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) = 1/16. Then the probability that at least one of the events occurs, is

- 7/32
- 7/16
- 7/64
- 3/16

**Solution**

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) - ∑ P (A ∩ B) = 3/8

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

= 3/8 + 1/16

= 7/16

**The correct option is B.**