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Bag $$B_{1}$$ contains 6 white and 4 blue balls, Bag $$B_{2}$$ contains 4 white and 6 blue balls, and Bag $$B_{3}$$ contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is drawn from Bag $$B_{2}$$, is :
P(B₁)=P(B₂)=P(B₃)=1/3
P(W|B₁)=6/10, P(W|B₂)=4/10, P(W|B₃)=5/10
P(W) = (1/3)(6/10+4/10+5/10) = (1/3)(15/10) = 1/2
P(B₂|W) = P(W|B₂)P(B₂)/P(W) = (4/10)(1/3)/(1/2) = (4/30)/(1/2) = 8/30 = 4/15
The correct answer is Option 1: 4/15.
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