# 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.

Last updated date: 16th Mar 2023

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Answer

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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.

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.

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