**Toolbox:**

- Bag 1has 3 Black and 2 White
- Bag 2has 2 Black and 4 White
- \(E_1\)=First bag is chosen
- \(E_2\)=secound bag is chosen
- A=Black ball drawn
- P(A)=\(P(E_1\))\(P(A/E_1\))+\(P(E_2\))\(p(B/E_2\))

Probability of choosing Bag 1=Probability of choosing Bag 2

\(P(E_1\))=\(P(E_2\))=\(\large\frac{1}{2}\)

\(P(A/E_1\))=P(getting a black ball if 1 bag is chosen)=\(\large\frac{3}{5}\)

\(p(B/E_2\))=P(getting a black ball if 2bag is chosen)=\(\large\frac{2}{6}\)

P(A)=\(\large\frac{1}{2}\times\)\(\large\frac{3}{5}\)+\(\large\frac{1}{2}\times\)\(\large\frac{2}{6}\)

=\(\large\frac{3}{10}\)+\(\large\frac{1}{6}\)

=\(\large\frac{7}{15}\)