Venn Diagrams

SPEC 3.1, 3.2 • PROBABILITY

Set notation, addition rule, mutually exclusive and independent events

Set Notation

Master these symbols for Venn diagrams:

ξ

Universal set — all possible outcomes

A \cap B

Intersection — "A and B" (both occur)

A \cup B

Union — "A or B" (at least one)

A^{'}

Complement — "not A"

Key Point

"Or" in probability means "and/or" — includes the case where both events happen.

Two-Set Venn Diagrams

A two-set Venn diagram has 4 regions:

1. Only A (inside A, outside B)

2. Both A and B (the overlap)

3. Only B (inside B, outside A)

4. Neither (outside both circles)

Tip

All 4 regions sum to 1 (or total frequency). Use this to find missing values!

Important

Venn diagrams can show probabilities (summing to 1) or frequencies (summing to total).

The Addition Rule

Formula

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

Use this to find: The probability of "A or B" (at least one event occurring).

Why Subtract P(A ∩ B)?

When we add P(A) + P(B), the intersection gets counted twice — once in P(A) and once in P(B). We subtract P(A ∩ B) to correct this double-counting.

Rearrangements

To find the intersection:

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

To find the union:

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

01Worked Example

In a class of 30 students, 18 study French, 12 study German, and 5 study both. Find:

(a) P(French only) (b) P(at least one language) (c) P(neither)

Draw and fill from the intersection outwards:

•

Both (intersection) = 5

•

French only = 18 − 5 = 13

•

German only = 12 − 5 = 7

•

Neither = 30 − 13 − 5 − 7 = 5

Answers: (a) 13/30 (b) 25/30 = 5/6 (c) 5/30 = 1/6

Worked Example02

P(A) = 0.4, P(B) = 0.5, P(A ∪ B) = 0.7. Find P(A ∩ B).

Rearrange the addition rule:

1

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

2

P (A \cap B) = 0.4 + 0.5 - 0.7 = 0.2

Answer: P(A ∩ B) = 0.2

Mutually Exclusive Events

Definition

Events are mutually exclusive if they cannot both occur:

P (A \cap B) = 0

For mutually exclusive events:

Simplified Rule

P (A \cup B) = P (A) + P (B)

(No overlap to subtract)

Examples:

Rolling 1 vs rolling 6 on one die

Drawing a heart vs drawing a club

Worked Example03

A bag has red, blue, and green counters. P(Red) = 0.35, P(Blue) = 0.4. The colours are mutually exclusive. Find P(Green).

Mutually exclusive and exhaustive → probabilities sum to 1:

1

P(R) + P(B) + P(G) = 1

2

0.35 + 0.4 + P(G) = 1

3

P(G) = 1 − 0.75 = 0.25

Answer: P(Green) = 0.25

Independent Events

Definition

Events are independent if one doesn't affect the other:

P (A \cap B) = P (A) \times P (B)

To test independence: Check if

P (A \cap B) = P (A) \times P (B)

Equal → independent

Not equal → not independent

Important

Independent ≠ Mutually Exclusive!

Mutually exclusive:

P (A \cap B) = 0

(can't both happen)

Independent:

P (A \cap B) = P (A) \times P (B)

(don't affect each other)

If P(A) > 0 and P(B) > 0, mutually exclusive events CANNOT be independent!

Worked Example04

P(A) = 0.6, P(B) = 0.5, P(A ∪ B) = 0.8. Are A and B independent?

First find P(A ∩ B) using the addition rule, then test for independence:

1

P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.6 + 0.5 − 0.8 = 0.3

2

Test: P(A) × P(B) = 0.6 × 0.5 = 0.3

3

P(A ∩ B) = 0.3 = P(A) × P(B) ✓

Answer: Yes, A and B are independent.

Three-Set Venn Diagrams

Three sets create 8 regions:

3 "only one set" regions

3 "exactly two sets" regions

1 "all three" region

1 "none" region

Strategy

Fill from centre outwards:

1. Centre (A∩B∩C) first

2. Two-set overlaps next

3. Single-set regions last

4. Outside (neither) to finish

Three-Set Regions Labelled

8 regions:

a = A ∩ B' ∩ C' (only A)

b = A ∩ B ∩ C' (A and B, not C)

c = A' ∩ B ∩ C' (only B)

d = A ∩ B' ∩ C (A and C, not B)

e = A ∩ B ∩ C (all three)

f = A' ∩ B ∩ C (B and C, not A)

g = A' ∩ B' ∩ C (only C)

h = (A ∪ B ∪ C)' (none)

Remember

All 8 regions sum to 1 (or total frequency). a + b + c + d + e + f + g + h = 1

Worked Example05

100 people surveyed: 50 like Football, 35 like Tennis, 30 like Cricket. 15 like F&T, 10 like F&C, 12 like T&C, 5 like all three. How many like none?

Fill from centre outwards:

1

Centre: 5 (all three)

2

F∩T only: 15−5=10, F∩C only: 10−5=5, T∩C only: 12−5=7

3

F only: 50−10−5−5=30, T only: 35−10−5−7=13, C only: 30−5−5−7=13

4

Total inside: 30+10+13+5+5+7+13=83

Answer: None = 100 − 83 = 17

Worked Example06

P(A) = 0.5, P(B) = 0.6, P(A ∩ B) = 0.2. Find P(A' ∩ B).

P(A' ∩ B) means "B but not A" — the B-only region:

1

P(B) = P(A ∩ B) + P(A' ∩ B)

2

P(A' ∩ B) = P(B) − P(A ∩ B)

3

P(A' ∩ B) = 0.6 − 0.2 = 0.4

Answer: P(A' ∩ B) = 0.4

Worked Example07

A and B are independent with P(A) = 0.3 and P(B) = 0.4. Find P(A ∪ B).

Use independence to find intersection, then addition rule:

1

Independent: P(A ∩ B) = P(A) × P(B) = 0.3 × 0.4 = 0.12

2

Addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

3

P(A ∪ B) = 0.3 + 0.4 − 0.12 = 0.58

Answer: P(A ∪ B) = 0.58

Set Algebra (No Diagrams)

For harder questions, use algebraic manipulation instead of Venn diagrams:

De Morgan's Laws

(A \cup B)^{'} = A^{'} \cap B^{'}

(A \cap B)^{'} = A^{'} \cup B^{'}

Key Result

P (A^{'} \cap B^{'}) = 1 - P (A \cup B)

"Neither A nor B" = complement of "A or B"

Strategy

Break complex expressions into known quantities using:

Complements: P(A') = 1 − P(A)

Partition: P(B) = P(A∩B) + P(A'∩B)

Addition rule variations

Worked Example08

P(A) = 0.4, P(B) = 0.5, P(A ∪ B) = 0.7. Find P(A' ∩ B').

Use De Morgan's Law — no Venn diagram needed:

1

Recognise: A' ∩ B' means "neither A nor B"

2

By De Morgan: A' ∩ B' = (A ∪ B)'

3

Therefore: P(A' ∩ B') = P((A ∪ B)') = 1 − P(A ∪ B)

4

P(A' ∩ B') = 1 − 0.7 = 0.3

Answer: P(A' ∩ B') = 0.3

Worked Example09

P(A) = 0.6, P(B) = 0.5, P(A ∩ B) = 0.2. Find P(A ∩ B') + P(A' ∩ B).

This is "exactly one of A or B" — use partition:

1

P(A ∩ B') = P(A) − P(A ∩ B) = 0.6 − 0.2 = 0.4

2

P(A' ∩ B) = P(B) − P(A ∩ B) = 0.5 − 0.2 = 0.3

3

P(A ∩ B') + P(A' ∩ B) = 0.4 + 0.3 = 0.7

Alternative: P(A ∪ B) − P(A ∩ B) = (0.6 + 0.5 − 0.2) − 0.2 = 0.7

Answer: P(exactly one) = 0.7

10

Worked Example

S and T are events with P(S) = 0.3, P(T) = 0.4, P(S but not T) = 0.18.

(a) Show that S and T are independent. (b) Find P(S ∩ T) and P(neither S nor T).

Key insight: P(S but not T) = P(S ∩ T') = P(S) − P(S ∩ T)

1

Find P(S ∩ T): 0.18 = 0.3 − P(S ∩ T) → P(S ∩ T) = 0.12

2

Test independence: P(S) × P(T) = 0.3 × 0.4 = 0.12 ✓

3

P(S ∩ T) = 0.12 = P(S) × P(T), so S and T are independent

4

P(S ∪ T) = 0.3 + 0.4 − 0.12 = 0.58

5

P(neither) = 1 − P(S ∪ T) = 1 − 0.58 = 0.42

Practice

Question 1: P(A) = 0.55, P(B) = 0.45, P(A ∩ B) = 0.15. Find P(A ∪ B) and determine whether A and B are independent.

Part 1: P(A ∪ B) = 0.55 + 0.45 − 0.15 = 0.85

Part 2: Test P(A) × P(B) = 0.55 × 0.45 = 0.2475

P(A ∩ B) = 0.15 ≠ 0.2475

Answer: P(A ∪ B) = 0.85. A and B are NOT independent.

Practice

Question 2: A and B are independent. P(A) = 0.7, P(A ∩ B) = 0.28. Find P(B).

For independent events: P(A ∩ B) = P(A) × P(B)

0.28 = 0.7 × P(B)

P(B) = 0.28 ÷ 0.7 = 0.4

Practice

Question 3: 80 students: 45 study Maths, 38 Physics, 20 Chemistry. 20 study M&P, 12 study M&C, 10 study P&C, 6 study all three. Find P(exactly one subject).

Fill from centre:

Centre: 6 | M∩P only: 14 | M∩C only: 6 | P∩C only: 4

M only: 45−14−6−6 = 19 | P only: 38−14−6−4 = 14 | C only: 20−6−6−4 = 4

Exactly one: 19 + 14 + 4 = 37

Answer: P(exactly one) = 37/80

Summary

Key Formulas

FormulaWhen to Use

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

Always valid for "A or B"

P (A \cup B) = P (A) + P (B)

Mutually exclusive only

P (A \cap B) = P (A) \times P (B)

Independent events / testing

P (A^{'}) = 1 - P (A)

Complement

Exam Skills

Three-set problems: Fill from centre outwards

Test independence: Check if P(A∩B) = P(A) × P(B)

Mutually exclusive: P(A∩B) = 0 (no overlap)

All regions sum to 1 (or total frequency)