Year 10 Extended · Unit 4 Sets, Venn diagrams and conditional probability · Lesson 2 of 6

How do you count each person exactly once?

Three overlapping sets need a rule for the size of their union. This lesson builds the rule step by step and proves that it counts every element exactly once.

Statistics and ProbabilityNo calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 10, lesson 1: simplify (A ∪ B)′ using De Morgan's law.
    AnswerA′ ∩ B′. Break the line and flip the sign.
  2. From Year 10, Standard: n(A) = 17, n(B) = 12 and n(A ∩ B) = 5. Find n(A ∪ B).
    Answer24. 17 + 12 − 5: the 5 in both were counted twice.
  3. From Year 7: how many different pairs can be made from the three letters X, Y, Z?
    Answer3: XY, XZ and YZ.
Words used in this lesson
union
all elements in at least one of the sets. For example: club members in robotics or choir or both.
intersection
elements in all the sets named. For example: members of both robotics and choir.
at least one
one or more of the sets. For example: a student in one, two or three clubs.
exactly two
in two of the sets and not in the third. For example: in robotics and choir only.
multiplicity
how many times an element has been counted so far. For example: an element counted twice so far has multiplicity 2.
pairwise disjoint
no two of the sets share an element. For example: three clubs with no student in two of them.

Three clubs, one announcement

At a school in Dhaka there are three clubs: robotics, debating and choir. Each club secretary reports the number of members. The head teacher adds the three numbers and announces that this many students belong to at least one club. The secretaries know that some students belong to two clubs, and a few belong to all three.

Predict first: is the head teacher's total too high, too low, or right? If it is wrong, what must be taken off? Test your prediction on the figure, one step at a time.

1
1
68running total
47direct count of the union
3 of 7regions counted once

Survey 1, step 1 (add the three set totals): the running total is 68 and the union holds 47. 3 of the 7 regions are counted exactly once.

Show me the steps
  1. Survey 1, step 1: n(A) + n(B) + n(C) = 24 + 21 + 23 = 68. The union holds only 47, because the three pair regions are counted twice and the centre three times. Only 3 of the 7 regions are counted once.
  2. Step 2, take off the three pair totals 8, 7 and 9, which sum to 24: 68 − 24 = 44. The pair-only regions are now counted once, but the centre was added three times and taken off three times, so it is counted 0 times and the total is 3 short of 47.
  3. Step 3, add the triple: 44 + 3 = 47. Every region is counted exactly once, 7 of 7, and the total equals the direct count.
  4. Choose survey 4, where no element is in all three sets. Step 1 gives 7 + 7 + 14 = 28 and step 2 gives 28 − 5 = 23, which already equals the direct count. Step 3 adds 0, so nothing changes.

The three-set rule. n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(A ∩ C) − n(B ∩ C) + n(A ∩ B ∩ C).

Why it works. Count how many times each kind of element is counted. In exactly one set: 1 time. In exactly two sets: 2 − 1 = 1 time. In all three: 3 − 3 + 1 = 1 time. Every element is counted once, so the total is the union.

Elements in exactly two sets = (sum of the three pair totals) − 3 × n(A ∩ B ∩ C), because each pair total includes the centre once.

Subtracting the pairs is not enough. Take A = {1, 2, 3}, B = {3, 4} and C = {3, 5}. n(A) + n(B) + n(C) = 3 + 2 + 2 = 7, and each pair intersection is {3}, so the pairs sum to 3. Then 7 − 3 = 4, but the union {1, 2, 3, 4, 5} has 5 elements. The element 3 was added three times and taken off three times, so it vanished and must be put back once.

Worked examples

1. n(A) = 40, n(B) = 35, n(C) = 28, n(A ∩ B) = 12, n(A ∩ C) = 10, n(B ∩ C) = 9 and n(A ∩ B ∩ C) = 5. Find n(A ∪ B ∪ C).

  1. Add the totals: 40 + 35 + 28 = 103.
  2. Take off the pairs: 12 + 10 + 9 = 31, and 103 − 31 = 72.
  3. Add the triple back: 72 + 5 = 77.

2. n(A ∪ B ∪ C) = 70, n(A) = 36, n(B) = 30, n(C) = 25 and the three pair totals sum to 26. Find n(A ∩ B ∩ C).

  1. Write the rule: 70 = (36 + 30 + 25) − 26 + t = 91 − 26 + t = 65 + t.
  2. So t = 70 − 65 = 5. Check that this is possible: each pair total contains the centre, so the pairs must add to at least 3 × 5 = 15, and 26 does. For example, pair totals 10, 8 and 8 give regions 23, 17, 14, 5, 3, 3 and 5, which add to 70.

3. Show that every element is counted once.

  1. An element in exactly one set is in one total and in no pair or triple: counted 1 time.
  2. An element in exactly two sets is in two totals (added twice) and in one pair (taken off once): 2 − 1 = 1.
  3. An element in all three is in three totals, three pairs and the triple: 3 − 3 + 1 = 1.

Your go. The first step is given. n(U) = 150, n(A) = 60, n(B) = 50, n(C) = 45, n(A ∩ B) = 20, n(A ∩ C) = 15, n(B ∩ C) = 18 and n(A ∩ B ∩ C) = 8. How many elements are in none of the three sets?

  1. The totals add to 60 + 50 + 45 = 155 and the pairs add to 20 + 15 + 18 = 53.
  2. Now find n(A ∪ B ∪ C), then take it from 150.
Check the answer

n(A ∪ B ∪ C) = 155 − 53 + 8 = 110, so the number in none of the sets is 150 − 110 = 40.

Check before you go on

1. The four-set rule adds the four totals, takes off every pair intersection, adds every triple intersection, then deals with n(A ∩ B ∩ C ∩ D). Before that last step, how many times has an element in all four sets been counted?

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. n(A) = 30, n(B) = 24, n(C) = 20, n(A ∩ B) = 9, n(A ∩ C) = 7, n(B ∩ C) = 6 and n(A ∩ B ∩ C) = 3. Find n(A ∪ B ∪ C).

3. n(U) = 120, n(A) = 50, n(B) = 45, n(C) = 40, n(A ∩ B) = 18, n(A ∩ C) = 15, n(B ∩ C) = 12 and n(A ∩ B ∩ C) = 8. How many elements are in none of the three sets?

Level 2 · Building

4. n(A ∩ B) = 15, n(A ∩ C) = 12, n(B ∩ C) = 10 and n(A ∩ B ∩ C) = 4. How many elements are in exactly two of the three sets?

5. n(A) = 22, n(B) = 19, n(C) = 16, n(A ∪ B ∪ C) = 40 and n(A ∩ B ∩ C) = 3. Find the sum n(A ∩ B) + n(A ∩ C) + n(B ∩ C).

Level 3 · Stretch

6. A club of 100 people, each in at least one of three teams, has n(A) + n(B) + n(C) = 165, and 15 people are in all three teams. How many are in exactly two teams?

7. The rule for n(A ∪ B ∪ C ∪ D) has the totals, then the pairs, then the triples, then the quadruple. How many terms does it have altogether?

Something to think about

When is n(A ∪ B ∪ C) exactly n(A) + n(B) + n(C)? And when is it as small as possible, equal to just one of the three totals? Describe the Venn diagram in each case.

One way to think about it

The union equals the sum of the totals only when nothing is counted twice, so the three sets are pairwise disjoint: all three pair intersections are empty, and the triple is empty too. It is as small as it can be when two of the sets sit inside the third, for example B ⊆ A and C ⊆ A, so the union is A and n(A ∪ B ∪ C) = n(A).

Exit ticket

8. n(A ∩ B) = 9, n(A ∩ C) = 7, n(B ∩ C) = 8 and n(A ∩ B ∩ C) = 3. How many elements are in at least two of the three sets?