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 of drawing a white ball now, is

• A. $\frac{km}{n}$
• B. $\frac{(m-n)}{k(m+1)}$
• C. $\frac{km}{m+n}$
• D. independent of $k$

Answer 1

Answer: (D)

Let $A$ : the first ball drawn is white.
Let $B$ : the first ball drawn is black.
Let $E$ : the second ball drawn is white.
Number of white ball $=m$,
number of black balls $=n$
Total number of balls $=m+n$
$\therefore P(A)=\frac{m}{m+n}$
$P(B)=\frac{n}{m+n}$
$P(E/A)=\frac{m+k}{m+n+k}$
$P(E/B)=\frac{m}{m+n+k}$
Now, $P(E)=P(A)\times P(E/A)+P(B)\times P(E/B)$
$=\frac{m}{m+n}\times \frac{m+k}{m+n+k}+\frac{n}{m+n}\times \frac{m}{m+n+k}$
$=\frac{m(m+k)+mn}{(m+n)(m+n+k)}$
$=\frac{m(m+k+n)}{(m+n)(m+n+k)}=\frac{m}{m+n}$
which is independent of $k$.
$\therefore$ The probability of drawing a white ball now does not depend on $k$.