An urn contains $m$ white and $n$ black balls. A ball is drawn at random and is put back into the urn along with $k$ additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Show that the probability of drawing a white ball now does not depend on $k$.

Let $U = \{ m$ white, $n$ black balls$\}$

${E_1} = \{$First ball drawn of white colour$\}$

${E_2} = \{$First ball drawn of black colour$\}$

and ${E_3} = \{$Second ball drawn of white colour$\}$

$\therefore$ $P\left( {{E_1}} \right) = \frac{m}{{m + n}}$

and $P\left( {{E_2}} \right) = \frac{n}{{m + n}}$

Also, $P\left( {{E_3}/{E_1}} \right) = \frac{{m + k}}{{m + n + k}}$

and $P\left( {{E_3}/{E_2}} \right) = \frac{m}{{m + n + k}}$

$\therefore$ $P\left( {{E_3}} \right) = P\left( {{E_1}} \right) \cdot P\left( {{E_3}/{E_1}} \right) + P\left( {{E_2}} \right) \cdot P\left( {{E_3}/{E_2}} \right)$

$= \frac{m}{{m + n}} \cdot \frac{{m + k}}{{m + n + k}} + \frac{n}{{m + n}} \cdot \frac{m}{{m + n + k}}$

$= \frac{{m(m + k) + nm}}{{(m + n + k)(m + n)}} = \frac{{{m^2} + mk + nm}}{{(m + n + k)(m + n)}}$

$= \frac{{m(m + k + n)}}{{(m + n + k)(m + n)}} = \frac{m}{{m + n}}$

Hence, the probability of drawing a white ball does not depend on $k$.

LONG ANWSER