According to the question, two unbiased dice are thrown.
By using the formula of probability, we get,
P (E) = favourable outcomes / total possible outcomes
So, we have to find the probability of getting the sum of digits on dice greater than 10
Total number of possible outcomes is $6^2=36$
n (S) = 36
Let E be the event of getting same number on all the three dice
E = {(5,6) (6,5) (6,6)}
n (E) = 3
P (E) = n (E) / n (S)
= 3 / 36
= 1/12