Question

A box B1 contains $1$ white ball, $3$ red balls and $2$ black balls. Another box B2 contains $2$ white balls, $3$ red balls and $4$black balls. A third box B3 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 B2, is

• A. $\frac{166}{181}$
• B. $\frac{126}{181}$
• C. $\frac{65}{181}$
• D. $\frac{55}{181}$

Answer 1

The correct option is D

$\frac{55}{181}$

Explanation for the correct answer:

Step 1: Given data and variable:

Box B1 contains white balls = $1$, red balls = $3$ and black balls = $2$

Box B2 contains white balls = $2$, red balls = $3$ and black balls = $4$

Box B3 contains white balls = $3$, red balls = $4$ and black balls = $5$

Total number of boxes = $3$

Let A be the event that is one white and the other ball is red.

Let E1 be the event that both the balls are drawn from box B1

Let E2 be the event that both the balls are drawn from box B2

Let E3 be the event that both the balls are drawn from box B3.

Step 2: Finding the probability values for possible events

The probability can be find by using Baye's theorem.

$P\left(A|B\right)=\frac{P\left(B|A\right)P\left(A\right)}{P\left(B\right)}$

The probability of E1 = E2 = E3, P(E) = $\frac{1}{3}$

$$\begin{gathered} P\left(A\right|E2)=\frac{{}^2{C}_1{\times }^3{C}_1}{{}^9{C}_2}, \\ P(A\left|E1\right)=\frac{{}^1{C}_1{\times }^3{C}_1}{{}^6{C}_2}, \\ P\left(A\right|E3)=\frac{{}^3{C}_1{\times }^4{C}_1}{{}^{12}{C}_2}, \end{gathered}$$

Step 3: Finding the probability $P\left(E2|A\right)$

According to Baye's theorem,

$P\left(E2|A\right)=\frac{P\left(A|E2\right)\times P\left(E2\right)}{[P\left(A|E1\right)\times P\left(E1\right)]+[P\left(A|E2\right)\times P\left(E2\right)]+[P\left(A|E3\right)\times P\left(E3\right)]}$

On substituting the values,

$$\begin{gathered} P\left(E2|A\right)=\frac{{\displaystyle \frac{{}^2{C}_1{\times }^3{C}_1}{{}^9{C}_2}}\times {\displaystyle \frac{1}{3}}}{{\displaystyle \frac{{}^1{C}_1{\times }^3{C}_1}{{}^6{C}_2}}\times {\displaystyle \frac{1}{3}}+{\displaystyle \frac{{}^2{C}_1{\times }^3{C}_1}{{}^9{C}_2}}\times {\displaystyle \frac{1}{3}}+{\displaystyle \frac{{}^3{C}_1{\times }^4{C}_1}{{}^{12}{C}_2}}\times {\displaystyle \frac{1}{3}}} \\ =\frac{{\displaystyle \frac{1}{6}}}{\left({\displaystyle \frac{1}{5}}\right)+\left({\displaystyle \frac{1}{6}}\right)+\left({\displaystyle \frac{2}{11}}\right)} \\ =\frac{{\displaystyle \frac{1}{6}}}{{\displaystyle \frac{181}{330}}} \\ =\frac{55}{181} \end{gathered}$$

Hence, option (D) is the correct answer.