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.

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
$\Rightarrow p\left( {{x_1}} \right) = \dfrac{{{\text{Favorable bag}}}}{{{\text{Total bag}}}} = \dfrac{1}{2} = p\left( {{x_2}} \right)$
Let ${A_1}$ be the event of getting a red ball.
Therefore probability of drawing a red ball from the first bag $\Rightarrow p\left( {\dfrac{{{A_1}}}{{{x_1}}}} \right) = \dfrac{{{\text{Favorable balls}}}}{{{\text{Total balls}}}} = \dfrac{4}{8} = \dfrac{1}{2}$
Therefore probability of drawing a red ball from the Second bag $\Rightarrow p\left( {\dfrac{{{A_1}}}{{{x_2}}}} \right) = \dfrac{{{\text{Favorable balls}}}}{{{\text{Total balls}}}} = \dfrac{2}{8} = \dfrac{1}{4}$
Therefore probability of drawing a ball from the first bag, given that the ball is red is given by$p\left( {\dfrac{{{x_1}}}{{{A_1}}}} \right)$.
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.
$p\left( {\dfrac{{{x_1}}}{{{A_1}}}} \right)$= Probability of ${x_1}$ given ${A_1}$is true.
$p\left( {\dfrac{{{A_1}}}{{{x_1}}}} \right)$ = Probability of ${A_1}$ given ${x_1}$ is true.
$p\left( {\dfrac{{{A_1}}}{{{x_2}}}} \right)$ = 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
$\Rightarrow p\left( {\dfrac{{{x_1}}}{{{A_1}}}} \right) = \dfrac{{p\left( {{x_1}} \right).p\left( {\dfrac{{{A_1}}}{{{x_1}}}} \right)}}{{p\left( {{x_1}} \right).p\left( {\dfrac{{{A_1}}}{{{x_1}}}} \right) + p\left( {{x_2}} \right).p\left( {\dfrac{{{A_1}}}{{{x_2}}}} \right)}}$

$\Rightarrow p\left( {\dfrac{{{x_1}}}{{{A_1}}}} \right) = \dfrac{{\dfrac{1}{2}.\dfrac{1}{2}}}{{\dfrac{1}{2}.\dfrac{1}{2} + \dfrac{1}{2}.\dfrac{1}{4}}} = \dfrac{{\dfrac{1}{4}}}{{\dfrac{1}{4} + \dfrac{1}{8}}} = \dfrac{{\dfrac{1}{4}}}{{\dfrac{3}{8}}} = \dfrac{2}{3} = 0.66$
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.