Question

A fair die is rolled. If face $1$ turns up, a ball is drawn from Bag $A$. If face $2$ or $3$ turns up, a ball is drawn from Bag $B$. If face $4$ or $5$ or $6$ turns up, a ball is drawn from Bag $C$. Bag $A$ contains $3$ red and $2$ white balls, Bag $B$ contains $3$ red and $4$ white balls and Bag $C$ contains $4$ red and $5$ white balls. The die is rolled, a Bag is picked up and a ball is drawn. If the drawn ball is red; what is the probability that it is drawn from Bag $B$?

Answer 1

Probability that ball is drawn from bag $A$ $=P\left(E_1\right)=\frac{1}{6}$

Probability that ball is drawn from bag $B$ $=P\left(E_2\right)=\frac{2}{6}=\frac{1}{3}$

Probability that ball is drawn from bag $C$ $=P\left(E_3\right)=\frac{3}{6}=\frac{1}{2}$

let $P\left(A\right)$ be the probability of drawing a red ball

Probability that it is drawn from bag $A$ $P\left({A|E}_1\right)=\frac{3}{5}$

Probability that it is drawn from bag $B$ $=P\left({A|E}_2\right)=\frac{3}{7}$

Probability that its is drawn from bag $C$ $=P\left({A|E}_3\right)=\frac{4}{9}$

Now we have to find $P\left(E_2|A\right)$ that is it is drawn from bag $B$

Using Baye's theorem

$P\left(E_2|A\right)=\frac{P\left(E_2\right)P\left({A|E}_2\right)}{P\left(E_1\right)P\left({A|E}_1\right)+P\left(E_2\right)P\left({A|E}_2\right)+P\left(E_3\right)P\left({A|E}_3\right)}P\left(E_2|A\right)=\frac{\frac{1}{3}\times \frac{3}{7}}{\frac{1}{6}\times \frac{3}{5}+\frac{1}{3}\times \frac{3}{7}+\frac{1}{2}\times \frac{4}{9}}P\left(E_2|A\right)=\frac{\frac{1}{7}}{\frac{1}{10}+\frac{1}{7}+\frac{2}{9}}=\frac{\frac{1}{7}}{\frac{293}{630}}=\frac{90}{293}$