**Solution:**

In throwing a dice, total possible outcomes = {1, 2, 3, 4, 5, 6}

So, n(S) = 6

For two dice, n(S) = 6 x 6 = 36

Favorable cases where the sum is 10 or more with 5 on 1^{st} die = {(5, 5), (5, 6)}

Event of getting the sum is 10 or more with 5 on 1^{st} die = n(E) = 2

Hence, the probability of getting a sum of 10 or more with 5 on 1^{st} die = n(E)/ n(S) = 2/ 36 = 1/18