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 = $\,{}^{5}C_{2} \times \frac{1}{6}\times \frac{1}{6}\times \frac{125}{216}$ 6th throw definitely gives six with probability = $\frac{1}{6}$ Required Probability = $\frac{125}{216 \times 36}$ × 10 × $\frac{1}{6}$ = $\frac{625}{23328}$