p = probability of success = $\frac{1}{6}$ , q = probability of failure = $\frac{5}{6}$
Third six comes at the 6th throw so the remaining two sixes can appear in any of the previous 5 throws.
Probability of obtaining 2 sixes in 5 throws
= $\stackrel{5}{}{\mathit{C}}_2\times \frac{1}{6}\times \frac{1}{6}\times \frac{125}{216}$
6th throw definitely gives six with probability = $\frac{1}{6}$
Required Probability
= $\frac{125}{216\times 36}$ × 10 × $\frac{1}{6}$
= $\frac{625}{23328}$