Three bags contain 3 red, 7 black; 8 red, 2 black, and 4 red & 6 black balls respectively. 1 of the bags is selected at random and a ball is drawn from it. If the ball drawn is red, find the probability that it is drawn from the third bag.

Let E1, E2, E3 and A are the events defined as follows.
E1 = First bag is chosen
E2 = Second bag is chosen
E3 = Third bag is chosen
A = Ball drawn is red
Since there are three bags and one of the bags is chosen at random, so P (E1) = P(E2) = P(E3) = 1 / 3
If E1 has already occurred, then first bag has been chosen which contains 3 red and 7 black balls. The probability of drawing 1 red ball from it is 3/10. So, P (A/E_{1}) = 3/10, similarly P(A/E_{2}) = 8/10, and P(A/E_{3}) = 4/10. We are required to find P(E_{3}/A) i.e. given that the ball drawn is red, what is the probability that the ball is drawn from the third bag by Baye’s rule

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