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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 :

(1) 7/32

(2) 7/16

(3) 7/64

(4) 3/16
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The correct answer is option B)7/16 

\(P(A) + P(B) – 2P (A ∩ B) \)

\( = 1/4 P(B) + P(C) – 2P(B ∩C) \)

\( = 1/4 P(A) + P(C) – 2P(A ∩ C) \)

\( = 1/4 P (A ∩ B ∩ C) \)

\( = 1/16 ∩ P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩C ) – P(A ∩C) + P(A ∩ B ∩ C) \)

\( = {3\over8}+{1\over16}={6\over16}+{1\over16}={7\over16} \)

 

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