Q. 43.7( 3 Votes )

The contents of the three urns are as follows:

Urn I contains 7 white and 3 black balls

Urn II contains 4 white and 6 black balls

Urn III contains 2 white and 8 black balls.

One of these urns are chosen at random probabilities 0.20,0.60 and 0.20 respectively. From the chosen urn two balls are drawn without replacement. If both these balls are white, what is the probability that these came from urn 3?

Answer :

Given:


Urn I has 7 white and 3 black balls


Urn II has 4 white and 6 black balls


Urn III has 2 white and 8 black balls


Let us assume U1, U2, U3 and A be the events as follows:


U1 = choosing Urn I


U2 = choosing Urn II


U3 = choosing Urn III


A = choosing 1 white and 1 red ball from urn


From the problem,





The Probability of choosing balls from each Urn differs from Urn to Urn and the probabilities are as follows:


P(A|U1) = P(Choosing required balls from Urn 1)






P(A|U2) = P(Choosing required balls from Urn 2)






P(A|U3) = P(Choosing required balls from Urn 3)






Now we find


P(U3|A) = P(The chosen balls are from Urn3)


Using Baye’s theorem:







The required probabilities is .


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