Question

Prove: P(A ∪ B ∪ C) = P(A) + P(Ac ∩ B) + P(Ac ∩ Bc ∩ C)

Answer 1

P(A U B U C) = P(A) + P(B) + P(C) - P(A B) - P(A C) - P(B C) + P(A B C)

P(A^c) = Everything other than A.

Therefore P(A^c B) = P(B) - P(A B) --------- (1)

P(A^c B^c) = P(A U B)^c, This is by De Morgans Law

Therefore P(A^c B^c C) = P(Only C) = P(C) - P(A C) - P(B C) + P(A B C) ------ (2)

We are adding back P(A B C), since P(B C) and P(A C) have P(A B C), which we are subtracting in each, therefore we need to add back once.

Therefore P(A U B U C) = P(A) + P(B) + P(C) - P(A B) - P(A C) - P(B C) + P(A B C) can be re written as P(A) + [P(B) - P(A B)] + [P(C) - P(A C) - P(B C) + P(A B C)]

Therefore P(A U B U C) = P(A) + 1 + 2 = P(A) + P(A^c B) + P(A^c B^c C)

The Diagrams below will be helpful