Explanation:

Here n(T)=6×6=36

Let A be the event "getting sum more than 8" I.e. sum of pair of dice =9,10,11,12

I.e A={ (3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6) }

Thus n(A)=10

Therefore required probability=n(A)/n(T)

 =10/36

 =5/18