Question

A box $B_1$ contains $1$ white ball, $3$ red balls, and $2$ black balls. Another box $B_2$ contains $2$ white balls, $3$ red balls, and $4$ black balls. A third box $B_3$ contains $3$ white balls, $4$ red balls, and $5$ black balls.
If $2$ balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these $2$ balls are drawn from box $B_2$ is

• A. $\frac{116}{182}$
• B. $\frac{126}{181}$
• C. $\frac{65}{181}$
• D. $\frac{55}{181}$

Answer 1

The correct option is D $\frac{55}{181}$

Let
$A$: one ball is white and other is red
$E_1$: both balls are from box $B_1$
$E_2$: both balls are from box $B_2$
$E_3$: both balls are from box $B_3$

Here, $P\left(\text{required}\right)=P\left(\frac{E_2}{A}\right)$
$=\frac{P\left(\frac{A}{E_2}\right).P\left(E_2\right)}{P\left(\frac{A}{E_1}\right).P\left(E_1\right)+P\left(\frac{A}{E_2}\right).P\left(E_2\right)+P\left(\frac{A}{E_3}\right).P\left(E_3\right)}$
$=\frac{\frac{{}^2{C}_1\times {}^3{C}_1}{{}^9{C}_2}\times \frac{1}{3}}{\frac{{}^1{C}_1\times {}^3{C}_1}{{}^6{C}_2}\times \frac{1}{3}+\frac{{}^2{C}_1\times {}^3{C}_1}{{}^9{C}_2}\times \frac{1}{3}+\frac{{}^3{C}_1\times {}^4{C}_1}{{}^{12}{C}_2}\times \frac{1}{3}}$
$=\frac{55}{181}$

Alternate Solution:

Probability of drawing a white and a red ball from bag $B_1$ is $P\left(B_1\right)=\frac{{}^1{C}_1\times 3C_1}{{}^6{C}_2}=\frac{1}{5}$
Probability of drawing a white and a red ball from bag $B_2$ is $P\left(B_2\right)=\frac{{}^2{C}_1\times 3C_1}{{}^9{C}_2}=\frac{1}{6}$
Probability of drawing a white and a red ball from bag $B_3$ is $P\left(B_3\right)=\frac{{}^3{C}_1\times 4C_1}{{}^{12}{C}_2}=\frac{2}{11}$
Therefore the probability that the red and the white ball drawn are from the bag $B_2$ is $\frac{P\left(B_2\right)}{P\left(B_1\right)+P\left(B_2\right)+P\left(B_3\right)}=\frac{55}{181}$