Question

Box 1 contains 2 white and 3 red balls and box ll contains 4 white and 5 red balls. One ball is drawn at random from one of the boxes and is found to be red. Then the percentage that it was from box ll is?
\[\begin{align}
  & A.\dfrac{54}{44} \\
 & B.\dfrac{54}{14} \\
 & C.\dfrac{54}{104} \\
 & D.\text{None of these} \\
\end{align}\]

Answer 1

Hint: For solving this, we will first define events of selecting box 1 as ${{E}_{1}}$, box ll as ${{E}_{2}}$, red balls by R, and white balls as W. Then we will use Bayes formula on finding probability that ball drawn is red and from box ll is given as,
$P\left( \dfrac{{{E}_{2}}}{R} \right)=\dfrac{P\left( {{E}_{2}} \right)\cdot P\left( \dfrac{R}{{{E}_{2}}} \right)}{P\left( {{E}_{1}} \right)\cdot P\left( \dfrac{R}{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right)\cdot P\left( \dfrac{R}{{{E}_{2}}} \right)}$.
Where $P\left( {{E}_{1}} \right)$ is the probability of box 1, $P\left( {{E}_{2}} \right)$ is the probability of picking box ll, $P\left( \dfrac{R}{{{E}_{2}}} \right)$ is the probability of drawing red ball if box ll is selected, $P\left( \dfrac{R}{{{E}_{1}}} \right)$ is the probability of drawing red balls if box 1 is selected.
So we will use all required probabilities and conclude our final answer.

Complete step-by-step answer:
Here, we are given two boxes, box 1 and box ll.
Let us suppose that ${{E}_{1}}$ represents that box 1 is chosen and ${{E}_{2}}$ represent that box ll is chosen.
Also let us suppose that, R represents that red ball is drawn and W represents that white ball is drawn.
Now we need to find that, if red ball is picked then what is the probability that, it is picked from box ll which means we need to find $P\left( \dfrac{{{E}_{2}}}{R} \right)$.
From Bayes formula we know that $P\left( \dfrac{{{E}_{2}}}{R} \right)=\dfrac{P\left( {{E}_{2}} \right)\cdot P\left( \dfrac{R}{{{E}_{2}}} \right)}{P\left( {{E}_{1}} \right)\cdot P\left( \dfrac{R}{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right)\cdot P\left( \dfrac{R}{{{E}_{2}}} \right)}$.
Where $P\left( {{E}_{1}} \right)$ is the probability of box 1, $P\left( {{E}_{2}} \right)$ is the probability of picking box ll, $P\left( \dfrac{R}{{{E}_{2}}} \right)$ is the probability of drawing red ball if box ll is selected, $P\left( \dfrac{R}{{{E}_{1}}} \right)$ is the probability of drawing red balls if box 1 is selected.
Let us now find all these probabilities.
1. $P\left( {{E}_{1}} \right)$
Here we have total boxes as 2 and we need to pick box 1 from 2 boxes. So probability becomes equal to $\dfrac{1}{2}$. Hence $P\left( {{E}_{1}} \right)=\dfrac{1}{2}$.
2. $P\left( {{E}_{2}} \right)$
Here, we have total boxes as 2 and we need to pick ll from 2 boxes so probability becomes equal to $\dfrac{1}{2}$. Hence $P\left( {{E}_{2}} \right)=\dfrac{1}{2}$.
3. $P\left( \dfrac{R}{{{E}_{1}}} \right)$
Here, we need to pick red balls from box 1. Since box 1 contains 2 white and 3 red balls so the number of red balls = 3.
Number of balls = 2+3 = 5.
So the probability of picking red balls is $\dfrac{3}{5}$.
Hence $P\left( \dfrac{R}{{{E}_{1}}} \right)=\dfrac{3}{5}$.
4. $P\left( \dfrac{R}{{{E}_{2}}} \right)$
Here we need to pick red balls from box ll.
Since box ll contains 4 white and 5 red balls, so number of balls = 5.
Total number of balls = 4+5 = 9.
So the probability of getting red balls from 9 balls becomes $\dfrac{5}{9}$.
Hence, $P\left( \dfrac{R}{{{E}_{2}}} \right)=\dfrac{5}{9}$.
Putting all the values in the formula we get:
\[\begin{align}
  & P\left( \dfrac{{{E}_{2}}}{R} \right)=\dfrac{\dfrac{3}{5}\cdot \dfrac{1}{2}}{\dfrac{5}{9}\cdot \dfrac{1}{2}+\dfrac{3}{5}\cdot \dfrac{1}{2}} \\
 & \Rightarrow \dfrac{\dfrac{3}{10}}{\dfrac{5}{18}+\dfrac{3}{10}} \\
 & \Rightarrow \dfrac{\dfrac{3}{10}}{\dfrac{50+54}{180}} \\
 & \Rightarrow \dfrac{\dfrac{3}{10}}{\dfrac{104}{180}} \\
 & \Rightarrow \dfrac{3}{10}\times \dfrac{180}{104} \\
 & \Rightarrow \dfrac{54}{104} \\
\end{align}\]
Hence required probability is $\dfrac{54}{104}$.

So, the correct answer is “Option C”.

Note: Students can get confused between $P\left( \dfrac{A}{B} \right)\text{ and }P\left( \dfrac{B}{A} \right)$. Here $P\left( \dfrac{A}{B} \right)$ means finding probability of A when B is the given condition and ... means finding probability of B when A is the given condition. Take care in calculation while finding probability. Make sure that probability lies between 0 and 1.