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 1st die = {(5, 5), (5, 6)}
Event of getting the sum is 10 or more with 5 on 1st die = n(E) = 2
Hence, the probability of getting a sum of 10 or more with 5 on 1st die = n(E)/ n(S) = 2/ 36 = 1/18