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.
Answer key: Sets, Venn diagrams and tree diagrams worksheet — GCSE Foundation
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- (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.
- (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) 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.
- (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) 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.
- (b) 40 — n(P ∪ Q) = n(P) + n(Q) − n(P ∩ Q) = 34 + 27 − 11 = 50. The complement is everyone outside both sets: n((P ∪ Q)′) = 90 − 50 = 40. Adding P and Q without subtracting the overlap gives 34 + 27 = 61, so 90 − 61 = 29 double-subtracts the 11 who are in both. Reporting n(P ∪ Q) itself, 50, forgets to take the complement at all. Subtracting only n(P) from the universal set, 90 − 34 = 56, ignores set Q altogether.
- (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.
- (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.
- (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) 12 — Method: each result of the first experiment can be paired with every result of the second, so the two counts are combined by multiplying. Working: the coin lands in 2 ways and the dice lands in 6 ways. Each of the 6 dice scores can appear with a head or with a tail, so the table has 6 × 2 rows. Answer: there are 12 different results. The distractors: 8 comes from adding the two counts, 6 + 2, instead of multiplying them; 6 comes from listing the dice scores only and treating the coin as making no difference to the table; 36 comes from working out 6 × 6, counting the coin as though it too had six equally likely results.
- (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.
- (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.
- (b) 3/10 — The number who use at least one app is 90 − 20 = 70. Since 55 + 42 double-counts the overlap, n(X ∩ Y) = 55 + 42 − 70 = 27, so P(both) = 27/90 = 3/10. Forgetting to subtract the 20 who use neither, and using the full 90 as the union, gives 55 + 42 − 90 = 7, so 7/90. Reporting the probability of using X or Y (or both), 70/90 = 7/9, answers a different question about the union, not the overlap. Reporting the probability of using neither app, 20/90 = 2/9, is the complement of the union, not the intersection.
- (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.
- (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.
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