Total number of sample space, n(S)=6×6=36
Let E= Event of getting maximum sum of two dices is 5 ={(1,1),(1,2),(2,1),(1,3),(3,1),(2,2)(1,3),(3,1),(1,4),(4,1),(2,3),(3,2)} ∴n(E)=12 ∴ Probability (the maximum sum of two dices is 5 ) =n(S)n(E)=3612=35 ∴ Probability (the sum of two dices is more than 5 ) =1− Probability (the sum of two dices is 5 ) =1−31=32