Solution
Event $\mathrm{E}_{1}$ : white ball is transferred
Event $\mathrm{E}_{2}$ : Black ball is transferred
\[
P\left(E_{1}\right)=\frac{6}{6+4}=\frac{3}{5}, P\left(E_{2}\right)=\frac{4}{6+4}=\frac{2}{5}
\]

Event X : From bag A , white ball is drawn
\[
\begin{aligned}
& P(X)=P\left(E_{1}\right) P\left(\frac{X}{E_{1}}\right)+P\left(E_{2}\right) P\left(\frac{X}{E_{2}}\right) \\
& =\frac{3}{5}\left(\frac{10}{18}\right)+\frac{2}{5}\left(\frac{9}{18}\right)=\frac{30+18}{5 \times 18}=\frac{8}{15}=\frac{p}{q} \\
& \therefore p+q=23 \text { Answer(2) }
\end{aligned}
\]