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

We can find the probability of selecting each box. Then we can consider 2 cases, one with the red ball is selected first and then the white ball and the second with white ball is selected first and then the red ball. We can find the probability of the 2 cases separately using conditional probability and Bayes’ theorem. For that we find the probability of selecting one colour ball at a time and take their product. Then we can take the sum of the probabilities of the 2 cases to get the required probability.

Complete step-by-step answer:
It is given that there are 3 boxes. Let the boxes be ${B_1}$ , ${B_2}$ and ${B_3}$ . The probability of selecting one box from the 3 will be equal.
So, the probability of selecting the 1st box is given by,
$\Rightarrow P\left( {{B_1}} \right) = \dfrac{1}{3}$
Similarly, the probability of selecting the 2nd box and 3rd box are given by,
$\Rightarrow P\left( {{B_2}} \right) = \dfrac{1}{3}$
$\Rightarrow P\left( {{B_3}} \right) = \dfrac{1}{3}$
Now we can consider the case where the first ball drawn is red and the second ball is white.
So, we can find the probability that the red ball was selected from the second box.
In the first box, there is 1 white ball, 2 red balls and 3 black balls.
So, the probability of selecting a red ball from it is given by the number of red balls in the box divided by the total number of balls in the box.
$\Rightarrow P\left( {\dfrac{R}{{{B_1}}}} \right) = \dfrac{2}{{1 + 2 + 3}}$
$\Rightarrow P\left( {\dfrac{R}{{{B_1}}}} \right) = \dfrac{2}{6}$
The second box has 2 white, 3 red and 1 black ball. Then the probability of selecting red ball from it is given by,
$\Rightarrow P\left( {\dfrac{R}{{{B_2}}}} \right) = \dfrac{3}{{2 + 3 + 1}}$ +
$\Rightarrow P\left( {\dfrac{R}{{{B_2}}}} \right) = \dfrac{3}{6}$
The third box has 3 white, 1 red and 2 black balls. Then the probability of selecting red ball from it is given by,
$\Rightarrow P\left( {\dfrac{R}{{{B_3}}}} \right) = \dfrac{1}{{3 + 1 + 2}}$
$\Rightarrow P\left( {\dfrac{R}{{{B_3}}}} \right) = \dfrac{1}{6}$
Now the probability that the red ball was selected from the second box is given by, $P\left( {\dfrac{{{B_2}}}{R}} \right)$
We know that, by Bayes’ theorem,
$P\dfrac{{{B_2}}}{R} = \dfrac{{P\left( {\dfrac{R}{{{B_2}}}} \right)P\left( {{B_2}} \right)}}{{P\left( {\dfrac{R}{{{B_1}}}} \right)P\left( {{B_1}} \right) + P\left( {\dfrac{R}{{{B_2}}}} \right)P\left( {{B_2}} \right) + P\left( {\dfrac{R}{{{B_3}}}} \right)P\left( {{B_3}} \right)}}$
On substituting the values, we get,
$\Rightarrow P\left( {\dfrac{{{B_2}}}{R}} \right) = \dfrac{{\dfrac{3}{6} \times \dfrac{1}{3}}}{{\dfrac{2}{6} \times \dfrac{1}{3} + \dfrac{3}{6} \times \dfrac{1}{3} + \dfrac{1}{6} \times \dfrac{1}{3}}}$
On simplification we get,
$\Rightarrow P\left( {\dfrac{{{B_2}}}{R}} \right) = \dfrac{3}{{2 + 3 + 1}}$
$\Rightarrow P\left( {\dfrac{{{B_2}}}{R}} \right) = \dfrac{3}{6}$
$\Rightarrow P\left( {\dfrac{{{B_2}}}{R}} \right) = \dfrac{1}{2}$
Therefore, the probability that the red ball was selected from the second box is $\dfrac{1}{2}$
Now we can find the probability that the white ball was selected 2nd time from the second box
In the first box, there is 1 white ball, 2 red balls and 3 black balls.
So, the probability of selecting white ball from it is given by the number of white balls in the box divided by the total number of balls in the box.
$\Rightarrow P\left( {\dfrac{W}{{{B_1}}}} \right) = \dfrac{1}{{1 + 2 + 3}}$
$\Rightarrow P\left( {\dfrac{W}{{{B_1}}}} \right) = \dfrac{1}{6}$
After drawing the 1st ball red from the second box, it has 2 white, 2 red and 1 black balls. Then the probability of selecting white ball from it is given by,
$\Rightarrow P\left( {\dfrac{W}{{{B_2}}}} \right) = \dfrac{2}{{2 + 2 + 1}}$
$\Rightarrow P\left( {\dfrac{W}{{{B_2}}}} \right) = \dfrac{2}{5}$
The third box has 3 white, 1 red and 2 black balls. Then the probability of selecting white ball from it is given by,
$\Rightarrow P\left( {\dfrac{W}{{{B_3}}}} \right) = \dfrac{3}{{3 + 1 + 2}}$
$\Rightarrow P\left( {\dfrac{W}{{{B_3}}}} \right) = \dfrac{3}{6}$
Now the probability that the white ball was selected from the second box is given by, $P\left( {\dfrac{{{B_2}}}{W}} \right)$
We know that, by Bayes’ theorem,
$P\left( {\dfrac{{{B_2}}}{W}} \right) = \dfrac{{P\left( {\dfrac{W}{{{B_2}}}} \right)P\left( {{B_2}} \right)}}{{P\left( {\dfrac{W}{{{B_1}}}} \right)P\left( {{B_1}} \right) + P\left( {\dfrac{W}{{{B_2}}}} \right)P\left( {{B_2}} \right) + P\left( {\dfrac{W}{{{B_3}}}} \right)P\left( {{B_3}} \right)}}$
On substituting the values, we get,
$\Rightarrow P\left( {\dfrac{{{B_2}}}{W}} \right) = \dfrac{{\dfrac{2}{5} \times \dfrac{1}{3}}}{{\dfrac{1}{6} \times \dfrac{1}{3} + \dfrac{2}{5} \times \dfrac{1}{3} + \dfrac{3}{6} \times \dfrac{1}{3}}}$
On simplification we get,
$\Rightarrow P\left( {\dfrac{{{B_2}}}{W}} \right) = \dfrac{{\dfrac{2}{5}}}{{\dfrac{1}{6} + \dfrac{2}{5} + \dfrac{3}{6}}}$
$\Rightarrow P\left( {\dfrac{{{B_2}}}{W}} \right) = \dfrac{{\dfrac{{12}}{{30}}}}{{\dfrac{5}{{30}} + \dfrac{{12}}{{30}} + \dfrac{{15}}{{30}}}}$
$\Rightarrow P\left( {\dfrac{{{B_2}}}{W}} \right) = \dfrac{{12}}{{32}}$
$\Rightarrow P\left( {\dfrac{{{B_2}}}{W}} \right) = \dfrac{3}{8}$
Therefore, the probability that the white ball was selected from the second box is $\dfrac{3}{8}$
Hence probability of getting the balls from second box if red ball selected first time and white ball second time $\dfrac{3}{8} \times \dfrac{1}{2} = \dfrac{3}{{16}}$ … (1)
Case 2:

Now we can consider the case where the first ball drawn is white and the second ball is red.
So, we can find the probability that the white ball was selected from the second box.

Now we can find the probability that the white ball was selected 2nd time from the second box
In the first box, there is 1 white ball, 2 red balls and 3 black balls.
So, the probability of selecting white ball from it is given by the number of white balls in the box divided by the total number of balls in the box.
$\Rightarrow P\left( {\dfrac{W}{{{B_1}}}} \right) = \dfrac{1}{{1 + 2 + 3}}$
$\Rightarrow P\left( {\dfrac{W}{{{B_1}}}} \right) = \dfrac{1}{6}$
The second box has 2 white, 3 red and 1 black balls. Then the probability of selecting white ball from it is given by,
$\Rightarrow P\left( {\dfrac{W}{{{B_2}}}} \right) = \dfrac{2}{{2 + 3 + 1}}$
$\Rightarrow P\left( {\dfrac{W}{{{B_2}}}} \right) = \dfrac{2}{6}$
The third box has 3 white, 1 red and 2 black balls. Then the probability of selecting white ball from it is given by,
$\Rightarrow P\left( {\dfrac{W}{{{B_3}}}} \right) = \dfrac{3}{{3 + 1 + 2}}$
$\Rightarrow P\left( {\dfrac{W}{{{B_3}}}} \right) = \dfrac{3}{6}$
Now the probability that the white ball was selected from the second box is given by, $P\left( {\dfrac{{{B_2}}}{W}} \right)$
We know that, by Bayes’ theorem,
$P\left( {\dfrac{{{B_2}}}{W}} \right) = \dfrac{{P\left( {\dfrac{W}{{{B_2}}}} \right)P\left( {{B_2}} \right)}}{{P\left( {\dfrac{W}{{{B_1}}}} \right)P\left( {{B_1}} \right) + P\left( {\dfrac{W}{{{B_2}}}} \right)P\left( {{B_2}} \right) + P\left( {\dfrac{W}{{{B_3}}}} \right)P\left( {{B_3}} \right)}}$
On substituting the values, we get,
$\Rightarrow P\left( {\dfrac{{{B_2}}}{W}} \right) = \dfrac{{\dfrac{2}{6} \times \dfrac{1}{3}}}{{\dfrac{1}{6} \times \dfrac{1}{3} + \dfrac{2}{6} \times \dfrac{1}{3} + \dfrac{3}{6} \times \dfrac{1}{3}}}$
On simplification we get,
$\Rightarrow P\left( {\dfrac{{{B_2}}}{W}} \right) = \dfrac{2}{{1 + 2 + 3}}$

$\Rightarrow P\left( {\dfrac{{{B_2}}}{W}} \right) = \dfrac{2}{6}$
$\Rightarrow P\left( {\dfrac{{{B_2}}}{W}} \right) = \dfrac{1}{3}$
Therefore, the probability that the red ball was selected from the second box is $\dfrac{1}{3}$
So, we can find the probability that the red ball was selected 2nd time from the second box.
In the first box, there is 1 white ball, 2 red balls and 3 black balls.
So, the probability of selecting a red ball from it is given by the number of red balls in the box divided by the total number of balls in the box.
$\Rightarrow P\left( {\dfrac{R}{{{B_1}}}} \right) = \dfrac{2}{{1 + 2 + 3}}$
$\Rightarrow P\left( {\dfrac{R}{{{B_1}}}} \right) = \dfrac{2}{6}$
After drawing the 1st ball white from the second box, it has 2 white, 3 red and 1 black ball. Then the probability of selecting red ball from it is given by,
$\Rightarrow P\left( {\dfrac{R}{{{B_2}}}} \right) = \dfrac{3}{{1 + 3 + 1}}$
$\Rightarrow P\left( {\dfrac{R}{{{B_2}}}} \right) = \dfrac{3}{5}$
The third box has 3 white, 1 red and 2 black balls. Then the probability of selecting red ball from it is given by,
$\Rightarrow P\left( {\dfrac{R}{{{B_3}}}} \right) = \dfrac{1}{{3 + 1 + 2}}$
$\Rightarrow P\left( {\dfrac{R}{{{B_3}}}} \right) = \dfrac{1}{6}$
Now the probability that the red ball was selected from the second box is given by, $P\left( {\dfrac{{{B_2}}}{R}} \right)$
We know that, by Bayes’ theorem,
$P\left( {\dfrac{{{B_2}}}{R}} \right) = \dfrac{{P\left( {\dfrac{R}{{{B_2}}}} \right)P\left( {{B_2}} \right)}}{{P\left( {\dfrac{R}{{{B_1}}}} \right)P\left( {{B_1}} \right) + P\left( {\dfrac{R}{{{B_2}}}} \right)P\left( {{B_2}} \right) + P\left( {\dfrac{R}{{{B_3}}}} \right)P\left( {{B_3}} \right)}}$
On substituting the values, we get,
$\Rightarrow P\left( {\dfrac{{{B_2}}}{R}} \right) = \dfrac{{\dfrac{3}{5} \times \dfrac{1}{3}}}{{\dfrac{2}{6} \times \dfrac{1}{3} + \dfrac{3}{5} \times \dfrac{1}{3} + \dfrac{1}{6} \times \dfrac{1}{3}}}$
On simplification we get,
$\Rightarrow P\left( {\dfrac{{{B_2}}}{R}} \right) = \dfrac{{\dfrac{{18}}{{30}}}}{{\dfrac{{10}}{{30}} + \dfrac{{18}}{{30}} + \dfrac{5}{{30}}}}$
$\Rightarrow P\left( {\dfrac{{{B_2}}}{R}} \right) = \dfrac{{18}}{{33}}$
$\Rightarrow P\left( {\dfrac{{{B_2}}}{R}} \right) = \dfrac{6}{{11}}$
Therefore, the probability that the white ball was selected from the second box is $\dfrac{6}{{11}}$ .
Hence probability of getting the balls from second box if white ball selected first time and red ball second time is $\dfrac{1}{3} \times \dfrac{6}{{11}} = \dfrac{2}{{11}}$ … (2)
From (1) and (2) the probability that one red and one white ball are selected from the second box is given by taking their sum.
$\Rightarrow P = \dfrac{3}{{16}} + \dfrac{2}{{11}}$
$\Rightarrow P = \dfrac{{33 + 32}}{{176}}$
$\Rightarrow P = \dfrac{{65}}{{176}}$
Therefore, the required probability is $\dfrac{{65}}{{176}}$.

Note: We must consider both the cases to find the required probability. While calculating the probabilities using Bayes’ theorem, the substitutions must be done carefully. While calculating the conditional probability of selecting the 2nd ball from the second box, we must consider the ball we took first while calculating the total number of balls in that box. We must take the product for finding the total probability of each case. For calculating the required probability, we must take the sum of the probabilities of each case, not the product.