Question

$A$ and $B$ each throw two dice$;$ if the sum of numbers on dice thrown by $A$ is $9,$ find $B$'s chance of throwing a higher number than $A$.

Answer 1

B needs to throw $10,11$ or $12$

Now, the number of cases for throwing $10=(5,5);(6,4);(4,6)=3$

Now, the number of cases for throwing $11=(5,6);(6,5)=2$

Now, the number of cases for throwing $12=(6,6)=1$

P(B throwing more than A)=$\frac{6}{36}=\frac{1}{6}$