Question

Bag $A$ contains $2$ white and $3$ red balls and bag $B$ contains $4$ white and $5$ red balls. One ball is drawn at random from one of the bag is found to be red. Find the probability that it was drawn from bag $B$.

• A. $\frac{3}{8}$
• B. $\frac{25}{52}$
• C. $\frac{1}{8}$
• D. $\frac{3}{14}$

Answer 1

The correct option is B $\frac{25}{52}$

Let $X$ be the probability of choosing bag $A$,$Y$ be the probability of choosing bag $B$

Let $E$ be the probability of ball drawn is red

Then $P\left(X\right)=\frac{1}{2}$

$P\left(Y\right)=\frac{1}{2}$

$P\left(E/X\right)=\frac{3}{5}$

$P\left(E/Y\right)=\frac{5}{9}$

Apply the Bayes theorem:

$P\left(Y/E\right)=\frac{P\left(Y\right)\times P\left(E/Y\right)}{P\left(X\right)\times P\left(E/X\right)+P\left(Y\right)\times P\left(E/Y\right)}$

$=\frac{\frac{1}{2}\times \frac{5}{9}}{\frac{1}{2}\times \frac{3}{5}+\frac{1}{2}\times \frac{5}{9}}$

$=\frac{50}{104}$

$=\frac{25}{52}$

Hence, the probability that the red ball is drawn from bag $B$ is $=\frac{25}{52}$