Question

A and B throw two dice each . Find the probability that B gets a higher sum than 9.

Answer 1

Here,$B$ will get sum $10,11,12$

No.of favourable cases are $(4,6),(5,5),(6,4),(5,6),(6,5)$

and $(6,6)=6$ cases $=6$ cases

Total no. of cases $=36$

The required probability =$\frac{6}{36}=\frac{1}{6}$