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Three boxes contain 6 white, 4 blue; 5 white, 5 blue and 4 white, 6 blue balls respectively. One of the box is selected at random and a ball is drawn from it. If the ball drawn is blue, find the probability that it is from the second box.

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Let A, B, C, D be the events defined as: 

A: Selecting the first box 

B: Selecting the second box 

C: Selecting the third box 

D: Event of drawing a blue ball. 

Since there are three boxes and each box has an equally likely chance of selection, P(A) = P(B) = P(C) = \(\frac{1}{3}\)

♦ If the first box is chosen, i.e., A has already occurred, then 

Probability of drawing a blue ball from A = \(\frac{4}{10}\)⇒ P(D/A) =\(\frac{4}{10}\)

♦ If the second box is chosen, i.e., B has already occurred, then 

Probability of drawing a blue ball from B =\(\frac{5}{10}\) ⇒ P(D/B) = \(\frac{5}{10}\)

Similarly P(D/C) = \(\frac{6}{10}\)

Now we are required to find the probability (B/D), i.e., given that the ball drawn is blue, we need to find the probability that it is drawn from the second box. 

By Baye’s Theorem,

 \(P(B/D) = \frac {P(B)\times P(D/B)}{P(A)\times P(D/A)+P(B)\times P(D/B) +P(C)\times P(D/C)}\)

 = \(\frac {\frac{1}{3}\times \frac{5}{10}}{\frac{1}{3}\times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}+ \frac{1}{3}\times \frac{6}{10}}\) = \(\frac {\frac{1}{3}\times \frac{5}{10}}{\frac{1}{3}\times \frac{15}{10}}\)\(\frac {1}{3}\)

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