Question

A bag contains 4 red and 4 black balls,another bag contains 2 red and 6 black balls.One of the two 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

The correct answer is: $\frac{2}{3}$
Let $E_1$ and $E_2$ be the events of selecting first bag and second bag respectively.
$P(E_1)=P(E_2)=\frac{1}{2},$
Let A be the event of getting a red ball.
$P(A|E_1)$=$P$(drawing a red ball from bag 1)$=\frac{4}{8}=\frac{1}{2}$
P(A|E_2)=$$P(drawing a red ball from bag II)$=\frac{2}{8}+\frac{1}{4}$
The probability of drawing a ball from the first bag, given that it is red, is given by $P (E_2|A).$
Therefore,by Bayes'theorem,
$P(E_1|A)=\frac{P(E_1).P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}$
$=\frac{\frac{1}{2}×\frac{1}{2}}{\frac{1}{2}×\frac{1}{2}+\frac{1}{2}×\frac{1}{4}}$
$=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{8}}$
$=\frac{1}{4}×\frac{8}{3}$
$=\frac{2}{3}$