Warm-up
Three quick ones from earlier work. Try each in your head or on paper before you open the answer.
- From Lesson 1: U = {1, …, 10}, A = {2, 4, 6, 8, 10} and B = {1, 2, 3, 4}. List A ∩ B.
Answer
{2, 4}. These are the numbers in both lists. - From Year 8: there are 28 students and 12 are left-handed. How many are not left-handed?
Answer
16. 28 − 12 = 16. - From Year 7: 7 + 9 − 3 = ?
Answer
13. Add first (16), then subtract 3.
Words used in this lesson
- overlap
- the elements that are in two sets at the same time. For example: students on both the drama and the debating list.
- only
- in this set and not in the others. For example: in drama only.
- neither
- in none of the sets being discussed. For example: on no list at all.
- at least one
- in one set or more than one. For example: on one list or on both.
- total
- the number of elements in a set, including any overlap. For example: n(A).
- region
- one of the separate parts of a Venn diagram. For example: the part where three circles meet.
Two lists that do not add up
A year group in Kampala has 60 students. The drama list has 38 names and the debating list has 29. Ravi adds the lists, 38 + 29, and gets 67, which is more students than the year group has. Nobody has been invented, so something has been counted twice.
Predict first: what must be true about some students? If every one of the 60 is on at least one list, what is the smallest number that could be on both? Then use the first figure to see how an overlap changes the count.
14 + 11 = 25 counts the 4 in both twice, so n(A ∪ B) = 25 − 4 = 21 and 40 − 21 = 19 are outside both sets.
Show me the steps
- n(A) = 14, n(B) = 11 and 4 in both: 14 + 11 = 25 counts the 4 twice, so n(A ∪ B) = 25 − 4 = 21 and 19 are outside both.
- Slide the overlap to 0: the circles share nothing, so nothing is counted twice and n(A ∪ B) = 14 + 11 = 25, with 15 outside.
- Slide the overlap to 6: n(A ∪ B) = 14 + 11 − 6 = 19, with 21 outside both sets. A bigger overlap means a smaller union.
- Set n(A) = 6, n(B) = 6 and the overlap to 6: A and B are the same set, so n(A ∪ B) = 6 and 34 are outside.
The overlap is counted twice when you add, so take it away once.
n(A ∪ B) = n(A) + n(B) − n(A ∩ B). The number outside both sets is n(U) − n(A ∪ B).
- In A only: n(A) − n(A ∩ B).
- In B only: n(B) − n(A ∩ B).
- For three sets, fill the diagram from the centre outwards: the region in all three first, then the regions in exactly two, then exactly one, then none.
Nothing is filled yet. Start with the region in all three sets, because every other count depends on it.
Show me the steps
- Layer 1: the region in all three sets is the given n(A ∩ B ∩ C) = 3.
- Layer 2: n(A ∩ B) = 8 already includes the 3 in the centre, so A and B only is 8 − 3 = 5. In the same way A and C only is 5 − 3 = 2 and B and C only is 7 − 3 = 4.
- Layer 3: n(A) = 17 includes the centre and two overlaps, so A only is 17 − 5 − 2 − 3 = 7. B only is 18 − 5 − 4 − 3 = 6 and C only is 17 − 2 − 4 − 3 = 8.
- Layer 4: the three sets hold 3 + 11 + 21 = 35 elements, so 40 − 35 = 5 are outside all three sets.
A total includes every region inside its circle. n(A ∩ B) = 8 counts the students in A and B, including those who are also in C. So the region “A and B only” is not 8: it is 8 minus the centre. Writing a given total straight into one region of the diagram is the commonest mistake.
Worked example: how many have neither?
In a class of 36 in Nairobi, 21 own a bike, 14 own a scooter and 5 own both. How many own neither?
- Put the overlap in first: 5 in both.
- Bike only: 21 − 5 = 16. Scooter only: 14 − 5 = 9.
- In at least one set: 16 + 5 + 9 = 30, which is 21 + 14 − 5.
- Outside both: 36 − 30 = 6 own neither.
Your go. The first step is given. In a three-set diagram, n(A) = 18, n(A ∩ B) = 7, n(A ∩ C) = 6 and n(A ∩ B ∩ C) = 2. How many are in A only?
- The centre is 2, and A and B only is 7 − 2 = 5.
- Now find A and C only, then take all three regions inside A away from 18.
Check the answer
A and C only is 6 − 2 = 4. The regions of A other than “A only” hold 2 + 5 + 4 = 11, so A only = 18 − 11 = 7. A common slip is 18 − 7 − 6 = 5, which takes the centre away twice.
Check before you go on
1. n(A) = 25, n(B) = 13 and n(A ∩ B) = 8. What is n(A ∪ B)?
Your turn
Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.
Level 1 · Getting started
2. In a group of 46 pupils in Tashkent, 24 play football, 15 play volleyball and 9 play both. How many play neither sport?
3. In a class of 40, 22 study art, 15 study music and 7 study both. How many study music only?
Level 2 · Building
4. n(A) = 28, n(A ∪ B) = 45 and n(A ∩ B) = 9. What is n(B)?
5. In a three-set Venn diagram, n(A ∩ B ∩ C) = 3, n(A ∩ B) = 10, n(A ∩ C) = 9 and n(B ∩ C) = 8. How many elements are in exactly two of the sets?
6. A two-set Venn diagram has 9 elements in A only, 7 in both, 11 in B only and 4 in neither. What is n(A′)?
Level 3 · Stretch
7. In a survey, 5 people are in all three of A, B and C. Also n(A ∩ B) = 12, n(A ∩ C) = 10, n(B ∩ C) = 9, n(A) = 31, n(B) = 35 and n(C) = 28, and 6 people are in none of the sets. How many people were surveyed?
8. A and B are any two sets. Which statement is always true?
Something to think about
Two sets have n(A) = 20 and n(B) = 15. What is the smallest n(A ∪ B) could be, and what is the largest? Say what the overlap looks like in each case.
One way to think about it
Largest: no overlap, so 20 + 15 = 35. Smallest: B sits entirely inside A, so the overlap is 15 and n(A ∪ B) = 20 + 15 − 15 = 20, the same as n(A). So n(A ∪ B) is always at least the larger of the two totals and at most their sum.
Exit ticket
9. Of 30 students, 18 have a brother, 14 have a sister and 5 have neither. How many have both a brother and a sister?