Suppose that we have two bags each containing black and white balls. The first b

Question

Suppose that we have two bags each containing black and white balls. The first bag contains 4 white balls and 1 black ball. The second bag contains 1 white ball and 4 black balls. An experiment consists of randomly choosing a bag and randomly picking one of the balls in that bag. Let us say that the ball picked is white. What is the probability that it came from the bag with more white balls? [Hint: Let A be the event that the first bag is chosen and let B be the event that a white ball is chosen. We need to compute P(A I B). From the data given,what is P(A), P( ), P(B I A), P(B I )? Now, if we know all these probabilities, how would you compute P(A I B)?]

Explanation / Answer

P(A/B) means probability of happening of A given B happenend.

This is conditional probability which can be re-written as

P(A/B) = (P(A n B)/P(B)) ----[1]

Now consider P(B/A) = probability that the ball is white , given it is picked from the bag with more white balls = no. of white balls/total no. of balls in that bag = 4/(4+1) = 4/5

also P(B/A) = P(A n B)/P(A) ==>  P(A n B) = P(B/A) x P(A) = P(A) x 4/5

So, [1] ==> P(A/B) = (4/5) x (P(A)/P(B))

P(A) = probability that 1st bag is choosen = 1/total number of bags = 1/2

P(B) = probability that white ball is choosen = (probability that 1st bag choosen)x(probability that white ball is choosen) + (Probability that 2nd bag is choosen)x(probability that white ball is choosen) = (1/2)x(4/5)+(1/2)x(1/5) = 1/2

==> P(A/B) = (4/5) x (P(A)/P(B)) = (4/5) x ((1/2)/(1/2)) = 4/5

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