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.Leo flips a fair coin twice. He makes a list of every possible result of the two flips, writing down what the first flip lands on and then what the second flip lands on. Work out how many of the results in his list contain at least one tail.
- 2.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.
- 3.A = {factors of 12} and B = {factors of 18}. Work out n(A ∩ B), the number of elements in both sets.
- 4.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.
- 5.80 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.
- 6.A school surveys 60 pupils. 32 say they walk to school, 24 say they cycle to school, and 12 say they do both, on different days. The rest travel only by car. One of the 60 pupils is chosen at random. Work out the probability that this pupil travels only by car.
- 7.150 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.
- 8.Two sets, A and B, sit inside a universal set. A region of a Venn diagram is described by the notation A ∩ B′. Work out which of these describes that region.
- 9.At a youth club, 65 members are asked which sports they play. 27 play football, 21 play basketball, and 9 play both. Work out the number of members who play neither sport.
- 10.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.
- 11.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.
- 12.Amir has 3 different coloured pens, red, blue and green, and 2 different types of paper, lined and plain. He wants to choose one pen and one type of paper, and draws a grid to list every combination systematically. Work out how many combinations are in his grid.
- 13.In a survey of 45 students, 22 said they like reading, 18 said they like gaming, and 6 said they like both. Work out how many students like exactly one of reading or gaming.
- 14.A bag contains one red counter and one blue counter. Josh takes a counter, notes its colour, puts it back, then takes a counter again. He draws a tree diagram to show every possible pair of colours. Work out how many outcomes are on his tree diagram.
- 15.The numbers 1 to 10 are each written on a card. Set A is the set of multiples of 3 from 1 to 10 and set B is the set of even numbers from 1 to 10. Work out how many of the numbers from 1 to 10 are in set A but not in set B.
Answer key
- (b) 3 — Method: build the list of possible results systematically, taking the first flip as a head and then as a tail, and keeping the two flips in order so that a head then a tail is a different result from a tail then a head; then count the results that match the description. Working: with a head at the first flip the results are HH and HT, and with a tail at the first flip they are TH and TT, so the list is HH, HT, TH, TT — four results in all. The results containing at least one tail are HT, TH and TT. Answer: 3 of the results in the list contain at least one tail. The distractors: 2 comes from treating a head and a tail as one result however they are ordered, which shortens the list to HH, HT, TT so that only two entries hold a tail; 1 comes from reading 'at least one tail' as a tail at both flips, which is the single result TT; 4 comes from counting the tails written across the list — one in HT, one in TH and two in TT — instead of counting the results that contain a tail.
- (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.
- (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) 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) 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.
- (a) 4/15 — Method: first find how many pupils travel only by car, then write that as a fraction of the 60 pupils surveyed. Working: pupils who walk or cycle or both = 32 + 24 − 12 = 44. Only by car = 60 − 44 = 16. P(only by car) = 16/60 = 4/15. Answer: 4/15. Watch out: writing down 1/5 takes the overlap of 12 pupils on its own, 12/60, mistaking the group who do both for the group who travel only by car. Writing down 4/5 comes from 60 − 12 = 48, subtracting only the overlap from the total instead of the whole walk-or-cycle count, so cyclists and walkers who are not in the overlap are wrongly swept into the only-car group. And writing down 7/15 comes from 60 − 32 = 28, subtracting the walkers alone and forgetting the cyclists altogether.
- (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.
- (c) The elements that are in A but not in B — The symbol ∩ means 'and', so A ∩ B′ means 'in A and also in the complement of B'. B′ means 'not in B'. So A ∩ B′ describes everything that is in A but not in B. The elements in both A and B describes A ∩ B, without the dash on B. The elements in B but not in A describes B ∩ A′, with the dash on A instead of B. The elements in neither A nor B describes (A ∪ B)′, the region outside both circles entirely.
- (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.
- (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) 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.
- (c) 6 — Method: to list every combination of one item from a group of 3 and one item from a group of 2 systematically, multiply the two numbers of choices together. Working: 3 × 2 = 6. Answer: 6. Watch out: adding the two totals instead of multiplying, 3 + 2 = 5, misses combinations that a full grid would show — a grid with 3 rows and 2 columns has 6 cells, not 5. Writing down 3 counts only the pens, and writing down 2 counts only the types of paper — neither one pairs every pen with every type of paper.
- (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.
- (d) 4 — Method: each of the first draw's 2 outcomes can be paired with each of the second draw's 2 outcomes, since the counter is put back before the second draw, so the tree has one branch for every combination. Working: 2 × 2 = 4 outcomes: red-red, red-blue, blue-red, blue-blue. Answer: 4. Watch out: writing down 2 lists only the colours of a single draw and never branches out to a second draw at all. Writing down 3 treats red-then-blue and blue-then-red as the same branch, when the tree diagram shows them as two separate paths, since the counter is put back and either colour could come first or second. And writing down 16 comes from working out 2 × 2 × 2 × 2, as though the counter were drawn four times instead of twice.
- (c) 2 — Set A, the multiples of 3, is {3, 6, 9}. Removing the numbers that are also in set B, the even numbers {2, 4, 6, 8, 10}, leaves {3, 9} — 2 numbers are in A but not B. Giving 3, the whole size of set A, forgets to remove the number 6, which is also even. Giving 7 counts every number in A or B combined, {2, 3, 4, 6, 8, 9, 10}, rather than only those in A but not B. Giving 4 counts the numbers in set B but not set A, {2, 4, 8, 10}, the wrong way round.
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