A fair die is thrown six times. The probability of getting a factor of 6 exactly four times is:

This is a binomial distribution problem with $n=6$.Factors of 6 on a die are $\{1,2,3,6\}$.Probability of success ($p$) = $\frac{4}{6}=\frac{2}{3}$.Probability of failure ($q$) = $1-p=\frac{1}{3}$.We need to find $P(X=4)$:$P(X=4)=\left(\genfrac{}{}{0ex}{}{6}{4}\right)p^4{q}^{6-4}$$P(X=4)=15\times (\frac{2}{3}{)}^4\times (\frac{1}{3}{)}^2$$P(X=4)=15\times \frac{16}{81}\times \frac{1}{9}=\frac{240}{729}=\frac{80}{243}$.