**Toolbox:**

- Let \(E_1\)=Whiteball transferred from bag1to bag2
- \(E_2\)=Blackball transferred from bag1 to bag2
- A=getting of white ball
- P(A)=P(\(E_1\))P(\(A/E_1\))+P(\(E_2\))P(\(A/E_2\))

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)=\(\large\frac{4}{9}\times\)\(\large\frac{10}{17}\)+\(\large\frac{5}{9}\times\)\(\large\frac{9}{17}\)

=\(\large\frac{85}{153}\)