Solution

To solve this question we will use Bayes Theorem to get reverse probability. 

Let first ball drawn is red and second ball drawn is white. Probability of choosing a box at random =  $\frac{1}{3}$

=> Probability of drawing a red ball (R) from Box-1 (B1) = p(R | B1) = $\frac{4}{9}$

=> Probability of drawing a red ball (R) from Box-2 (B2) = p(R | B2) =  $\frac{2}{7}$

=> Probability of drawing a red ball (R) from Box-3 (B3) = p(R | B3) =  $\frac{3}{8}$

=> Probability of getting the ball from Box-1 if red ball is selected is given by - \(P(B1 | R) = \frac{P(B1) \times p(R | B1)}{P(B1) \times p(R | B1)+P(B2) \times p(R | B2)+P(B3) \times p(R | B3)} = \frac{\frac{1}{3} \times \frac{4}{9}}{\frac{1}{3} \times \frac{4}{9} + \frac{1}{3} \times \frac{2}{7} + \frac{1}{3} \times \frac{3}{8}} = \frac{224}{557}\) ................ (1)

Now let us consider white ball is drawn second time.

=> Probability of drawing a white ball (W) from Box-1 = p(W | B1) =  $\frac{2}{8} = \frac{1}{4}$ (Note - one ball is taken out already from Box-1)

=> Probability of drawing a white ball (W) from Box-2 = p(W | B2) = $\frac{3}{7}$

=> Probability of drawing a white ball (W) from Box-3 = p(W | B3) = $\frac{4}{8} = \frac{1}{2}$

=> Probability of getting the ball from Box−1 if white ball is selected is given by - \(P(B1 | W) = \frac{P(B1) \times p(W | B1)}{P(B1) \times p(W | B1)+P(B2) \times p(W | B2)+P(B3) \times p(W | B3)} = \frac{\frac{1}{3} \times \frac{1}{4}}{\frac{1}{3} \times \frac{1}{4} + \frac{1}{3} \times \frac{3}{7} + \frac{1}{3} \times \frac{1}{2}} = \frac{7}{33}\) ..........(2)

• Hence probability of getting the balls from Box-1 if red ball selected first time and white ball second time = $\frac{224}{557} \times \frac{7}{33} = \frac{1568}{18381}$ ​ ............................................(3)

Let us consider white ball is drawn the first time.

=> Probability of drawing a white ball (W) from Box-1 = p(W | B1) =  $\frac{2}{9}$

=> Probability of drawing a white ball (W) from Box-2 = p(W | B2) =  $\frac{3}{7}$

=> Probability of drawing a white ball (W) from Box-3 = p(W | B3) =  $\frac{4}{8} = \frac{1}{2}$

=> Probability of getting the ball from Box−1 if white ball is selected is given by - \(P(B1 | W) = \frac{P(B1) \times p(W | B1)}{P(B1) \times p(W | B1)+P(B2) \times p(W | B2)+P(B3) \times p(W | B3)} = \frac{\frac{1}{3} \times \frac{2}{9}}{\frac{1}{3} \times \frac{2}{9} + \frac{1}{3} \times \frac{3}{7} + \frac{1}{3} \times \frac{1}{2}} = \frac{28}{145}\) ..........(4)

Now let us consider red ball is drawn second time.

=> Probability of drawing a red ball (R) from Box-1 (B1) = p(R | B1) =  $\frac{4}{8} = \frac{1}{2}$ (Note - one ball is taken out already from Box-1)

=> Probability of drawing a red ball (R) from Box-2 (B2) = p(R | B2) =  $\frac{2}{7}$

=> Probability of drawing a red ball (R) from Box-3 (B3) = p(R | B3) =  $\frac{3}{8}$

=> Probability of getting the ball from Box-1 if red ball is selected is given by - \(P(B1 | R) = \frac{P(B1) \times p(R | B1)}{P(B1) \times p(R | B1)+P(B2) \times p(R | B2)+P(B3) \times p(R | B3)} = \frac{\frac{1}{3} \times \frac{1}{2}}{\frac{1}{3} \times \frac{1}{2} + \frac{1}{3} \times \frac{2}{7} + \frac{1}{3} \times \frac{3}{8}} = \frac{28}{65}\) ................ (5)

Hence probability of getting the balls from Box-1 if white ball selected first time and red ball second time =  $\frac{28}{145} \times \frac{28}{65} = \frac{784}{9425}$ ​ ...........................................................(6)

Using (3) and (6), probability of getting two balls from Box-2 if one of then red and other one is white = \(\frac{1568}{18381} + \frac{784}{9425} = 0.0863 + 0.08418\) ~ 0.18