Q. 225.0( 1 Vote )

# Bag I contains 1

Answer :

let P(B1), P(B2) and P(B3) be the probability of choosing Bag I, Bag II and Bag III respectively

As there are three bags probability of choosing one bag would be P(B1) = P(B2) = P(B3) = …(a)

Let A be the event of choosing 1 red and 1 white ball

We have to find P(B3|A) which means that probability of choosing bag III given that the ball chosen are 1 red and 1 white (that is event A has occurred)

By Bayes’ theorem Now we have to find P(A|B1), P(A|B2) and P(A|B3)

P(A|B1) means the probability of selecting 1 red and 1 white ball (with replacement) given that bag I is selected

Now bag I has 1 white, 2 black and 3 red balls total 6 balls

Probability of selecting 1 white ball will be and then we are putting that ball again in the bag I (this is the meaning of the term with replacement) and hence bag I now again contain total 6 balls hence probability of selecting 1 red ball will be  We multiplied by two because we have two cases either the first ball selected can be red or white (meaning first red then white OR first white then red)

Similarly writing P(A|B2) and P(A|B3)  Substitute values from equation (a), (b), (c) and (d) in (i) Cancel out and 2          Hence probability that one red and one white ball chosen is from bag III is Rate this question :

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