Question

A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Answer 1

Hint: Use Bayes’ theorem and probability is the ratio of favorable number of outcomes to the total number of outcomes.

Given data
First bag contains 4 red and 4 black ball
Therefore total ball in first bag $=4+4=8$
Second ball contains 2 red and 6 black ball
Therefore total ball in second bag $=2+6=8$
Let $x_1$ and $x_2$ be the events of selecting first and second bag respectively.
Therefore probability of selecting one bag
$\Rightarrow p\left(x_1\right)={\displaystyle \frac{\text{Favorable bag}}{\text{Total bag}}}={\displaystyle \frac{1}{2}}=p\left(x_2\right)$
Let $A_1$ be the event of getting a red ball.
Therefore probability of drawing a red ball from the first bag $\Rightarrow p\left({\displaystyle \frac{A_1}{x_1}}\right)={\displaystyle \frac{\text{Favorable balls}}{\text{Total balls}}}={\displaystyle \frac{4}{8}}={\displaystyle \frac{1}{2}}$
Therefore probability of drawing a red ball from the Second bag $\Rightarrow p\left({\displaystyle \frac{A_1}{x_2}}\right)={\displaystyle \frac{\text{Favorable balls}}{\text{Total balls}}}={\displaystyle \frac{2}{8}}={\displaystyle \frac{1}{4}}$
Therefore probability of drawing a ball from the first bag, given that the ball is red is given by$p\left({\displaystyle \frac{x_1}{A_1}}\right)$.
Now we have to use the Bayes’ theorem to find out the total probability of drawing a ball from the first bag, given that the ball is red.
Bayes’ Theorem - In probability theory and statistics, Bayes' theorem describes the probability of an event, based on prior knowledge of conditions that might be related to the event.
Here,
$x_1,x_2,A_1$ = events.
$p\left({\displaystyle \frac{x_1}{A_1}}\right)$= Probability of $x_1$ given $A_1$is true.
$p\left({\displaystyle \frac{A_1}{x_1}}\right)$ = Probability of $A_1$ given $x_1$ is true.
$p\left({\displaystyle \frac{A_1}{x_2}}\right)$ = Probability of $A_1$ given $x_2$ is true.
$p\left(x_1\right)$, $p\left(x_2\right)$ = independent probabilities of $x_1$ and $x_2$.
So, by Bayes’ theorem we have
$\Rightarrow p\left({\displaystyle \frac{x_1}{A_1}}\right)={\displaystyle \frac{p\left(x_1\right).p\left({\displaystyle \frac{A_1}{x_1}}\right)}{p\left(x_1\right).p\left({\displaystyle \frac{A_1}{x_1}}\right)+p\left(x_2\right).p\left({\displaystyle \frac{A_1}{x_2}}\right)}}$

$\Rightarrow p\left({\displaystyle \frac{x_1}{A_1}}\right)={\displaystyle \frac{{\displaystyle \frac{1}{2}}.{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{2}}.{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}.{\displaystyle \frac{1}{4}}}}={\displaystyle \frac{{\displaystyle \frac{1}{4}}}{{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{8}}}}={\displaystyle \frac{{\displaystyle \frac{1}{4}}}{{\displaystyle \frac{3}{8}}}}={\displaystyle \frac{2}{3}}=0.66$
Therefore the required probability of drawing a ball from the first bag, given that the ball is red is 0.66.

Note: In such types of questions first find out the probability of selecting a bag then find out the probability of drawing a red ball from each bag then apply Bayes’ theorem we easily calculate the required probability of drawing a ball from the first bag, given that the ball is red.