Let S be the sample space associated with the random experiment of throwing a die twice.
Then, n(S) = 36
∴ A B= first and second throw show 5, i.e., getting 5 in each throw.
We have, A = (5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
And B = (1,5)(2,5)(3,5)(4,5)(5,5)(6,5)
∴ P(A) = P(B) = 636 and P(A ∩ B) = 136
∴ Required probability= Probability that at least one of the two throws shows 5.
= P(A∪B) = P(A) + P(B) - P(A∩B)
= 636+636−136=1136