Question

An urn contains m white and n black balls. A ball is drawn at randow 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 randow. Show that the probability of drawing a white ball now does not depend on k.

Answer 1

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}

$\therefore P\left(E_1\right)=\frac{m}{m+n}andP\left(E_2\right)=\frac{n}{m+n}$
$Also,P\left(E_3/E_1\right)=\frac{M+k}{m+n+k}andP\left(E_3/E_2\right)=\frac{M+k}{m+n+k}$
=$\therefore P\left(E_3\right)=P\left(E_1\right).P\left(E_3/E_1\right)+P\left(E_2\right).P\left(E_3/E_2\right)$
$=\frac{m}{m+n}.\frac{M+K}{M+n+k}+\frac{n}{M+n}.\frac{M}{M+n+k}$
$=\frac{m(m+k)+nm}{(m+n+k)(m+n)}=\frac{m^2+mk+mn}{(m+n+k)(m+n)}$
$=\frac{m(m+k+n)}{(m+n+k)(m+n)}=\frac{m}{m+n}$

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