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

Possible outcomes for a score of $9$ with two dice: $(3,6),(4,5),(5,4),(6,3)$.Total outcomes = $36$.Probability of success (score of $9$) in one throw, $p=\frac{4}{36}=\frac{1}{9}$.Probability of failure, $q=1-p=\frac{8}{9}$.We use the Binomial Probability formula: $P(X=k)={}^n{C}_k{p}^k{q}^{n-k}$.Here $n=3$, $k=2$.$P(X=2)={}^3{C}_2{\left(\frac{1}{9}\right)}^2{\left(\frac{8}{9}\right)}^{3-2}$ $
$=3\times \frac{1}{81}\times \frac{8}{9}=\frac{24}{729}=\frac{8}{243}$ $