**Toolbox:**

- Baye's Theorem :$P(E_i/E)=\large\frac{P(E_i).P(E/E_i)}{P(E_1).P(E/E_1)+P(E_2).P(E/E_2)+P(E_3).P(E/E_3)}$

This question is from Baye's Theorem.

Let $E_1 $ be event of choosing Urn $A$, $E_2 $ be event of choosing Urn $B$ and

$E_3 $ be event of choosing Urn $C$.

Let $E$ be the event of drawing one white and one red ball from the selected urn.

$P(E_1)=P(E_2)=P(E_3)=\large\frac{1}{3}$ (since the selection of any urn is equally likely.)

Given: In Urn $A$ there are $W-1,\:B-2,\:and\:R-3$ balls

In Urn $B$ there are $W-2,\:B-1,\:and\:R-1$ balls and

In Urn $C$ there are $W-4,\:B-5,\:and\:R-3$ balls

$P(E/E_1)=P$(drawing a white and a red ball from urn $A)=\large\frac{^1C_1.^3C_1}{^6C_2}=\frac{1}{5}$

$P(E/E_2)=P$(drawing a white and a red ball from urn $B)=\large\frac{^2C_1.^1C_1}{^4C_2}=\frac{1}{3}$

$P(E/E_1)=P$(drawing a white and a red ball from urn $C)=\large\frac{^4C_1.^3C_1}{^{12}C_2}=\frac{2}{11}$

According to Baye's Theorem,

$P(E_1/E)=\large\frac{P(E_1).P(E/E_1)}{P(E_1).P(E/E_1)+P(E_2).P(E/E_2)+P(E_3).P(E/E_3)}$

$\Rightarrow\:$ P(That the balls (1 W, 1 R) are drawn from Urn $A$) is

$P(E_1/E)=\large\frac{1/3\times1/5}{1/3\times1/5+1/3\times1/3+1/3\times2/11}$

$=\large\frac{1}{5}.\frac{165}{118}=\frac{33}{118}$

$P(E_2/E)=\large\frac{P(E_2).P(E/E_2)}{P(E_1).P(E/E_1)+P(E_2).P(E/E_2)+P(E_3).P(E/E_3)}$

$\Rightarrow\:$ P(That the balls (1 W, 1 R) are drawn from Urn $B$) is

$P(E_2/E)=\large\frac{1/3\times1/3}{1/3\times1/5+1/3\times1/3+1/3\times2/11}$

$=\large\frac{1}{3}.\frac{165}{118}=\frac{55}{118}$

and P(That the balls (1 W, 1 R) are drawn from Urn $C) =1-P(E_1/E)-P(E_2/E)$

$=\large\frac{15}{118}$