Total number of balls $=25$,
Number of balls marked ' $\mathrm{X}$ ' $=10$
Let $\mathrm{X}$ denote the event of getting a ball marked $\mathrm{X}, \mathrm{Y}=$ event of getting a ball $Y$
$\therefore \quad \mathrm{P}(\mathrm{X})=\frac{10}{25}=\frac{2}{5}=\mathrm{p}$
$\therefore \quad \mathrm{P}(\mathrm{Y})=1-\frac{2}{5}=\frac{3}{5}=\mathrm{q}$
Now 6 balls are drawn.
(i) $\mathrm{P}($ all the balls are marked $\mathrm{X})=\left(\frac{2}{5}\right)^6$
(ii) Event that not more than 2 will bear ' $\mathrm{Y}$ ' mark)
$\Rightarrow$ (6X, 0Y), $(5 \mathrm{X}, 1 \mathrm{Y}),(4 \mathrm{X}, 2 \mathrm{Y})$
$\therefore \mathrm{P}$ (not more than two will bear ' $\mathrm{Y}$ ' mark)
$=\mathrm{P}(6)+\mathrm{P}(5)+\mathrm{P}(4)$
$=\left(\frac{2}{5}\right)^6+{ }^6 \mathrm{C}_5\left(\frac{3}{5}\right)\left(\frac{2}{5}\right)^5+{ }^6 \mathrm{C}_4\left(\frac{3}{5}\right)^2\left(\frac{2}{5}\right)^4$
$=\left(\frac{2}{5}\right)^4\left[\frac{4}{25}+\frac{6 \times 3 \times 2}{25}+\frac{15 \times 9}{25}\right]$
$$
\begin{aligned}
&=\left(\frac{2}{5}\right)^4\left[\frac{4+36+135}{25}\right] \\
&=\left(\frac{2}{5}\right)^4\left[\frac{175}{25}\right]=7\left(\frac{2}{5}\right)^4
\end{aligned}
$$
(iii) Event that at last 1 ball will bear ' $\mathrm{Y}$ ' $=\{(5 \mathrm{X}, 1 \mathrm{Y}),(4 \mathrm{X}, 2 \mathrm{Y}),(3 \mathrm{X}, 3 \mathrm{Y})$, $(2 \mathrm{X}, 4 \mathrm{Y}),(\mathrm{X}, 5 \mathrm{Y}),(6 \mathrm{X}, 6 \mathrm{Y})\}$
$\mathrm{P}$ (at least 1 ball will bear mark $\mathrm{Y}$ $=1-\mathrm{P}$ (no ball will bear mark $\mathrm{Y})$ $=1-\mathrm{P}($ all balls bear mark $\mathrm{X})=1-\left(\frac{2}{5}\right)^6$
(iv) $\mathrm{P}(3$ balls each are marked $\mathrm{X}$ and $\mathrm{Y})$
$$
\begin{aligned}
P(3) &={ }^6 C_3\left(\frac{3}{5}\right)^3\left(\frac{2}{5}\right)^3 \\
&=\frac{6 \times 5 \times 4}{6} \times \frac{27}{125} \times \frac{8}{125}=\frac{864}{3125}
\end{aligned}
$$