According to Addition Theorem on Probability, for any two elements A, B

**P(A∪B) = P(A) + P(B) – P(A∩B)**

#### Addition Theorem on Probability Proof :-

Expressing A∪B as the union of two mutually exclusive events we get

(A∪B) = A ∪ (B-A)

P(A∪B) = P(A ∪ (B-A))

By applying axiom of union we get,

P(A∪B) = P(A) + P(B-A) since A, B-A are exclusive

= P(A) + P(B – (A∩B))

= P(A) + P(B) – P(A∩B) since P(B-A) = P(B) – P(A) when every A⊆B

**note:** If you find any errors then do correct in comment section below.

Therefore P(A∪B) = P(A) + P(B) – P(A∩B)