conditional probability

Question

Let A and B be events. Prove that 𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵).

Answer 1

AB, AB', and A'B are mutually exclusive.

A = A(B∪B') = AB U AB' P(A) = P(AB) + P(AB'), as they are mutually exclusive

B = B(A∪A') = AB ∪ A'B P(B) = P(AB) + P(A'B), as they are mutually exclusive

Then, A ∪ B = AB U AB' ∪ AB ∪ A'B = AB U AB' ∪ A'B

As these 3 sets are mutually exclusive,

P(A ∪ B) = P(AB) + P(AB') + P(A'B)

P(A) + P(B) - P(AB) = P(AB) + P(AB') + P(AB) + P(A'B) - P(AB) = P(AB) + P(AB') + P(A'B)

Thus, P(A ∪ B) = P(A) + P(B) - P(AB)

Answer 2

𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)

is the formula for non-mutual exclusive sets or events A and B. If A and B are mutual exclusive,

then 𝑃(𝐴 ∩ 𝐵)=0

Proof:

Let say:

𝐴 ∪ 𝐵 = 𝐴∪ (𝐴' ∩ 𝐵) =>1st eq.

𝑃 (𝐴 ∪ 𝐵) = 𝑃 (𝐴) + 𝑃 (𝐴' ∩ 𝐵) =>2nd eq.

𝑃 (𝐵) = 𝑃 (𝐴 ∩ 𝐵) + 𝑃 (𝐴' ∩ 𝐵) =>3rd eq.

Subtract both sides by 𝑃 (𝐴 ∩ 𝐵) from the 3rd eq.

𝑃 (𝐵) - 𝑃 (𝐴 ∩ 𝐵) = 𝑃 (𝐴' ∩ 𝐵)

Substitute 𝑃 (𝐴' ∩ 𝐵) from the 2nd eq.

𝑃 (𝐴 ∪ 𝐵) = 𝑃 (𝐴) + 𝑃 (𝐵) - 𝑃 (𝐴 ∩ 𝐵).