results are recorded on a frequency tree. The first pair of branches splits the 80 people into 50 tea drinkers and 30 non-tea-drinkers. Of the 50 tea drinkers, 32 also drink coffee. Of the 30 non-tea-drinkers, 6 drink coffee. Work out n(T ∩ C), the number of people who drink both tea and coffee.
- 15.The universal set is {1, 2, 3, ..., 30}. Set A is the set of multiples of 4 between 1 and 30. Work out n(A′), the number of elements not in A.
Answer key
- (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.
- (b) 10 — Method: the number using at least one machine is the running total plus the weights total minus the overlap, since the overlap would otherwise be added in twice; the number using neither is the survey total minus that. Working: at least one machine = 24 + 20 − 9 = 35. Neither = 45 − 35 = 10. Answer: 10. Watch out: subtracting 24 + 20 from 45 without adding the 9 back, 45 − 24 − 20 = 1, removes the overlap twice over instead of once. Writing down 15, which is 24 − 9, gives the number who use ONLY the running machines, not the number who use neither. And writing down 9 mistakes the overlap region for the region outside both circles altogether.
- (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.
- (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.
- (a) 2/3 — Method: 'in A or in B' means every score that belongs to at least one of the two sets; a score that belongs to both is still only one outcome, so it is listed once. Working: the even scores are 2, 4 and 6; the scores greater than 4 are 5 and 6. Listing the scores that appear in either set gives 2, 4, 5 and 6, with 6 written once. That is 4 of the 6 faces, or 4/6. Answer: the probability is 2/3. The distractors: 5/6 comes from adding the sizes of the two sets, 3 + 2, so that the score 6 is counted in both and appears twice; 1/2 comes from using the even scores alone; 1/6 comes from giving the probability that the score is in both sets, which is the single score 6, rather than in either of them.
- (c) 4/5 — Add the counts of everyone who owns a cat, a dog, or both: 12 + 15 + 5 = 32 out of 40 people, which simplifies to 4/5. Leaving out the 5 people who own both, and adding only the two only-groups, gives 27/40. Using the 8 people who own neither, instead of everyone who owns at least one pet, gives 8/40 = 1/5. Counting the 5 people who own both twice, once alongside each only-group as well as on their own, gives 12 + 15 + 5 + 5 = 37 out of 40, or 37/40.
- (b) 26 — Method: n(G ∪ H) = n(G) + n(H) − n(G ∩ H), taking off the overlap once so the pupils who study both subjects are not counted twice. Working: 19 + 15 − 8 = 26. Answer: 26. Watch out: adding 19 and 15 without taking off the overlap gives 34, which counts the 8 pupils who study both subjects twice. Taking the 8 off both totals before adding, 19 − 8 + 15 − 8 = 18, counts only the pupils who study exactly one of the two subjects and leaves out the 8 who study both. And writing down 11, which is 19 − 8, gives the number who study geography only, not the number who study geography or history or both.
- (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) 1/3 — List the outcomes for the two spinners systematically in a 3 × 3 grid: 1-1, 1-2, 1-3, 2-1, 2-2, 2-3, 3-1, 3-2, 3-3, where the first number is the score on spinner A and the second is the score on spinner B — 9 equally likely outcomes in total. The pairs where the two numbers are the same are 1-1, 2-2 and 3-3, so there are 3 favourable outcomes. P(same number) = 3/9 = 1/3. 2/3 comes from working out the probability that the two numbers are different and then forgetting to take the complement the right way round, so the probability of "different" is given instead of the probability of "same". 1/2 comes from listing only the 6 unordered pairs 1-1, 2-2, 3-3, 1-2, 1-3, 2-3 instead of all 9 ordered outcomes in the grid, then taking 3 out of that 6. 1/9 comes from spotting only one of the three matching pairs, such as 1-1, and missing 2-2 and 3-3.
- (d) 21 — Method: 'yoga only' means yoga but not pilates, so subtract the number who do both from the total who do yoga. Working: 32 − 11 = 21. Answer: 21. Watch out: writing down 32, the total who do yoga, answers 'how many do yoga' rather than 'how many do yoga only' — it still includes the 11 who also do pilates. Adding the overlap instead of subtracting it, 32 + 11 = 43, moves in the wrong direction entirely. And writing down 11 gives the number who do both activities, which is the opposite of yoga only.
- (c) 9/25 — The frequency tree already shows the swim-and-cycle branch directly: 54 out of 150, which simplifies to 9/25. Adding both cycling branches together, 54 + 18 = 72, gives the total number of cyclists, so 72/150 = 12/25, not just those who also swim. Using the non-swimmers' cycling figure, 18, gives 18/150 = 3/25, the probability of cycling WITHOUT swimming. Using the swimmers' total of 90, before splitting by cycling, gives 90/150 = 3/5, the probability of swimming on its own.
- (d) 1/5 — Method: list the pairs systematically, work out each total, then count the pairs that meet the condition and compare that count with the length of the list. Working: the pairs and their totals are 1 and 2 giving 3, 1 and 3 giving 4, 1 and 4 giving 5, 1 and 5 giving 6, 2 and 3 giving 5, 2 and 4 giving 6, 2 and 5 giving 7, 3 and 4 giving 7, 3 and 5 giving 8, and 4 and 5 giving 9. That is 10 pairs, of which 2 have a total of more than 7. Answer: the probability is 1/5. The distractors: 2/5 comes from counting the totals of exactly 7 as well, reading 'more than 7' as '7 or more'; 1/10 comes from finding only the pair 4 and 5 and missing that 3 and 5 also beat 7; 4/5 comes from counting the pairs on the wrong side of the condition, the 8 pairs whose total is 7 or less.
- (d) 32 — The frequency tree already shows the tea-and-coffee branch directly: of the 50 tea drinkers, 32 also drink coffee, so n(T ∩ C) = 32. Adding both coffee branches together, 32 + 6 = 38, gives n(C), the total number of coffee drinkers, not just those who also drink tea. Using the non-tea branch's figure, 6, describes people who drink coffee but NOT tea. Subtracting to get 50 − 32 = 18 finds the tea drinkers who do NOT drink coffee, the opposite region to the one asked for.
- (c) 23 — Multiples of 4 from 1 to 30 are 4, 8, 12, 16, 20, 24 and 28 — 7 numbers, so n(A) = 7. The universal set has 30 elements, so n(A′) = 30 − 7 = 23. Reporting n(A) itself, 7, without subtracting it from the universal set forgets what the complement means. Estimating the count of multiples of 4 as 30 ÷ 4 = 7.5, rounded to 8, instead of listing them exactly, gives 30 − 8 = 22. Miscounting the universal set as having 31 elements instead of 30, an off-by-one slip, gives 31 − 7 = 24.
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