A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is  

A pair of fair dice is thrown, the sample space _{$S=\left(1,1\right),\left(1,2\right)\left(1,3\right)\dots \dots .=36$}

_$\therefore$ Possibility of getting score 9 in a single throw = _{$\frac{4}{36}=\frac{1}{9}$}

_$\therefore$ probability of getting score 9 exactly twice _{$=3_{C_2}\times {\left(\frac{1}{9}\right)}^2.\left(1-\frac{1}{9}\right)=\frac{3!}{2!}\times \frac{1}{9}\times \frac{1}{9}\times \frac{8}{9}$}  

_{$=\frac{3.2!}{2!}\times \frac{1}{9}\times \frac{1}{9}\times \frac{8}{9}=\frac{8}{243}$}