Question

For $k=1,2,3$ the box ${{B}_{k}}$ contains k red balls and $\left( k+1 \right)$ white balls, let $P\left( {{B}_{1}} \right)=\dfrac{1}{2}$ , $P\left( {{B}_{2}} \right)=1$ and $P\left( {{B}_{3}} \right)=\dfrac{1}{6}$ . 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:
$\text{A}\text{. }\dfrac{35}{78}$
 $\text{B}\text{. }\dfrac{14}{39}$
 $\text{C}\text{. }\dfrac{10}{13}$
 $\text{D}\text{. }\dfrac{12}{13}$

Answer 1

Hint: Here, the question is given in probability. In this question we are asked to find the probability of the red ball chosen from the box ${{B}_{2}}$ . To find this first we need to find the No’s of balls contain in ${{B}_{1}},{{B}_{2}}\And {{B}_{3}}$ , and probability of red balls of each box. Then we will find the answer with the help of Bayes’ theorem which say’s:\[\text{Required Probability =}\dfrac{P\left( {{B}_{2}} \right)\times P\left( {{R}_{2}} \right)}{\left( P\left( {{B}_{1}} \right)\times P\left( {{R}_{1}} \right) \right)+\left( P\left( {{B}_{2}} \right)\times P\left( {{R}_{2}} \right) \right)+\left( P\left( {{B}_{3}} \right)\times P\left( {{R}_{3}} \right) \right)}\] .
Complete step by step solution:
It is given that, in a Box
${{B}_{k}}$Contain k red balls and $\left( k+1 \right)$ white balls
If $k=1$ , then –
${{B}_{1}}$ Contain 1 red ball and 2 white balls.
If $k=2$ , then –
${{B}_{2}}$ Contains 2 red balls and 3 white balls.
If $k=3$ , then –
${{B}_{3}}$ Contains 3 red balls and 4 white balls.
Now, we will find the probability of drawing red ball from each Box i.e. $P\left( {{R}_{1}} \right),P\left( {{R}_{2}} \right)\And P\left( {{R}_{3}} \right)$ , by taking the formula of probability.
I.e. $\text{Probability=}\dfrac{\text{No}\text{. of favorable outcomes}}{\text{Total No}\text{. of outcomes}}$ .
For ${{B}_{1}}$ :
Favorable outcomes of red ball is 1 and total No. of outcomes is \[1+2=3\]
$\therefore P\left( {{R}_{1}} \right)=\dfrac{1}{3}$
For ${{B}_{2}}$ :
Favorable outcomes of red ball is 2 and total No. of outcomes is \[2+3=5\]
$\therefore P\left( {{R}_{2}} \right)=\dfrac{2}{5}$
For ${{B}_{3}}$ :
Favorable outcomes of red ball is 3 and total No. of outcomes is \[3+4=7\]
$\therefore P\left( {{R}_{3}} \right)=\dfrac{3}{7}$
It is also given that –
$P\left( {{B}_{1}} \right)=\dfrac{1}{2}$ , $P\left( {{B}_{2}} \right)=1$ and $P\left( {{B}_{3}} \right)=\dfrac{1}{6}$
Now, we will find the probability of red balls that have come from ${{B}_{2}}$ by taking the help of Bayes’ theorem.
We know that –
\[\text{Required Probability =}\dfrac{P\left( {{B}_{2}} \right)\times P\left( {{R}_{2}} \right)}{\left( P\left( {{B}_{1}} \right)\times P\left( {{R}_{1}} \right) \right)+\left( P\left( {{B}_{2}} \right)\times P\left( {{R}_{2}} \right) \right)+\left( P\left( {{B}_{3}} \right)\times P\left( {{R}_{3}} \right) \right)}\] .
$=\dfrac{\left( 1\times \dfrac{2}{5} \right)}{\left( \dfrac{1}{2}\times \dfrac{1}{3} \right)+\left( 1\times \dfrac{2}{5} \right)+\left( \dfrac{1}{6}\times \dfrac{3}{7} \right)}$
By simplifying the above equation, we get –
$=\dfrac{\dfrac{2}{5}}{\dfrac{1}{6}+\dfrac{2}{5}+\dfrac{3}{42}}$
By multiplying the L.C.M of 6, 5 & 42 in the denominator, we get –
$=\dfrac{\dfrac{2}{5}}{\dfrac{1}{6}\times 210+\dfrac{2}{5}\times 210+\dfrac{3}{42}\times 210}$
By simplifying the above equation we get –
$=\dfrac{\dfrac{2}{5}}{\dfrac{35+84+15}{210}}$
$=\dfrac{\dfrac{2}{5}}{\dfrac{134}{210}}$
$\Rightarrow \dfrac{2}{5}\div \dfrac{134}{210}=\dfrac{2}{5}\times \dfrac{210}{134}$
$=\dfrac{420}{670}$
By canceling the common factors from numerator and denominator, we get –
$=\dfrac{42}{67}$
Therefore, probability of getting red ball that comes from \[{{B}_{2}}\] is $\dfrac{42}{67}$
Hence, option (E) is the correct answer.
Note: Students should be very careful while understanding this question as they may do mistakes while finding the probability. They may find the probability of getting red ball that comes from ${{B}_{2}}$ directly by taking the formula of a $\text{Probability=}\dfrac{\text{No}\text{. of favorable outcomes}}{\text{Total No}\text{. of outcomes}}$ which is wrong. They should know that Bayes’ theorem is a way of finding a probability when we know a certain other probability.