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- 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}$

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