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A bag contains 4 red balls and 4 black balls. Another bag contains 2 red and 6 black. One of the 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.

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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: \(\;P(E_i/A)\)=\(\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}\)
Step 1:
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) = $\large \frac{2}{8} = \frac{1}{4}$ = $P(A/E_2)$
Step 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}$$\times \large\frac{1}{4}$
$\qquad\qquad\qquad\qquad\qquad\qquad\;\;= \large\frac{2+1}{8}$
$\qquad\qquad\qquad\qquad\qquad\qquad\;\;=\large \frac{3}{8}$
Therefore $P(E_1/A) = \large \frac{\Large \frac{1}{4}}{\Large \frac{3}{8}}$$ =\large \frac{2}{3}$
answered Oct 3, 2013 by sreemathi.v

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