Question

Sol. [Key Concept: Hypergeometric / Bayes on the 10-ball (4 black) bag.]

$E_{i} \rightarrow$ the event that ' i ' white are drawn from the first bag

$A \rightarrow$ one ball drawn from second bag is white.

$$\begin{matrix} P\left( E_{2}/A \right) & \ = \frac{P\left( \text{ }A/E_{2} \right)P\left( E_{2} \right)}{\sum_{i = 1}^{5}\mspace{2mu}\mspace{2mu} P\left( E_{i} \right)P\left( \text{ }A/E_{i} \right)} \\ & \ = \frac{\frac{\ ^{4}C_{3}\ ^{6}C_{2}}{\ ^{10}C_{5}} \times \frac{2}{5}}{\frac{4}{5} \times \frac{\ ^{4}C_{1}\ ^{6}C_{4}}{\ ^{10}C_{5}} + \frac{3}{5}\frac{\ ^{4}C_{2}\ ^{6}C_{3}}{\ ^{10}C_{5}} + \frac{2}{5}\frac{\ ^{4}C_{3}\ ^{6}C_{2}}{\ ^{10}C_{5}} + \frac{\ ^{4}C_{4}\ ^{6}C_{1}}{\ ^{10}C_{5}} \times \frac{1}{5}} \\ & \ = \frac{4 \times 15 \times 2}{4 \times 4 \times 15 + 3 \times 6 \times 20 + 2 \times 4 \times 15 + 6} = \frac{20}{121} \end{matrix}$$