Let A,B,C be three events. If the probability of occurring exactly one event out of A and B is 1−a, out of B and C is 1−2a, out of C and A is 1−a and that of occurring three events simultaneously is a2, then the probability  that at least one out of A,B,C will occur is

P( exactly one event out of A and B occurs )=PA∩B′∪A′∩B=P(A∪B)−P(A∩B)=P(A)+P(B)−2P(A∩B)∴ P(A)+P(B)−2P(A∩B)=1−a----1 Similarly, P(B)+P(C)−2P(B∩C)=1−2a----2P(C)+P(A)−2P(C∩A)=1−a----3P(A∩B∩C)=a2---4now ,P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(C∩A)+P(A∩B∩C)=12[P(A)+P(B)−2P(B∩C)+P(B)+P(C)−2P(B∩C)+P(C)+P(A)−2P(C∩A)]+P(A∩B∩C)=12[1−a+1−2a+1−a]+a2 [ using (1),(2),(3) and (4)] =32−2a+a2=12+(a−1)2>12