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 : 7 64 (2) 3 16 (3) 7 32 (4) 7 16 A. 3p+2p2 2 B. p+3p2 2 C. 3p+p2 2 D. 3p+2p2 4 Select the correct answer from above options

Answer 1

Correct Answer - A We have, P(A)+P(B)-2P(A∩B)=P P(B)+P(C)-2P(B∩C)=p P(C)+P(A)-2P(C∩A)=p and, P(A∩B∩C)=p2 Adding (i),(ii) and (iii), we get 2[P(A)+P(B)+P(C)+P(A∩B)-P(B∩C)-P(A∩C)]=3p ⇒P(A)+P(B)+P(C)-P(A∩B)-P(B∩C) P(A∩C)=3p/2 ∴ Required probability =P(A∪B∪C) =P(A)+P(B)+P(C)-P(A∩B)-P(B∩C)-P(A∩C)+P(A∩B∩C) = P ( A ) + P ( B ) + P ( C ) − P ( A ∩ B ) − P ( B ∩ C ) − P ( A ∩ C ) + P ( A ∩ B ∩ C ) = 3p 2 +p2= 3p+2p2 2