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 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.

Answer 1

Let $E_1$ : red ball is transferred from bag I to bag II

and $E_2$ : black ball is transferred from bag I to bag II

$\therefore E_{1}and{E}_2$ are mutually exclusive and exhaustive events.

$\therefore$ $P\left(E_1\right)=\frac{3}{3+4}=\frac{3}{7}andP\left(E_2\right)=\frac{4}{3+4}=\frac{4}{7}$

Let E be the event that the ball drawn is red. When a red ball is transferred from bag I to II.

$P\left(\frac{E}{E_1}\right)=\frac{4+1}{(4+1)+5}=\frac{5}{10}=\frac{1}{2}$

When a black ball is transferred from bag I to II

$P\left(\frac{E}{E_2}\right)=\frac{4}{4+(5+1)}=\frac{4}{10}=\frac{2}{5}$

$\therefore$ Required probability

$P\left(\frac{E_2}{E}\right)=\frac{P\left(\frac{E}{E_2}\right)P\left(E_2\right)}{P\left(\frac{E}{E_1}\right)P\left(E_1\right)+P\left(\frac{E}{E_2}\right)P\left(E_2\right)}=\frac{\frac{2}{5}\times \frac{4}{7}}{\frac{1}{2}\times \frac{3}{7}+\frac{2}{5}\times \frac{4}{7}}=\frac{\frac{8}{35}}{\frac{3}{14}+\frac{8}{35}}=\frac{\frac{8}{35}}{\frac{105+112}{14\times 35}}=\frac{8\times 14}{217}=\frac{16}{31}$