Given an urn contains 25 balls of which 10 balls bear a mark 'X' and the remaining 15 bear a mark 'Y'.

P (drawing an X) $ = p = \large\frac{10}{25} = \frac{2}{5} \rightarrow$$ q = 1- p = \large\frac{3}{5}$

Given that 6 balls are to be drawn at random, it is a case of Bernoulli trials $(n=6)$, with $p(X=r)=C^{6}_{r} \large(\frac{2}{5})^{r} (\frac{3}{5})^{6-r}$

(i) all will bear the X mark

$P (X = 6) = C^{6}_{6} \large(\frac{2}{5})^{6} (\frac{3}{5})^{6-6} = (\frac{2}{5})^6$

(ii) not more than 2 balls will bear the Y mark.

P (not more than 2 balls with Y mark = not less than 4 bear X mark) $ = P (X \geq 4) = P (X=4) + P (X=5) + P(X=6)$

=\(\large\;c^6_4\;(\frac{2}{5})^4\;(\frac{3}{5})^2+c^6_4\;(\frac{2}{5})^5\;(\frac{3}{5})^1\;+\;c^6_6\;(\frac{2}{5})^6\;(\frac{3}{5})^0\)

=\(\;15\;\large \frac{2^4x3^2}{5^6}\)\(+\;6\;\large\frac{2^5 \times 3}{5^6}+\frac{2^6}{5^6}\)

=\(\;7\large (\frac{2}{5})^4\)

(iii) at least 1 ball will bear Y mark.

P (at least 1 Y mark) = P (at most 5 X marks) $= 1 - P (X = 6)$

=\(\;1-\large c^6_6\;(\frac{2}{5})^6\;(\frac{3}{5})^0\)

=\(\;1-\large( \frac{2}{5})^6\)

(iv) number of balls w/ X mark and Y mark will be equal.

P ( same number of X and Y out of 6) = P (exactly 3 successes and 3 failures) $ = P (3)$

$P (X = 3) = C^{6}_{3} \large(\frac{2}{5})^{3} (\frac{3}{5})^{6-3} $$= 20 \large (\frac{2}{5})^3 (\frac{3}{5})^3 = \frac{864}{3125}$