Question

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. Then the probability that the ball drawn now is white, is:

• A. $\frac{n}{m+n}$
• B. $\frac{m}{m+n}$
• C. $\frac{2n}{m+n}$
  
• D. $\frac{m-n}{m+n}$

Answer 1

The correct option is B $\frac{m}{m+n}$

Let $E_1$ be the event first ball drawn is white colour and $E_2$ be the event first ball drawn is black colour.
And A be the event second ball drawn is white
$P\left(E_1\right)=\frac{m}{m+n},P\left(E_2\right)=\frac{n}{m+n}$
$P\left(\frac{A}{E_1}\right)=\frac{m+k}{m+n+k}\text{and}P\left(\frac{A}{E_2}\right)=\frac{m}{m+n+k}$
Using total probability theorem,
$P\left(A\right)=P\left(E_1\right)\cdot P\left(A/E_1\right)+P\left(E_2\right)\cdot P\left(A/E_2\right)$
$\Rightarrow P\left(A\right)=\frac{m}{m+n}.\frac{m+k}{m+n+k}+\frac{n}{m+n}.\frac{m}{m+n+k}\Rightarrow P\left(A\right)=\frac{m^2+mk+nm}{(m+n)(m+n+k)}=\frac{m}{m+n}$