Paragraph: Let U_1 and U_2 be two urns such that U_1 contains 3...

Now, probability of the drawn ball from $\mathrm{V}_2$ being white is $$\Rightarrow P \text{ (white/ } V_2 \text{ ) }$$\begin{aligned} & \left.\Rightarrow P \text{ (white/ } V_2\right) \\ & =P(H) \cdot\left\{\frac{{ }^3 C_1}{{ }^5 C_1} \times \frac{{ }^2 C_1}{{ }^2 C_1}+\frac{{ }^2 C_1}{{ }^5 C_1} \times \frac{{ }^1 C_1}{{ }^2 C_1}\right\} \\ & +P(T)\left\{\frac{{ }^3 C_2}{{ }^5 C_2} \times \frac{{ }^3 C_2}{{ }^3 C_2}+\frac{{ }^2 C_2}{{ }^5 C_2} \times \frac{{ }^1 C_1}{{ }^3 C_2}\right. \\ & \left.+\frac{{ }^3 C_1 \cdot{ }^2 C_1}{{ }^5 C_2} \times \frac{{ }^2 C_1}{{ }^3 C_2}\right\} \\ & \text{ Now, } P\left(\text{ white } / V_2\right)=\frac{1}{2}\left\{\frac{3}{5} \times 1+\frac{2}{5} \times \frac{1}{2}\right\} \\ & +\frac{1}{2}\left\{\frac{3}{10} \times 1+\frac{1}{10} \times \frac{1}{3}+\frac{6}{10} \times \frac{2}{3}\right\}=\frac{23}{30} \\ & \end{aligned}