Venn diagram problems · Sets & Venn diagrams

Most WAEC sets questions are about counting: a survey gives how many people are in each group and in the overlaps, and you find how many are in some other part. Write the number of elements in each region of a Venn diagram, and every question becomes adding and taking away.

Two sets

A two-set diagram has four regions. Each total in a question covers more than one region: “20 like rice” means rice only plus both.

The four regionsn(A) = A only + both

n(A ∪ B)n(A) + n(B) counts 'both' twice, so subtract it once

n(A \cup B) = n(A) + n(B) - n(A \cap B)

and everyone else is outside both circles:

n(U) = n(A \cup B) + \text{neither}

More: two-set counting

When the overlap is unknown

If the question doesn’t give the overlap, call it $x$ , write every region in terms of $x$ , and make them add up to the total.

Worked example · WAEC 2021

WAEC 2021 · Paper 2 · Q6 (a)

In a class of 80 students, $\frac34$ study Biology and $\frac35$ study Physics. If each student studies at least one of the subjects: (i) draw a Venn diagram to represent this information; (ii) how many students study both subjects? (iii) find the fraction of the class that study Biology but not Physics.

Turn the fractions into numbers
Biology: $\frac34 \times 80 = 60$ . Physics: $\frac35 \times 80 = 48$ .
Think first. What is ¾ of 80? And ⅗ of 80?
(i) Draw the diagram
Let $x$ study both. Biology only is $60 - x$ and Physics only is $48 - x$ . Every student studies at least one subject, so nobody is outside the circles.
Class of 80
Think first. If x study both, how many study Biology only?
(ii) Both subjects
$\begin{aligned} (60 - x) + x + (48 - x) &= 80 \\ 108 - x &= 80 \\ x &= 28 \end{aligned}$
28 students study both.
Think first. The three regions make 80. Write the equation.
(iii) Biology but not Physics
Biology only $= 60 - 28 = 32$ , which is $\frac{32}{80} = \frac25$ of the class.

More: an unknown overlap

Three sets: fill from the centre

With three sets there are eight regions. Every number in the question includes the centre, so start there and work outwards.

All three goes in the centre.
Two only: take the centre off each “A and B” total.
One only: take everything already inside a circle off that circle’s total.
Outside: take everything inside the circles off the overall total.

Fill a Venn diagram from the inside outStep through

The totalsstep 0 of 4

P = 30, B = 50, M = 44 (each includes its overlaps). P and M = 12, B and M = 9, P and B = 7 (each includes the centre). All three = 4. Total 120.

More: three-set counting

Three sets with an unknown

When the centre is the unknown, call it $x$ and write every other region in terms of $x$ , still working from the centre outwards. Then add all eight regions and set them equal to the total.

Worked example · WAEC 2022

WAEC 2022 · Paper 2 · Q9

In a class of 50 students, 18 play football, 20 volleyball and 21 handball. 9 play football and volleyball, 6 football and handball, 11 volleyball and handball, while 15 play none of the 3 games.

Represent the information in a Venn diagram.

How many students play: (i) all the three games? (ii) only one game? (iii) only football?

If a student is chosen at random from the class, find the probability that the student played only two games.

Two games only
Let $x$ play all three. Football and volleyball only: $9 - x$ . Football and handball only: $6 - x$ . Volleyball and handball only: $11 - x$ .
Think first. If x play all three, how many play football and volleyball only?
One game only
Football only: $18 - (9 - x) - (6 - x) - x = 3 + x$ . Volleyball only: $20 - (9 - x) - (11 - x) - x = x$ . Handball only: $21 - (6 - x) - (11 - x) - x = 4 + x$ .
Class of 50
Think first. Football only is 18 minus everything else inside the football circle.
Solve for x
The regions inside the circles hold $50 - 15 = 35$ students:
$\begin{aligned} &(3 + x) + x + (4 + x) + (9 - x) \\ &\quad + (6 - x) + (11 - x) + x = 35 \\ &33 + x = 35, \text{ so } x = 2 \end{aligned}$
So the regions are: F only 5, V only 2, H only 6, F and V only 7, F and H only 4, V and H only 9, all three 2.
Think first. 50 students, and 15 are outside. What do the regions inside add up to?
(b) Counting
(i) All three: 2. (ii) Only one game: $5 + 2 + 6 = 13$ . (iii) Only football: 5.
(c) Probability
Only two games: $7 + 4 + 9 = 20$ students, so $P = \frac{20}{50} = \frac25$ .
Think first. How many play exactly two games?

Probability from a Venn diagram

Once the diagram is filled, a probability is a region’s number over the total:

P(\text{region}) = \frac{\text{number in the region}}{n(U)}

More: probability from a Venn diagram

Your turn

WAEC 2014 · Paper 2 · Q9 (a)

(a)
In the Venn diagram, $P$ , $Q$ and $R$ are subsets of the universal set $U$ . The regions are: $P$ only $16 - 2x$ , $P \cap Q$ only $5x$ , $Q$ only $6 + x$ , $P \cap R$ only $8x$ , $Q \cap R$ only $7x$ , $P \cap Q \cap R$ $4x$ , $R$ only $19 - 3x$ , and $4$ outside. If $n(U) = 125$ , find: (i) the value of $x$ ; (ii) $n[(P \cup Q) \cap R']$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

The eight regions together hold all 125 elements: $(16 - 2x) + 5x + (6 + x) + 8x + 7x + 4x + (19 - 3x) + 4 = 125$ .
Collect the numbers and the $x$ terms: $45 + 20x = 125$ , so $20x = 80$ and $x = 4$ .

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