discrete math ee

Question

. Events A, B, and C in a sample space have P(A)=0.2, P(B)=0.4, P(C)=0.5, P(~B ∪ ~C)=0.9, and P(A ∪ C)=0.6. Find P(A ∪ B ∪ C) if A and B are mutually exclusive.

Answer 1

Using De Morgan's Laws:

~B ∪ ~C = ~(B ∩ C), then P(B ∩ C) = 1 - P(~(B ∩ C) = 1 - 0.9 = 0.1

P(A ∪ C) = P(A) + P(C) - P(A ∩ C), then P( A ∩ C) = P(A) +P(C) - P( A ∪ C) = 0.2 + 0.5 - 0.6 = 0.1

Since A and B are mutually exclusive, P(A ∩ B) = 0, then P(A ∪ B) = P(A) + P(B) = 0.2 + 0.4 = 0.6

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

Using De Morgan's Laws

(A∪B)∩C = (A∩C) ∪ (B∩C) ,

A ∩ C and B ∩ C are mutually exclusive, since A ∩ C ⊂A and B ∩ C ⊂ B,

Therefore, P( (A∪B)∩C ) = P(A∩C) + P(B∩C) = 0.1 + 0.1 = 0.2

Then,

P( A ∪ B ∪ C) = P( (A ∪ B) ) + P(C) - P( (A∪B)∩C ) = 0.6 + 0.5 - 0.2 = 0.9