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:

$\frac{2}{5}$

The correct answer is Option (2) → $\frac{2}{5}$

case 1 when first ball is white

Remaining → 2 white, 2 Red ⇒ P(red) = $\frac{2}{4}$

case 2 when first ball is red

Remaining → 3 white, 1 Red ⇒ P(red) = $\frac{1}{4}$

probability to go for case 1 is $\frac{2}{5}$

for case 2 is $\frac{3}{5}$

total probability = $\frac{2}{5}×\frac{2}{4}+\frac{3}{5}×\frac{1}{4}=\frac{2}{5}$