If A, B, C are any three events, then prove that
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(B ∩ C) - P(C ∩ A) + P(A ∩ B ∩ C)
Let (B ∪ C) = D
∴  P(A ∪ B ∪ C) = P(A ∪ D) = P(A) + P(D) - P(A ∩ D)
             = P(A) + P(B ∪ C) - P[A ∩ (B ∪ C)]
             = P(A) + P(B) + P(C) - P(B ∩ C) - P[A ∩ (B ∪ C)]
                                        [∵P(A ∪ B) = P(A) + P(B) - P(A ∩ B)]
             =  P(A) + P(B) + P(C) - P(B ∩ C) - P[(A ∩ B) ∪ (A ∩ C)]
                                [By using A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)]
             = P(A) + P(B) + P(C) - P(B ∩ C) - {P(A ∩ B)
                                  + P(A ∩ C) - P[(A ∩ B) ∩ (A ∩ C]}
             =  P(A) + P(B) + P(C) - P(B ∩ C) - P(A ∩ B)
                                         - P(C ∩ A) + P(A ∩ B ∩ C)
Hence, P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(B ∩ C) - P(C ∩ A) + P(A ∩ B ∩ C)