According to Addition Theorem on Probability, for any two elements A, B
P(A∪B) = P(A) + P(B) - P(A∩B)
(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)
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)
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