Question

For$k=1,2,3$ the box,$Bk$ contains $k$ red balls and $(k+1)$white balls. Let $P\left(B_1\right)=\frac{1}{2},P\left(B_2\right)=\frac{1}{3}$and $P\left(B_3\right)=\frac{1}{6}$.

A box is selected at random and a ball is drawn from it.

If a red ball is drawn, then the probability that it has come from box $B_2$ is

• A. $\frac{35}{78}$
• B. $\frac{14}{39}$
• C. $\frac{10}{13}$
• D. $\frac{12}{13}$

Answer 1

The correct option is B

$\frac{14}{39}$

The explanation for the correct answer:

Step1. Find the possibility for each box:

$\text{'}R\text{'}$ represents the red balls

$\text{'}W\text{'}$represents the white balls

Number of balls in each box

$$\begin{gathered} B_1:1R,2W \\ {B}_2:2R,3W \\ {B}_3:3R,4W \end{gathered}$$

Given,

$$\begin{gathered} P\left(B_1\right)=\frac{1}{2} \\ P\left(B_2\right)=\frac{1}{3} \\ P\left(B_3\right)=\frac{1}{6} \end{gathered}$$

Step2. Find the required probability:

We know that $\begin{array}{rcl}\mathit{P}\left(\frac{B_2}{R}\right)& \mathbf{=}& \frac{\mathbf{P}\mathbf{\left(}{\mathbf{B}}_{\mathbf{2}}\mathbf{\right)}\mathbf{P}\left({\displaystyle \frac{R}{B_2}}\right)}{\mathbf{P}\mathbf{\left(}{\mathbf{B}}_{\mathbf{1}}\mathbf{\right)}\mathbf{P}\left(\frac{R}{B_1}\right)\mathbf{+}\mathbf{P}\mathbf{\left(}{\mathbf{B}}_{\mathbf{2}}\mathbf{\right)}\mathbf{P}\left(\frac{R}{B_2}\right)\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{3}\mathbf{\right)}\mathbf{P}\left(\frac{R}{B_3}\right)}\end{array}$

Now substitute the values

$\begin{array}{rcl}& \Rightarrow & \mathit{P}\mathbf{\left(}\frac{{\mathbf{B}}_{\mathbf{2}}}{\mathbf{R}}\mathbf{\right)}\mathbf{=}\frac{\left({\displaystyle \frac{1}{3}}\right)\left({\displaystyle \frac{2}{5}}\right)}{\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{3}}\right)+\left({\displaystyle \frac{1}{3}}\right)\left({\displaystyle \frac{2}{5}}\right)+\left({\displaystyle \frac{1}{6}}\right)\left({\displaystyle \frac{3}{7}}\right)}=\frac{\left({\displaystyle \frac{2}{15}}\right)}{{\displaystyle \frac{78}{210}}}\\ & =& \frac{\mathbf{14}}{\mathbf{39}}\end{array}$

Hence, option(B) is the correct answer.