An urn contains 25 balls of which 10 balls bear a mark $X$ and the remaining 15 bear a mark
$Y$ . A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that

(i) all will bear $X$ mark.

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

(iii) at least one ball will bear $Y$ mark.

(iv) the number of balls with $X$ and $Y$ mark will be equal.

.: Let $p$ be the probability of drawing a ball with mark $X = \cfrac{{10}}{{25}} = \cfrac{2}{5}$

Let $q$ be the probability of drawing a ball with mark $Y = \cfrac{{15}}{{25}} = \cfrac{3}{5}$

Let $X$ has a binomial distribution with $n = 6,p = \cfrac{2}{5},q = \cfrac{3}{5}$

$\therefore$ $P(X = r){ = ^n}{C_r}{(q)^{n - r}}{p^r}$

(i) $P$ ( all bear $X$ mark)
$= P(X = 6){ = ^6}{C_6}{q^0}{p^6} = (1)(1){\left( {\cfrac{2}{5}} \right)^6} = {\left( {\cfrac{2}{5}} \right)^6}$

(ii) $P$ (Not more than 2 bear $Y$ mark) $= P$ (means at least 4 bear $X$ mark)

$\therefore$ Required probability $= P(X = 4) + P(X = 5) + P(X = 6)$

${ = ^6}{C_4}{q^2}{p^4}{ + ^6}{C_5}q{p^5}{ + ^6}{C_6}{q^0}{p^6}$

$= 15{\left( {\cfrac{3}{5}} \right)^2}{\left( {\cfrac{2}{5}} \right)^4} + 6\left( {\cfrac{3}{5}} \right){\left( {\cfrac{2}{5}} \right)^5} + (1)(1){\left( {\cfrac{2}{5}} \right)^6}$

$= {\left( {\cfrac{2}{5}} \right)^4}\left[ {\cfrac{{135}}{{25}} + \cfrac{{36}}{{25}} + \cfrac{4}{{25}}} \right]$

$= \cfrac{{175}}{{25}}{\left( {\cfrac{2}{5}} \right)^4} = 7{\left( {\cfrac{2}{5}} \right)^4}$

(iii) $P$ (at least one will bear $Y$ mark) $= P$

(at most 5 will bear $X$ mark) Required probability $= P(X \le 5) = 1 - P(X = 6)$

$= 1{ - ^6}{C_6}{q^0}{p^6} = 1 - (1)(1){\left( {\cfrac{2}{5}} \right)^6} = 1 - {\left( {\cfrac{2}{5}} \right)^6}$

(iv) $P$ (No. of balls with ${\rm{X}}$ mark)

$= P$ (No. of balls with $Y$ mark)
$\Rightarrow P(X = 3) = P(Y = 3){ = ^6}{C_3}{q^3}{p^3}$

$= 20{\left( {\cfrac{3}{5}} \right)^3}{\left( {\cfrac{2}{5}} \right)^3} = 20{\left( {\cfrac{6}{{25}}} \right)^3} = \cfrac{{864}}{{3125}}$