Question

Find the chance of throwing $9$ at least in a single throw with two dice.

Answer 1

$9$ can be made up in $4$ ways, and thus the chance of throwing $9$ is $\frac{4}{36}$.
$10$ can be made up in $3$ ways, and thus the chance of throwing $10$ is $\frac{3}{36}$.
$11$ can be made up in $2$ ways, and thus the chance of throwing $11$ is $\frac{2}{36}$.
$12$ can be made up in $1$ way, and thus the chance of throwing $12$ is $\frac{1}{36}$.
Now the chance of throwing a number not less than $9$ is the sum of these separate chances$;$
$\therefore$ The required chance $=\frac{4+3+2+1}{36}=\frac{5}{18}$.