Two dice are thrown simultaneously. What is the probability of obtaining sum of the numbers less than 11?

Favourable cases to get the sum not less than 11 are {(5, 6),(6, 6),(6, 5)} = 3

Hence favourable cases to get the sum less than 11 are (36- 3) = 33. So required probability $=\frac{33}{36}=\frac{11}{12}$.