A box contains 3 white and 2 red balls. A balls is drawn and without replacing it, another ball is drawn. The probability that the second drawn ball is red, is

2/5

Let the first ball be white and second be red.

The probability of such a situation = $\frac{3}{5}.\frac{2}{4}=\frac{3}{10}$

Let the first ball be red and second also red.

The probability of the occurrence = $\frac{2}{5}.\frac{1}{4}=\frac{1}{10}$

The required probability = $\frac{3}{10}+\frac{1}{10}=\frac{2}{5}$

Hence (B) is the correct answer.