Question

A bag X contains 4 white balls and 2 black balls, while another bag Y contains 3 white balls and 3 black balls. Two balls are drawn (without replacement) at random from one of the bags and were found to be one white and one black. Find the probability that the balls were drawn from bag Y.

Answer 1

Let us consider the events as:
$E_1:$ Bag X is selected
$E_2:$ Bag Y is selected

$A$ : Drawn balls are 1 Black and 1 white

$P\left(E_1\right)=\frac{1}{2}$ and $P\left(E_2\right)=\frac{1}{2}$

Here, $P\left(\frac{A}{E_1}\right)=\frac{{}^4{C}_1{\times }^2{C}_1}{{}^6{C}_2}=\frac{4\times 2}{15}=\frac{8}{15}$

$P\left(\frac{A}{E_2}\right)=\frac{{}^3{C}_1{\times }^3{C}_1}{{}^6{C}_2}=\frac{9}{15}$

Now, we have to find $P\left(\frac{E_2}{A}\right)$

Using Bayes theorem

$P\left(\frac{E_2}{A}\right)=\frac{P\left(E_2\right)\times P\left(\frac{A}{E_2}\right)}{P\left(E_1\right)\times P\left(\frac{A}{E_1}\right)+P\left(E_2\right)\times P\left(\frac{A}{E_2}\right)}$

$\therefore$ Required probability

$=\frac{\frac{1}{2}.\frac{9}{15}}{\frac{1}{2}\times \frac{8}{15}+\frac{1}{2}\times \frac{9}{15}}=\frac{\frac{9}{30}}{\frac{8}{30}+\frac{9}{30}}$

$=\frac{9}{17}$