Question

$A$ and $B$ throw 2 dice. If $A$ throws 9, then $B$'s chance of throwing a higher number is $\frac{1}{6}$.

If this is true enter 1, else enter 0.

Answer 1

$n=$total no.of ways$=$ $6\times 6=36.$
The numbers higher than 9 are 10,11,12 in the case of two dice,
$\therefore$ $m=$ favourable no.of ways$=3+2+1=6$ ways
Hence $p=\frac{m}{n}=\frac{6}{36}=\frac{1}{6}.$