On a throw of three dice, $[\mathrm{n}(\mathrm{S})]=6^3=216$
Let $\mathrm{E}_1$ is the event when the sum of numbers on the dice was six and $E_2$ is the event when three two's occurs.
$$
\begin{aligned}
&\Rightarrow \mathrm{E}_1=\{(1,1,4),(1,2,3,),(1,3,2),(1,4,1),(2,1,3), \\
&(2,2,2),(2,3,1),(3,1,2),(3,2,1),(4,1,1,)\} \\
&\Rightarrow \mathrm{n}\left(\mathrm{E}_1\right)=10 \text { and } \mathrm{E}_2=\{2,2,2\} \\
&\Rightarrow \mathrm{n}\left(\mathrm{E}_2\right)=1 \\
&\text { Also, }\left(\mathrm{E}_1 \cap \mathrm{E}_2\right)=1 \\
&\therefore \quad \mathrm{P}\left(\mathrm{E}_2 / \mathrm{E}_1\right)=\frac{\mathrm{P} \cdot\left(\mathrm{E}_1 \cap \mathrm{E}_2\right)}{\mathrm{P}\left(\mathrm{E}_1\right)}=\frac{1 / 216}{10 / 216}=\frac{1}{10}
\end{aligned}
$$