Given A pair of dice is thrown
To find: Probability that the total of numbers on the dice is greater than $9$
Le t’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 $9$ are
$(5,5)$, $(5,6)$, $(6,4)$, $(4,6)$, $(6,5)$ and $(6,6)$.
Thus, the total number of favorable events i.e. getting the total of numbers on the dice greater than $9$ is $6$
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 $9=6/36=1/6$