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,
\(P(E_1|A)=\frac{P(E_1).P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}\)
\(=\frac{\frac{1}{2}×\frac{1}{2}}{\frac{1}{2}×\frac{1}{2}+\frac{1}{2}×\frac{1}{4}}\)
\(=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{8}}\)
\(=\frac{1}{4}×\frac{8}{3}\)
\(=\frac{2}{3}\)