We need to find the P (obtaining the third six in 6th throw of the die) = P (2 sixes in first 5 throws) + P (1 six in sixth throw)

Since P (getting a six) = $\large\frac{1}{6} \rightarrow$$ P (\text{obtaining the third six in 6th throw of the die}) = P (\text{2 sixes in first 5 throws}) \times \large\frac{1}{6}$

P (getting 2 sixes in first 5 throws) $= C^{5}_{2} \large\frac{1}{6}^{2} \large\frac{5}{6}^{5-2} = \frac{5 \times 4}{1 \times 2} \frac{5^3}{6^5} = \frac{625}{3888}$

$P (\text{obtaining the third six in 6th throw of the die}) = \large \frac{625}{3888} \times \frac{1}{6} = \frac{625}{23328}$