Question

A bag $A$ contains $2$ white and $3$ red balls and a bag $B$ contains $4$ white and $5$ red balls. One ball is drawn at random from one of the bags and is found to be red. The probability that it was drawn from bag $B$ is $\frac{25}{X}$, then find the value of X.

Answer 1

Let $E_1$ be the event that the ball is drawn from bag $A$, $E_2$ the event that it is drawn from bag $B$ and $E$ that the ball is red.

We have to find $P(E_2\mid E).$
Since both the bags are equally likely to be selected, we have
$P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2}.$

Also $P\left(E/E_1\right)=\frac{3}{5}$ and $P\left(E/E_2\right)=\frac{5}{9}$.

Hency by Baye's theorem, we have
$P\left(E_2/E\right)=\frac{P\left(E_2\right)P\left(E/E_2\right)}{P\left(E_1\right)P\left(E/E_1\right)+P\left(E_2\right)P\left(E/E_2\right)}$

$=\frac{\frac{1}{2}.\frac{5}{9}}{\frac{1}{3}.\frac{3}{5}+\frac{1}{2}.\frac{5}{9}}=\frac{25}{52}.$

Now comparing the given answer with $\frac{25}{X}$

$\therefore X=52$

Note. If in the problem, you read' is found to be' then it is the problem on Baye's Theorem.