Hard probability proof problem

Question

If A, B, C are three events, show that P(A^c ∩ (B ∪ C)) = P(B) + P(C) − P(B ∩ C) − P(C ∩ A) − P(A ∩ B) + P(A ∩ B ∩ C).

(The complement of an event A is denoted as A^c)

Answer 1

This problem uses some preliminary facts about sets and probability. Let R, S, and T be sets.

Fact (1)

You can divide S into the disjoint union of the part of S that is in R, and the part of S that is not in R:

(R ∩ S) ∪ (R^c ∩ S) = S

Fact (2)

The distributive property for sets gives:

R ∩ (S ∪ T) = (R ∩ S) ∪ (R ∩ T)

Fact (3)

The additive property for the probability of a union is:

P(R ∪ S) = P(R) + P(S) - P(R ∩ S)

Note that when R and S are disjoint, this reduces to:

P(R ∪ S) = P(R) + P(S)

Proof

Using (1), we have (A ∩ (B ∪ C)) ∪ (A^c ∩ (B ∪ C)) = B ∪ C. This union is disjoint, so using (3) we have

P(A ∩ (B ∪ C)) + P(A^c ∩ (B ∪ C)) = P(B ∪ C).

Then

P(A^c ∩ (B ∪ C)) = P(B ∪ C) - P(A ∩ (B ∪ C))

= P(B ∪ C) - [P((A ∩ B) ∪ (A ∩ C))] using (2)

= P(B ∪ C) - [P(A ∩ B) + P(A ∩ C) - P((A ∩ B) ∩ (A ∩ C))] using (3)

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

= P(B) + P(C) - P(B ∩ C) - P(A ∩ B) - P(A ∩ C) + P(A ∩ B ∩ C) using (3)