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

there are two cases arise.
Case I: If 1st ball is white, then probability $=\frac{{ }^3{\mathrm{C}}_1}{{ }^5{\mathrm{C}}_1}\times \frac{{ }^2{\mathrm{C}}_1}{{ }^4{\mathrm{C}}_1}=\frac{6}{20}=\frac{3}{10}$
Case II: if 1st ball is red, then probability  $=\frac{{ }^2{\mathrm{C}}_1}{{ }^5{\mathrm{C}}_1}\times \frac{{ }^1{\mathrm{C}}_1}{{ }^4{\mathrm{C}}_1}=\frac{2}{20}=\frac{1}{10}$

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