Three urns contain 2 white and 3 black balls; 3 white and 2 black balls, and 4 white and 1 black ball respectively. One ball is drawn from an urn chosen at random, and it was found to be white. Find the probability that it was drawn from the first urn.

Home » Three urns contain 2 white and 3 black balls; 3 white and 2 black balls, and 4 white and 1 black ball respectively. One ball is drawn from an urn chosen at random, and it was found to be white. Find the probability that it was drawn from the first urn.

Three urns contain 2 white and 3 black balls; 3 white and 2 black balls, and 4 white and 1 black ball respectively. One ball is drawn from an urn chosen at random, and it was found to be white. Find the probability that it was drawn from the first urn.

let $\mathrm{A}:$ Ball drawn from bag $\mathrm{A}$
$B:$ Ball is drawn from bag $B$
$C:$ Ball is drawn from bag $C$
BB : Black ball
WB : White ball
Assuming, selecting bags is of equal probability i.e. $\frac{1}{3}$
We want to find $P(A \mid W)$ , i.e. probability of selected White ball is from bag A
$\begin{array}{l} \mathrm{P}(\mathrm{A} \mid \mathrm{W})=\frac{\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{W} \mid \mathrm{A})}{\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{W} \mid \mathrm{A})+\mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{W} \mid \mathrm{B})+\mathrm{P}(\mathrm{C}) \cdot \mathrm{P}(\mathrm{W} \mid \mathrm{C})} \\ =\frac{\left(\frac{1}{3}\right)\left(\frac{2}{5}\right)}{\left(\frac{1}{3}\right)\left(\frac{2}{5}\right)+\left(\frac{1}{3}\right)\left(\frac{3}{5}\right)+\left(\frac{1}{3}\right)\left(\frac{4}{5}\right)} \\ =\frac{2}{9} \end{array}$
Therefore, the probability of selected white ball is from bag A is $\frac{2}{9}$