$\begin{array}{1 1}(A)\;\large\frac{2}{3}\\(B)\;\large\frac{1}{3}\\(C)\;\large\frac{4}{3}\\(D)\;\large\frac{5}{3}\end{array} $

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Given :

Urn $U_1:B_1,B_2,W$

Urn $U_2:B,W_1,W_2$

Case I:

Selection from first urn

$P=\big[\large\frac{2C_1}{3C_1}\times \large\frac{1}{2C_1}\big]$

$\;\;\;=\big[\large\frac{2}{3}\times \frac{1}{2}\big]$

$\;\;\;=\large\frac{1}{3}$

Step 2:

Case II:

Selection from second urn

$P=\big[\large\frac{1}{3C_1}\times \large\frac{2C_1}{2C_1}\big]$

$\;\;\;=\big[\large\frac{1}{3}\times $$1\big]$

$\;\;\;=\large\frac{1}{3}$

Step 3:

P(Two balls are of opposite colour)=$\large\frac{1}{3}+\frac{1}{3}$

$\Rightarrow \large\frac{2}{3}$

Hence (A) is the correct answer.

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