P(\(E_1\))=P(transferring white ball)
=\(\large\frac{4}{9}\)
P(\(E_2\))=P(transferring black ball)
=\(\large\frac{5}{9}\)
P(\(A/E_1\))=P(getting white from bag 2 when white ball is transfered)
=\(\large\frac{10}{17}\)
P(\(A/E_2\))=P(getting white from bag 2 when black ball is transfered)
=\(\large\frac{9}{17}\)
P(A)=\(\large\frac{4}{9}\times\)\(\large\frac{10}{17}\)+\(\large\frac{5}{9}\times\)\(\large\frac{9}{17}\)
=\(\large\frac{85}{153}\)