Bag I contains 3 red and 4 black balls
Bag II contains 4 red and 5 black balls.
Let $\mathrm{E}_1=$ Event that a red ball is drawn from Bag I
$\mathrm{E}_2=$ Event that a black ball is drawn from Bag I
$\therefore \quad \mathrm{P}\left(\mathrm{E}_1\right)=\frac{3}{7}, \mathrm{P}\left(\mathrm{E}_2\right)=\frac{4}{7}$
After transferring a red ball from Bog I to Bag II, the Bag II will have 5 red and 5 black balls.
Let A be the event of drawing red ball
$\therefore \quad \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)=\frac{5}{10}=\frac{1}{2}$
Further when a black ball is transfered from Bag I to Bag II, it will contain 4 red and 6 black balls.
$$
\mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)=\frac{4}{10}=\frac{2}{5}
$$
$\therefore \quad \mathrm{P}\left(\mathrm{E}_2 / \mathrm{A}\right)$
$$
=\frac{P\left(E_2\right) P\left(A / E_2\right)}{P\left(E_1\right) P\left(A / E_1\right)+P\left(E_2\right) P\left(A / E_2\right)}
$$
$$
=\frac{\frac{4}{7} \times \frac{2}{5}}{\frac{3}{7} \times \frac{1}{2}+\frac{4}{7} \times \frac{2}{5}}=\frac{\frac{8}{35}}{\frac{3}{14}+\frac{8}{35}}=\frac{16}{31}
$$