Venn Diagrams

This lesson covers set notation, constructing Venn diagrams, finding probabilities from Venn diagrams, and extending to three sets for Higher tier students. Venn diagrams are a major topic in the Edexcel GCSE Mathematics (1MA1) specification and regularly appear in 4–6 mark exam questions.

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Set Notation

Before working with Venn diagrams, you need to understand set notation:

Symbol	Name	Meaning
ξ (or U)	Universal set	The set of all elements being considered
A	Set A	A particular group of elements
A ∪ B	A union B	Elements in A or B (or both)
A ∩ B	A intersection B	Elements in both A and B
A'	Complement of A	Elements not in A
n(A)	Number in A	The count of elements in set A
∅	Empty set	A set with no elements

Key Relationships

• A ∪ B includes everything in A, everything in B, and everything in both.
• A ∩ B includes only the elements that belong to both A and B.
• A' includes all elements in the universal set that are not in A.
• (A ∪ B)' = elements in neither A nor B.
• (A ∩ B)' = elements not in the intersection (i.e. not in both A and B simultaneously).

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Two-Set Venn Diagrams

Structure

A two-set Venn diagram has two overlapping circles inside a rectangle. The rectangle represents the universal set ξ.

The four regions are:

1. Only A (in A but not in B)
2. A ∩ B (in both A and B) — the overlap
3. Only B (in B but not in A)
4. (A ∪ B)' — outside both circles (in neither A nor B)

Region Map for a Two-Set Venn Diagram

Although a real Venn diagram uses circles, we can picture the four regions as nodes inside the universal set:

graph TD
    U["ξ (Universal Set)"] --> R1["A only<br/>(A ∩ B’)"]
    U --> R2["A ∩ B<br/>(both)"]
    U --> R3["B only<br/>(A’ ∩ B)"]
    U --> R4["Neither<br/>(A’ ∩ B’) = (A ∪ B)’"]

    style R2 fill:#27ae60,color:#fff
    style R4 fill:#95a5a6,color:#fff

Important: Always fill in the intersection first, then work outwards.

Worked Example 1

In a class of 30 students:

• 18 study French (F)
• 12 study Spanish (S)
• 5 study both French and Spanish

Draw a Venn diagram and find: (a) the number who study French only (b) the number who study neither language (c) P(a randomly chosen student studies Spanish only)

Solution:

Start with the intersection: n(F ∩ S) = 5

French only = 18 − 5 = 13 Spanish only = 12 − 5 = 7 Neither = 30 − 13 − 5 − 7 = 5

The Venn diagram regions:

• Only F: 13
• F ∩ S: 5
• Only S: 7
• Neither: 5
• Total: 13 + 5 + 7 + 5 = 30 ✓

(a) French only = 13

(b) Neither = 5

(c) Spanish only = 7, total = 30. P(Spanish only) = 7/30 = 7/30

Edexcel Exam Tip: Edexcel awards marks for the process of filling in the Venn diagram. Always start from the intersection and work outwards. Label each region clearly.

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Finding Probabilities from Venn Diagrams

Once the Venn diagram is complete, probabilities are found by dividing the relevant region total by the grand total.

Worked Example 2

Using the Venn diagram from Worked Example 1, find:

(a) $P(F \cup S)$P(F∪S) — probability of French or Spanish (or both) (b) $P(F')$P(F′) — probability of not French (c) $P(F \cap S')$P(F∩S′) — probability of French but not Spanish

Solution:

(a) $P(F \cup S)$P(F∪S) = (13 + 5 + 7) / 30 = 25/30 = 5/6

(b) $P(F')$P(F′) = (7 + 5) / 30 = 12/30 = 2/5

(c) $P(F \cap S') = \frac{13}{30}$P(F∩S′)=3013​ = 13/30

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Using Venn Diagrams with Algebra

Worked Example 3

80 students are asked about sports. Let T = {students who play tennis} and B = {students who play badminton}.

• n(T) = 45
• n(B) = 38
• n(T ∪ B)' = 12

Find n(T ∩ B).

Solution:

Students in T ∪ B = 80 − 12 = 68

Using the addition rule: n(T ∪ B) = n(T) + n(B) − n(T ∩ B)

68 = 45 + 38 − n(T ∩ B) 68 = 83 − n(T ∩ B) n(T ∩ B) = 83 − 68 = 15

Check: T only = 45 − 15 = 30, B only = 38 − 15 = 23, both = 15, neither = 12. Total = 30 + 15 + 23 + 12 = 80 ✓

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The Addition Rule for Probability

For any two events A and B:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$P(A∪B)=P(A)+P(B)−P(A∩B)

This avoids double-counting the intersection.

Worked Example 4

$P(A)$P(A) = 0.6, $P(B)$P(B) = 0.5, $P(A \cap B)$P(A∩B) = 0.2. Find $P(A \cup B)$P(A∪B).

Solution: $P(A \cup B)$P(A∪B) = 0.6 + 0.5 − 0.2 = 0.9

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Three-Set Venn Diagrams (Higher)

For three sets A, B and C, there are eight regions:

1. Only A
2. Only B
3. Only C
4. A ∩ B only (not C)
5. A ∩ C only (not B)
6. B ∩ C only (not A)
7. A ∩ B ∩ C (all three)
8. (A ∪ B ∪ C)' (none)

Strategy: Always fill in the centre region (all three) first, then the two-set intersections, then the single-set only regions, and finally the outside.

Worked Example 5

100 students are surveyed about the subjects they enjoy.

• 50 enjoy Maths (M)
• 42 enjoy Science (S)
• 35 enjoy English (E)
• 20 enjoy Maths and Science
• 15 enjoy Maths and English
• 10 enjoy Science and English
• 5 enjoy all three

Find the number who enjoy none of the three subjects.

Solution:

Start from the centre:

• M ∩ S ∩ E = 5
• M ∩ S only = 20 − 5 = 15
• M ∩ E only = 15 − 5 = 10
• S ∩ E only = 10 − 5 = 5
• M only = 50 − 15 − 10 − 5 = 20
• S only = 42 − 15 − 5 − 5 = 17
• E only = 35 − 10 − 5 − 5 = 15

Total in at least one set = 20 + 15 + 10 + 5 + 17 + 5 + 15 = 87

None = 100 − 87 = 13

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Mutually Exclusive Events

Two events are mutually exclusive if they cannot both happen at the same time.

If A and B are mutually exclusive: $P(A \cap B)$P(A∩B) = 0 and $P(A \cup B) = P(A) + P(B)$P(A∪B)=P(A)+P(B)

On a Venn diagram, mutually exclusive sets do not overlap.

Worked Example 6

A card is drawn from a standard pack. Let A = {card is a king} and B = {card is a queen}.

A and B are mutually exclusive because a card cannot be both a king and a queen.

P(king or queen) = P(king) + P(queen) = 4/52 + 4/52 = 8/52 = 2/13

Worked Example 7 — Finding $P(A \cup B)$P(A∪B), $P(A \cap B)$P(A∩B), $P(A')$P(A′)

A two-set Venn diagram is filled in as follows for 40 people surveyed:

• Only A: 15
• A ∩ B: 8
• Only B: 12
• Neither: 5

Find: (a) $P(A \cup B)$P(A∪B) (b) $P(A \cap B)$P(A∩B) (c) $P(A')$P(A′) (d) $P(A' \cap B)$P(A′∩B)

Solution:

Total = 15 + 8 + 12 + 5 = 40 ✓

(a) n(A ∪ B) = 15 + 8 + 12 = 35. $P(A \cup B) = \frac{35}{40}$P(A∪B)=4035​ = 7/8

(b) n(A ∩ B) = 8. $P(A \cap B) = \frac{8}{40}$P(A∩B)=408​ = 1/5

(c) n(A') = "not in A" = B only + Neither = 12 + 5 = 17. $P(A') = \frac{17}{40}$P(A′)=4017​ = 17/40

(d) A' ∩ B means "in B but not in A" — that is Only B = 12. $P(A' \cap B) = \frac{12}{40}$P(A′∩B)=4012​ = 3/10

Worked Example 8 — Combined Events and the Addition Rule

Events A and B have $P(A)$P(A) = 0.55, $P(B)$P(B) = 0.40 and $P(A \cup B)$P(A∪B) = 0.75.