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

A
8/25
B
2/5
C
3/5
D
21/25

Solution:

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

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

Solution:

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