Question

There are 3 boxes, the first one containing 1 white, 2 red and 3 black balls; the second one containing 2 white, 3 red and 1 black ball and the third one containing 3 white, 1 red and 2 black balls. A box is chosen at random and from it two balls are drawn at random. One ball is red and the other, white. What is the probability that they come from the second box?

Answer 1

To solve this question, We have to 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 $=\frac{1}{3}$.

probabilty of drawing a red ball from box$-1=p\left(R|B_1\right)=\frac{1}{3}$

probabilty of drawing a red ball from box$-2=p\left(R|B_2\right)=\frac{1}{2}$

probabilty of drawing a red ball from box$-3=\left(R|B_3\right)=\frac{1}{6}$

probability of getting the ball from box$-2$ if red ball is selected is given by

$P\left(B_2|R\right)=\frac{P\left(B_2\right)\times p\left(R|B_2\right)}{P\left(B_1\right)\times p\left(R|B_1\right)+P\left(B_2\right)\times p\left(R|B_2\right)+P\left(B_3\right)\times p\left(R|B_3\right)}$

$=\frac{\frac{1}{3}\times \frac{1}{2}}{\frac{1}{3}\times \frac{1}{3}+\frac{1}{3}\times \frac{1}{2}+\frac{1}{3}\times \frac{1}{6}}$

$=\frac{\frac{1}{6}}{\frac{1}{9}+\frac{1}{6}+\frac{1}{18}}$

$=\frac{\frac{1}{6}}{\frac{2+3+1}{18}}$

$=\frac{1}{6}\times \frac{18}{6}=\frac{1}{2}$ ......$\left(1\right)$

Now let us consider white ball is drawn second time

probabilty of drawing a white ball from box$-1=p\left(W|B_1\right)=\frac{1}{6}$

probabilty of drawing a white ball from box$-2=p\left(W|B_2\right)=\frac{2}{5}$ ( note:- one ball is taken out already from box$-2$)

probabilty of drawing a white ball from box$-3=p\left(W|B_3\right)=\frac{1}{2}$

probability of getting the ball from box$-2$ if white ball is selected is given by

$P\left(B_2|W\right)=\frac{P\left(B_2\right)\times p\left(W|B_2\right)}{P\left(B_1\right)\times p\left(W|B_1\right)+P\left(B_2\right)\times p\left(W|B_2\right)+P\left(B_3\right)\times p\left(W|B_3\right)}$

$=\frac{\frac{1}{3}\times \frac{2}{5}}{\frac{1}{3}\times \frac{1}{6}+\frac{1}{3}\times \frac{2}{5}+\frac{1}{3}\times \frac{1}{2}}$

$=\frac{\frac{2}{15}}{\frac{1}{18}+\frac{2}{15}+\frac{1}{6}}$

$=\frac{3}{8}$ ......$\left(2\right)$

Hence probability of getting the balls from box$-2$ if red ball selected first time and white ball second time $=\frac{1}{2}\times \frac{3}{8}=\frac{3}{16}...................$(3)$

Now we consider the event, first selected ball is white and the second one red

probabilty of drawing a white ball from box$-1=p\left(W|B_1\right)=\frac{1}{6}$

probabilty of drawing a white ball from box$-2=p\left(W|B_2\right)=\frac{1}{3}$

probabilty of drawing a white ball from box$-3=p\left(W|B_3\right)=\frac{1}{2}$

probability of getting the ball from box$-2$ if white ball is selected is given by

$P\left(B_2|W\right)=\frac{P\left(B_2\right)\times p\left(W|B_2\right)}{P\left(B_1\right)\times p\left(W|B_1\right)+P\left(B_2\right)\times p\left(W|B_2\right)+P\left(B_3\right)\times p\left(W|B_3\right)}$

$=\frac{\frac{1}{3}\times \frac{1}{3}}{\frac{1}{3}\times \frac{1}{6}+\frac{1}{3}\times \frac{1}{3}+\frac{1}{3}\times \frac{1}{2}}$

$=\frac{\frac{1}{9}}{\frac{1}{18}+\frac{1}{9}+\frac{1}{6}}$

$=\frac{1}{3}$ ......$\left(4\right)$

Now let us consider red ball is drawn second time

probabilty of drawing a red ball from box$-1=p\left(R|B_1\right)=\frac{1}{3}$

probabilty of drawing a red ball from box$-2=p\left(R|B_2\right)=\frac{3}{5}$

probabilty of drawing a red ball from box$-3=p\left(R|B_3\right)=\frac{1}{6}$

probability of getting the ball from box$-2$ if red ball is selected second time is given by

$P\left(B_2|R\right)=\frac{P\left(B_2\right)\times p\left(R|B_2\right)}{P\left(B_1\right)\times p\left(R|B_1\right)+P\left(B_2\right)\times p\left(R|B_2\right)+P\left(B_3\right)\times p\left(R|B_3\right)}$

$=\frac{\frac{1}{3}\times \frac{3}{5}}{\frac{1}{3}\times \frac{1}{3}+\frac{1}{3}\times \frac{1}{5}+\frac{1}{3}\times \frac{1}{6}}$

$=\frac{\frac{3}{15}}{\frac{1}{9}+\frac{1}{5}+\frac{1}{18}}$

$=\frac{6}{11}$ ......$\left(5\right)$

Hence probability of getting the balls from box$-2$ if white ball is selected first time and red ball is selected second time $=\frac{1}{3}\times \frac{6}{11}=\frac{2}{11}$ ...................$\left(6\right)$

Using $\left(3\right)$ and $\left(6\right)$, probability of getting two balls from box$-2$ if one of then red and other one is white $=\frac{3}{16}+\frac{2}{11}=\frac{65}{176}=0.37$