Urn A contains 2 white and 2 black balls while urn B contains 3...

From the given information
$\mathrm{E}_1=$ One white ball transferred from A to B then urn A contains White $=1$, Black $=2$ and urn B contains White $=4$, Black $=2\mathrm{E}_2=$ One Black ball transferred from A to B then urn A contains White $=2$, Black $=1$ and urn B contains White $=3$, Black $=3\mathrm{F}=$ One white ball is drawn from urn B . $$\begin{aligned} & \therefore \quad \mathrm{P}(\mathrm{ F})=\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\mathrm{ F} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\mathrm{ F} / \mathrm{E}_2\right) \\ & =\frac{2}{4} \times \frac{4}{6}+\frac{2}{4} \times \frac{3}{6}=\frac{8}{24}+\frac{6}{24}=\frac{14}{24}=\frac{7}{12} \end{aligned}$$