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# One urn contains two black balls (labelled $B_1,B_2$) and one white ball.A second urn contains one black ball and two white balls (labelled $W_1,W_2$).Suppose the following experiment is performed.One of the two urns is chosen at random.Next a ball is randomly chosen from the urn.Then a second ball is chosen at random from the same urn without replacing the first ball.What is the probability that two black balls are chosen?

$\begin{array}{1 1}(A)\;\large\frac{1}{6}\\(B)\;\large\frac{1}{4}\\(C)\;\large\frac{1}{5}\\(D)\;\text{None of these}\end{array}$

Given :
Urn $U_1: B_1,B_2,W$
Urn $U_2: B,W_1,W_2$
P(Two black balls are selected)
(i.e) We have to select in first urn =$\large\frac{1}{6}$
Hence (A) is the correct answer.