Question

A bag contains $5$ white, and $3$ black balls and four balls are sucessively drawn out and not replaced. The probability that they are alternately of different colours, is

• A. $\frac{1}{196}$
• B. $\frac{2}{7}$
• C. $\frac{1}{7}$
• D. $\frac{13}{56}$

Answer 1

The correct option is C

$\frac{1}{7}$

Explanation for correct option:

Calculate the probability :

There are two cases :

Case $1.$ : When first ball is white, second is black, third is white and fourth is black. (WBWB)

Case $2.$ : When first ball is black, second is white, third is black and fourth is white. (BWBW)

And bag contains $5$ white, and $3$ black balls

In case $\left(1\right)$ the probability of getting alternate colours (without replacement ) $=\frac{5}{8}\times \frac{3}{7}\times \frac{4}{6}\times \frac{2}{5}=\frac{1}{14}$

In case $\left(2\right)$ the probability of getting alternate colours (without replacement ) $=\frac{3}{8}\times \frac{5}{7}\times \frac{2}{6}\times \frac{4}{5}=\frac{1}{14}$

Therefore, the required probability $=\frac{1}{14}+\frac{1}{14}=\frac{2}{14}=\frac{1}{7}$

Hence the correct option is (C).