Three dice are thrown at the same time.

Given that the dice is thrown three times

So, the sample space n(S) = 63 = 216

Let E1 be the event when the sum of number on the dice was 6 and E2 be the event when three 2’s occur.

E1 = {(1, 1, 4), (1, 2, 3), (1, 4, 1), (2, 1, 3), (2, 2, 2), (2, 3, 1), (3, 1, 2), (3, 2, 1), (4, 1, 1)}

n(E1) =10 and n(E2) = 1 [Since, E2 = (2, 2, 2)]

Thus, P(E2/E1) = P(E1 ⋂ E2)/P(E1) = (1/216)/(10/216) = 1/10.