Question

A bag contains $3$ black, $3$ white and $2$ red balls. One by one, $3$ balls are drawn without replacement. The probability that the third ball is red, is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $\frac{1}{4}$

Answer 1

The correct option is D

$\frac{1}{4}$

Explanation for the correct option:

Finding the probability that the third ball is red:

Given, A bag contains $3$ black, $3$ white and $2$ red balls

Therefore, total number of ball $8$

Let ‘W’ be the white ball, ‘R’ be the red ball & ‘B’ be the black ball

Three balls are drawn without replacement, & the third ball should be red.

Therefore, the required probability

$$\begin{gathered} =P\left(WWR\right)+P\left(BBR\right)+P\left(WBR\right)+P\left(BWR\right)+P\left(WRR\right)+P\left(BRR\right)+P\left(RWR\right)+P\left(RBR\right) \\ =\left[\frac{3\times 2\times 2}{8\times 7\times 6}\right]+\left[\frac{3\times 2\times 2}{8\times 7\times 6}\right]+\left[\frac{3\times 3\times 2}{8\times 7\times 6}\right]+\left[\frac{3\times 3\times 2}{8\times 7\times 6}\right]+\left[\frac{3\times 2\times 1}{8\times 7\times 6}\right]+\left[\frac{3\times 2\times 1}{8\times 7\times 6}\right]+\left[\frac{2\times 3\times 1}{8\times 7\times 6}\right]+\left[\frac{2\times 3\times 1}{8\times 7\times 6}\right] \\ =\frac{2}{56}+\frac{2}{56}+\frac{3}{56}+\frac{3}{56}+\frac{1}{56}+\frac{1}{56}+\frac{1}{56}+\frac{1}{56} \\ =\frac{14}{56} \\ =\frac{1}{4} \end{gathered}$$

Hence, correct option is $\mathbf{\left(}\mathit{D}\mathbf{\right)}$.