# 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.

This question has appeared in model paper 2012

Toolbox:
• Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability P(E_i|A) for any event A associated w/ $E_i$ using the Bayes theorem as follows: $$\;P(E_i/A)$$=$$\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}$$
Let $E_1$ be the event that the first bag is selected, and $E_2$ the event that the second bag is selected.
Since $E_1$ and $E_2$ are mutually exclusive and exhaustive, $P(E_1) = P(E_2) = \large\frac{1}{2}$
Let A be the event of drawing a red ball. Given that Bag1 has 4 red and 4 black balls, n(Bag1) = 8, and n(R1) = 4. Similarly for Bag2, n(Bag2) = 8 and n(R2) = 2.
$\Rightarrow$ P (drawing a red from Bag 1) = $\large \frac{4}{8} = \frac{1}{2}$ =$P (A|E_1)$
$\Rightarrow$ P (drawing a red from Bag 2) = $\large \frac{2}{8} = \frac{1}{4}$ = $P(A|E_2)$
To find the probability of that the ball is drawn from the first bag, let's use Bayes theorem: $$\;P(E_i/A)$$=$$\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}$$
Therefore the probability that the ball is drawn from the first bag = $P(E_1|A) = \large\frac{P(E_1)(P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)+P(A|E_2)}$
$P(E_1)(P(A|E_1) = \large\frac{1}{2}$$\times$$ \large \frac{1}{2} = \frac{1}{4}$
$P(E_1)P(A|E_1) + P(E_2)+P(A|E_2) = \large\frac{1}{2}$$\times$$\large\frac{1}{2} + \frac{1}{2}$$\times$$\large\frac{1}{4} = \frac{2+1}{8} = \frac{3}{8}$
Therefore $P(E_1|A) = \Large \frac{\large \frac{1}{4}}{\large \frac{3}{8}}$$=\large \frac{2}{3}$