Question

A bag contains $4$ red and $4$ black balls, another bag contains $2$ red and $6$ black balls. One of the bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Answer 1

Let $E_1$ is the event of drawing from bag I and $E_2$ of drawing from bag II
and $A$ represent the event of drawing red ball
Probability of chosing one bag$=\frac{1}{2}$
or $P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2}$
There are $4$ red and $4$ black balls in bag I
Probability of drawing red ball from it
$=\frac{4}{8}=\frac{1}{2}$
$P\left(\frac{A}{E_1}\right)=\frac{1}{2}$
In second bag, there are $2$ red and $6$ black balls
Probability of drawing red ball from it
$P\left(\frac{A}{E_2}\right)=\frac{2}{8}=\frac{1}{4}$
Now probability of drawing red ball from first bat
$P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right).P\left(\frac{A}{E_1}\right)}{P\left(E_1\right)\times P\left(\frac{A}{E_1}\right)+P\left(E_2\right)\times P\left(\frac{A}{E_2}\right)}=\frac{\frac{1}{2}\times \frac{1}{2}}{\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{4}}=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{8}}=\frac{\frac{1}{4}}{\frac{3}{8}}=\frac{2}{3}$
Hence probability that ball is drawn from the first bag
$=\frac{2}{3}$