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