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, $P(E)=P(A) \times P(E / A)+P(B) \times P(E / B)$
$=\frac{m}{m+n} \times \frac{m+k}{m+n+k}+\frac{n}{m+n} \times \frac{m}{m+n+k}$
$=\frac{m(m+k)+m n}{(m+n)(m+n+k)}$
$=\frac{m(m+k+n)}{(m+n)(m+n+k)}=\frac{m}{m+n}$
which is independent of $k$.
$\therefore$ The probability of drawing a white ball now does not depend on $k$.