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

Let $E_1$ and $E_2$ be the events of selecting first bag and second bag respectively.
$\therefore P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2}$
Let $A$ be the event of getting a red ball.
$\Rightarrow P\left(A|E_1\right)=P\left(\text{drawing a red ball from first bag}\right)=\frac{4}{8}=\frac{1}{2}$
$\Rightarrow P\left(A|E_2\right)=P\left(\text{drawing a red ball from second bag}\right)=\frac{2}{8}=\frac{1}{4}$
The probability of drawing a ball from the first bag, given that it is red, is given by $P\left(E_2|A\right)$.
By using Baye's theorem, we obtain
$P\left(E_1|A\right)=\frac{P\left(E_1\right)\cdot P\left(A|E_1\right)}{P\left(E_1\right)\cdot P\left(A|E_1\right)+P\left(E_2\right)\cdot P\left(A|E_2\right)}$
$=\frac{\frac{1}{2}\cdot \frac{1}{2}}{\frac{1}{2}\cdot \frac{1}{2}+\frac{1}{2}\cdot \frac{1}{4}}$
$=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{8}}$
$=\frac{4}{\frac{3}{8}}$
$=\frac{2}{3}=0.66$