Question

A bag $A$ contains $4$ green and $3$ red balls and bag $B$ contains $4$ red and $3$ green balls. One bag is taken at random and a ball is drawn and noted to be green. The probability that it comes from bag $B$ is

• A. $\frac{2}{7}$
• B. $\frac{2}{3}$
• C. $\frac{3}{7}$
• D. $\frac{1}{3}$

Answer 1

The correct option is C

$\frac{3}{7}$

Explanation for correct option:

Step-1: Finding probability using classical approach.

Given, bag $\text{'}A\text{'}$ contains $4$ Green and $3$ red balls and bag $B$ contains $4$ red and $3$ green balls.

Let $E_G$ be the event in which the green ball is selecting, $E_A$be the event in which the ball selecting from bag $A$, $E_b$be the event in which the ball selecting from bag $B$.

Probability of ball drawn from the bag $A$, $P\left(A\right)=\frac{1}{2}$

Probability of ball drawn from the bag $B$, $P\left(B\right)=\frac{1}{2}$

Probability of green ball drawn from the bag $\left(A\right)$, $P\left(\frac{G}{A}\right)=\frac{4}{7}$

Probability of green ball drawn from the bag $\left(B\right)$, $P\left(\frac{G}{B}\right)=\frac{3}{7}$

Step-2: Applying Bayes theorem,

Probability of green ball drawn from bag $\left(B\right)$,

$$\begin{gathered} P\left(\frac{B}{G}\right)=\frac{P\left({\displaystyle \frac{G}{B}}\right)P\left(B\right)}{P\left({\displaystyle \frac{G}{A}}\right)P\left(A\right)+P\left({\displaystyle \frac{G}{B}}\right)P\left(B\right)} \\ =\left(\frac{{\displaystyle \frac{3}{7}}\times {\displaystyle \frac{1}{2}}}{{\displaystyle \frac{4}{7}}\times {\displaystyle \frac{1}{2}}+{\displaystyle \frac{3}{7}}\times {\displaystyle \frac{1}{2}}}\right) \\ =\frac{{\displaystyle \frac{3}{14}}}{{\displaystyle \frac{4+3}{14}}} \\ =\frac{3}{7} \end{gathered}$$

Therefore, the probability that the green ball comes from the bag $B$is $\frac{3}{7}$

Hence, correct answer is option $\left(C\right)$