Question

Three boxes ${{B}_{1}},{{B}_{2}}$ and ${{B}_{3}}$ contain balls with different colors as shown below

Box	White	Black	Red
${{B}_{1}}$	2	1	2
${{B}_{2}}$	3	2	4
${{B}_{3}}$	4	3	2

A die is thrown, ${{B}_{1}}$ is chosen if either $1$ or $2$ turns up. ${{B}_{2}}$ is chosen if either $3$ or $4$ turns up and ${{B}_{3}}$ is chosen if either $5$ or $6$ turns up. Having chosen a box in this way, a ball is chosen at random from this box. If the ball is bound to be red, find the probability it is from box ${{B}_{2}}$.

Answer 1

Hint: In this question we have been given with $3$ boxes which have balls of $3$ colors in them. We have to find the probability of finding a red ball, given that it is from box ${{B}_{2}}$. We will solve this question by dividing the probability of the red ball being from box ${{B}_{2}}$ with the probability of getting a red ball from all the boxes ${{B}_{1}}$, ${{B}_{2}}$ and ${{B}_{3}}$.

Complete step-by-step solution:
We know that there are a total of $6$ sides on a dice. ${{B}_{1}}$ is chosen if either $1$ or $2$ turns up. ${{B}_{2}}$ is chosen if either $3$ or $4$ turns up and ${{B}_{3}}$ is chosen if either $5$ or $6$ turns up. Since for all the $3$ boxes there are two numbers for it to be chosen out of the total of $6$, we get the probability of getting the boxes as:
$\Rightarrow {{B}_{1}}=\dfrac{2}{6}$, ${{B}_{2}}=\dfrac{2}{6}$ and ${{B}_{3}}=\dfrac{2}{6}$
Now the total number of balls in boxes ${{B}_{1}}$, ${{B}_{2}}$ and ${{B}_{3}}$ are $5,9$ and $9$ respectively and the number of red balls in the boxes are $2,4$ and $2$ respectively. Now we need to find the probability that the ball is red given that it is from box ${{B}_{2}}$.
We get the probability of a ball from ${{B}_{2}}$ given it is red as:
$= \dfrac{\text{Probability of red from }{{\text{B}}_{\text{2}}}}{\text{Probability of red from }{{\text{B}}_{1}}+\text{Probability of red from }{{\text{B}}_{\text{2}}}+\text{Probability of red from }{{\text{B}}_{3}}}$
Now we know before choosing a ball from the box, the dice has to be rolled for choosing the number of the box therefore, we have:
$\Rightarrow\text{Probability of red from }{{\text{B}}_{1}}=\dfrac{2}{6}\times \dfrac{2}{5}$
$\Rightarrow \text{Probability of red from }{{\text{B}}_{\text{2}}}=\dfrac{2}{6}\times \dfrac{4}{9}$
$\Rightarrow\text{Probability of red from }{{\text{B}}_{3}}=\dfrac{2}{6}\times \dfrac{2}{9}$
Now, on substituting the values in the formula, we get:
$= \dfrac{\dfrac{2}{6}\times \dfrac{4}{9}}{\dfrac{2}{6}\times \dfrac{2}{5}+\dfrac{2}{6}\times \dfrac{4}{9}+\dfrac{2}{6}\times \dfrac{2}{9}}$
On simplifying the terms in the numerator, we get:
$= \dfrac{\dfrac{4}{27}}{\dfrac{2}{6}\times \dfrac{2}{5}+\dfrac{2}{6}\times \dfrac{4}{9}+\dfrac{2}{6}\times \dfrac{2}{9}}$
On multiplying the terms in the denominator, we get:
$= \dfrac{\dfrac{4}{27}}{\dfrac{2}{15}+\dfrac{4}{27}+\dfrac{2}{27}}$
On adding the terms with the same denominator, we get:
$= \dfrac{\dfrac{4}{27}}{\dfrac{2}{15}+\dfrac{6}{27}}$
On simplifying the terms, we get:
$=\dfrac{\dfrac{4}{27}}{\dfrac{2}{15}+\dfrac{2}{9}}$
On taking the lowest common multiple of the fractions in the denominator, we get:
$= \dfrac{\dfrac{4}{27}}{\dfrac{6+10}{45}}$
On adding the terms, we get:
$= \dfrac{\dfrac{4}{27}}{\dfrac{16}{45}}$
On rearranging the terms in the fraction, we get:
$= \dfrac{4\times 45}{27\times 16}$
On simplifying the terms, we get:
$= \dfrac{5}{12}$, which is the required probability of the ball being red from box ${{B}_{2}}$, which is the required solution.

Note: It is to be noted that in this question we had to multiply the probability of getting a specific box from the rolling of a die with the probability of getting a red ball from all the balls in the box. It is to be remembered that when fractions with dissimilar denominators are to be added, the lowest common multiple of the fraction should be taken.