# Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls.

Question:

Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Solution:

Let $E_{1}$ and $E_{2}$ respectively denote the events that a red ball is transferred from bag I to II and a black ball is transferred from bag I to II.

$P\left(E_{1}\right)=\frac{3}{7}$ and $P\left(E_{2}\right)=\frac{4}{7}$

Let A be the event that the ball drawn is red.

When a red ball is transferred from bag I to II,

$P\left(A \mid E_{1}\right)=\frac{5}{10}=\frac{1}{2}$

When a black ball is transferred from bag I to II,

$P\left(A \mid E_{2}\right)=\frac{4}{10}=\frac{2}{5}$

$\therefore \mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)}$

$=\frac{\frac{4}{7} \times \frac{2}{5}}{\frac{3}{7} \times \frac{1}{2}+\frac{4}{7} \times \frac{2}{5}}$

$=\frac{16}{31}$