Question

For three events A, B and C, P(Exactly one of A or B occurs) = P(Exactly one of B or C occurs) = P(Exactly one of C or A occurs) = 1/4 and P (All the three events occur simultaneously) = 1/6.
Then the probability that at least one of the events occurs, is :

(1) 7/32
(2) 7/16
(3) 7/64
(4) 3/16

Answer 1

P(A) + P(B) – 2P (A ∩ B) = 1/4
P(B) + P(C) – 2P(B ∩ C) = 1/4
P(A) + P(C) – 2P(A ∩ C) = 1/4
P (A ∩ B ∩ C) = 1/16
Therefore, P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C ) – P(A ∩ C) + P(A ∩ B ∩ C) = 3/8 + 1/16
= (6+1)/16
= 7/16