A bag contains 4 white and 5 black balls. Another bag contains 9 white and 7 black balls. A ball is transferred from the fisrt bag to the second and then a ball is drawn at random from the second bag. Find the probability that the ball drawn is white.

Let, ${\mathrm{W}}_1=\{4 whiteballs\mathrm{\}} and {\mathrm{B}}_1=\{5 blackballs\mathrm{\}}$

And ${\mathrm{W}}_2=\{9 \mathrm{whiteballs}\mathrm{\}} \mathrm{and} {\mathrm{B}}_2=\{7 \mathrm{blackballs}\mathrm{\}}$

Let ${\mathrm{E}}_1$ is the event that ball transferred from the first bag is white and ${\mathrm{E}}_2$ is the event that the ball transferred from the first bag is black.

Also, E is the event that the ball drawn from the second bag is white.

$\therefore \mathrm{P}\left(\frac{\mathrm{E}}{{\mathrm{E}}_1}\right)=\frac{10}{17},\mathrm{P}\left(\frac{\mathrm{E}}{{\mathrm{E}}_2}\right)=\frac{9}{17}$

And $\mathrm{P}\left({\mathrm{E}}_1\right)=\frac{4}{9}$ and $\mathrm{P}\left({\mathrm{E}}_2\right)=\frac{5}{9}$

$\therefore \mathrm{P}\left(\mathrm{E}\right)=\mathrm{P}\left({\mathrm{E}}_1\right)\cdot \mathrm{P}\left(\frac{\mathrm{E}}{{\mathrm{E}}_1}\right)+\mathrm{P}\left({\mathrm{E}}_2\right)\cdot \mathrm{P}\left(\frac{\mathrm{E}}{{\mathrm{E}}_2}\right)$

$=\frac{4}{9}\cdot \frac{10}{17}+\frac{5}{9}\cdot \frac{9}{17}$

$=\frac{40+45}{153}=\frac{\frac{85}{153}}{153}=\frac{5}{9}$

Therefore, the probability of the event that the ball drawn from the second bag is white is  $\frac{5}{9}$