Given that: A pair of dice is thrown

To find: Probability that the total of numbers on the dice is greater than $10$

Let’s write the all possible events that can occur

$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)$,

$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)$,

$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$,

$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$,

$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$,

$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$,

It’s clear that the total number of events is ${{6}^{^{2}}}=36$

Favorable events which means getting the total of numbers on the dice greater than $10$ are $(5,6)$,$(6,5)$ and $(6,6)$.

Thus, the total number of favorable events

i.e. getting the total of numbers on the dice greater than $10$ is $3$.

As We know that,

Probability = Number of favorable outcomes/ Total number of outcomes

Therefore, the probability of getting the total of numbers on the dice greater than $10=3/36=1/12$