# Three boxes contain 6 white, 4 blue; 5 white, 5 blue and 4 white, 6 blue balls respectively

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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,