# A box contains 3 white and 2 red balls. If we draw one ball and without replacing the first ball, the probability of drawing red ball in the second draw is

$\begin {array} {1 1} (A)\;\large\frac{8}{25} & \quad (B)\;\large\frac{2}{5} \\ (C)\;\large\frac{3}{5} & \quad (D)\;\large\frac{21}{25} \end {array}$

Case (i) The first ball is white and second is red.
Its probability = $\large\frac{3}{5} \times \large\frac{2}{4}$
$= \large\frac{3}{10}$
Case (ii) The first ball is red and second is red.
Its probability = $\large\frac{ \not 2}{5} \times \large\frac{1}{\not 4\: 2}$
$= \large\frac{1}{10}$
$\therefore$ Required probability = $\large\frac{3}{10} +\large\frac{1}{10}$
$= \large\frac{4}{10} = \large\frac{2}{5}$
Ans : (B)