In a single throw of two dice find the probability that neither a doublet nor a total of 8 will appe...........

Explanation:

n(S) = 36

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) }

n(A)=6, n(B)=5, n(A∩B)=1

Required probability = P(A∪B)

 = P(A)+P(B)−P(A∩B)

=  6/36+5/36−1/36 = 5/18

Hence, option 4 is the correct answer.