Let $$E$$ be the event of one white and one red balls $$A_1, A_2, A_3$$ be the events of both balls are from box $$\mathrm{B}_1, \mathrm{~B}_2, \mathrm{~B}_3$$ respectively

Now, the probability of one white and one red both balls are drawn from box $$B_2$$ is $$P\left(\frac{A_2}{E}\right)$$.

$$\Rightarrow P\left(\frac{A_2}{E}\right)=\frac{P\left(A_2 \cap E\right)}{P(E)}$$

$$\begin{aligned} & \Rightarrow P\left(\frac{A_2}{E}\right)=\frac{P\left(A_2\right) \cdot P\left(\frac{E}{A_2}\right)}{P\left(A_1\right) \cdot P\left(\frac{E}{A_1}\right)+P\left(A_2\right).\mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{A}_2}\right)+\mathrm{P}\left(\mathrm{A}_3\right) \cdot \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{A}_3}\right)} \\ & \Rightarrow \mathrm{P}\left(\frac{\mathrm{A}_2}{\mathrm{E}}\right)=\frac{\frac{1}{3} \times \frac{{ }^2 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_1}{{ }^9 \mathrm{C}_2}}{\frac{1}{3} \times \frac{{ }^1 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_1}{{ }^6 \mathrm{C}_2}+\frac{1}{3} \times \frac{{ }^2 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_1}{{ }^9 \mathrm{C}_2}+\frac{1}{3} \times \frac{{ }^3 \mathrm{C}_1 \cdot{ }^4 \mathrm{C}_1}{{ }^{12} \mathrm{C}_2} } \\ & \Rightarrow \mathrm{P}\left(\frac{\mathrm{A}_2}{\mathrm{E}}\right)=\frac{\frac{1}{6}}{\frac{1}{5}+\frac{1}{6}+\frac{2}{11}} \\ & \Rightarrow \mathrm{P}\left(\frac{\mathrm{A}_2}{\mathrm{E}}\right)=\frac{55}{181} \\ \end{aligned}$$

Hints:

$$\Rightarrow$$ Recall the Bay's theorem

$$ \begin{gathered} P\left(\frac{A_2}{E}\right)=\frac{P\left(A_2\right) \cdot P\left(\frac{E}{A_2}\right)}{P\left(A_1\right) \cdot P\left(\frac{E}{A}\right)+P\left(A_2\right) \cdot P\left(\frac{E}{A_2}\right)+P\left(A_3\right) \cdot P\left(\frac{E}{A_3}\right)}\\ \end{gathered}$$