Question

A bag $\text{'}A\text{'}$ contains $2$ white and $3$ red balls and bag $\text{'}B\text{'}$ contains $4$ white and $5$ red balls. One ball is drawn at random from a randomly chosen bag and is found to be red. The probability that it was drawn from bag $\text{'}B\text{'}$ was

• A. $\frac{5}{14}$
• B. $\frac{5}{16}$
• C. $\frac{5}{18}$
• D. $\frac{25}{52}$

Answer 1

The correct option is D

$\frac{25}{52}$

Explanation for correct option:

Step-1: Finding probability with respect to classical approach.

Given, bag $\text{'}A\text{'}$ contains $2$ white and $3$ red balls and bag $\text{'}B\text{'}$ contains $4$ white and $5$ red balls.

Let $E_R$ be the event in which the red 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 red ball drawn from the bag $\left(A\right)$, $P\left(\frac{R}{A}\right)=\frac{3}{5}$

Probability of red ball drawn from the bag $\left(B\right)$, $P\left(\frac{R}{B}\right)=\frac{5}{9}$

Step-2: Applying Bayes theorem,

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

$$\begin{gathered} P\left(\frac{B}{R}\right)=\frac{P\left({\displaystyle \frac{R}{B}}\right)P\left(B\right)}{P\left({\displaystyle \frac{R}{A}}\right)P\left(A\right)+P\left({\displaystyle \frac{R}{B}}\right)P\left(B\right)} \\ =\left(\frac{{\displaystyle \frac{5}{9}}\times {\displaystyle \frac{1}{2}}}{{\displaystyle \frac{3}{5}}\times {\displaystyle \frac{1}{2}}+{\displaystyle \frac{5}{9}}\times {\displaystyle \frac{1}{2}}}\right) \\ =\frac{{\displaystyle \frac{5}{18}}}{{\displaystyle \frac{27+25}{90}}} \\ =\frac{25}{52} \end{gathered}$$

Therefore, correct answer is option $\left(D\right)$