A die is thrown again and again
Probability of getting a 'six' in a throw $=\frac{1}{6}$
Probability of getting not a 'six' in a throw
$$
=1-\frac{1}{6}=\frac{5}{6}
$$
There are two 'sixes' in first five throws and one 'six' at the sixth throw.
Probability of getting two sixes in 5 throws
$$
={ }^5 \mathrm{C}_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^3
$$
Probability of getting a 'six' in sixth throw $=\frac{1}{6}$
$\therefore$ Probability of getting a third 'six' in the sixth throw $=$
${ }^5 C_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^3 \times \frac{1}{6}$
$$
=10 \times \frac{125}{6^6}=\frac{1250}{46656}=\frac{625}{23338}
$$