Q. 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) $\frac{8}{25}$
B) $\frac{2}{5}$
C) $\frac{3}{5}$
D) $\frac{21}{25}$
Explanation
(b)
There are two cases arise.
Case I If Ist ball is white, then
$P = \frac{{\ ^{3}C}_{1}}{{\ ^{5}C}_{1}} \times \frac{{\ ^{2}C}_{1}}{{\ ^{4}C}_{1}} = \frac{6}{20} = \frac{3}{10}$
Case II If Ist ball is red, then
$P = \frac{{\ ^{2}C}_{1}}{{\ ^{5}C}_{1}} \times \frac{{\ ^{1}C}_{1}}{{\ ^{4}C}_{1}} = \frac{2}{20} = \frac{1}{10}$
$\therefore$ Required probability $= \frac{3}{10} + \frac{1}{10} = \frac{2}{5}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.