Question

A box contains $2$ black, $4$ white and $3$ red balls. One ball is drawn at random from the box and kept aside. From the remaining balls in the box, another ball is drawn at random and kept beside the first. The process is repeated, find the probability that the balls drawn are in the sequence of $2$ black, $4$ white and $3$ red.

• A. $\frac{1}{1260}$
• B. $\frac{8}{1260}$
• C. $\frac{1}{1360}$
• D. $\frac{3}{1360}$

Answer 1

Answer: (A)

$\frac{1}{1260}$

Answer 2

In total, the number of balls $= 2 + 4 + 3 = 9$ and the balls are being drawn one by one without replacement. Now $P$ (first drawing gives a black ball) $= \frac{2}{9}$ $P$ (second drawing gives a black ball) $= \frac{1}{8}$ $P$ (third drawing gives a white ball) $=\frac{4}{7}$, ... and so on. Hence the required probability $= P$ (the balls drawn are in of $2$ black, $4$ white and $3$ red) $= P (BBWWWWRRR)$ $= \frac{2}{9}\times\frac{1}{8}\times \frac{4}{7}\times \frac{3}{6}\times \frac{2}{5}\times \frac{1}{4}\times \frac{3}{3}\times \frac{2}{2}\times \frac{1}{1}$ $= \frac{1}{1260}$