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 $I$ and Bag $II$ and then a ball is drawn from Bag $II$. The ball so drawn is found to be red in colour. Find $310$ times the probability that the transferred ball is black.

Answer 1

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,
$P\left(A|E_1\right)=\frac{5}{10}=\frac{1}{2}$
When a black ball is transferred from bag I to II,
$P\left(A|E_2\right)=\frac{4}{10}=\frac{2}{5}$
$\therefore P\left(E_2|A\right)=\frac{P\left(E_2\right)P\left(A|E_2\right)}{P\left(E_1\right)P\left(A|E_1\right)+P\left(E_2\right)P\left(A|E_2\right)}$
$=\frac{\frac{4}{7}\times \frac{2}{5}}{\frac{3}{7}\times \frac{1}{2}+\frac{4}{7}\times \frac{2}{5}}$
$=\frac{16}{31}=0.51$