Q. 15.0( 1 Vote )

The contents of the urns I,II,III are as follows:

Urn I: 1 white, 2 black and 3 red balls

Urn II: 2 white, 1 black and 1 red balls

Urn III: 4 white, 5 black and 3 red balls.

One urn is chosen at random and two balls are drawn. They happen to be white and red. What is the probability that they come from Urns I,II,III?

Answer :

Given:


Urn I has 1 white, 2 black and 3 red balls


Urn II has 2 white, 1 black and 1 red balls


Urn III has 4 white, 5 black and 3 red 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


We know that each urn is most likely to choose. So, probability of choosing a urn will be same for every Urn.





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(U1|A) = P(The chosen balls are from Urn1)


P(U2|A) = P(The chosen balls are from Urn2)


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


Using Baye’s theorem:

















The required probabilities are .


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