Q. A bag contains 6 red and 3 white balls. Four balls are drawn one by one and not replaced. The probability that they are alternatively of different colours, is
A) $\frac{4}{42}$
B) $\frac{5}{42}$
C) $\frac{7}{42}$
D) $\frac{8}{42}$
Explanation
(b)
Four balls can be drawn alternatively in the following two ways $RWRW\ or\ WRWR$
If red ball is drawn first, then probability of drawing the balls alternatively
$= \frac{6}{9} \times \frac{3}{8} \times \frac{5}{7} \times \frac{2}{6} = \frac{5}{84}$
If white ball is drawn first, then probability of drawing the balls alternatively
$= \frac{3}{9} \times \frac{6}{8} \times \frac{2}{7} \times \frac{5}{6} = \frac{5}{84}$
Since, these two are mutually exclusive ways.
$\therefore Required\ probability = \frac{5}{84} + \frac{5}{84} = \frac{10}{84} = \frac{5}{42}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.