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                                    Um A contains 7 white and 3 black balls, um B contains 4 white and 6 black balls, um C contains 2 white and 8 black balls. One of these urns is chosen at random with probabilities 0.2, 0.6 and 0.2 respectively. From the chosen urn, two balls are drawn at random without replacement. Both the balls happen to be white. Find the probability that the balls drawn are from the urn C. | 
                            
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Answer»  Let E be the event that 2 white BALLS are DRAWN. Then, `P(E//A)=(.^7C_2)/(.^10C_2),P(E//B)=(.^4C_2)/(.^10C_2),P(E//C)=(.^2C_2)/(.^10C_2)` `:.` required probability `=P(C//E)=(P(E//C).P(C))/(P(E//A).P(A)+P(E//B).P(B)+P(E//C).P(C))`.  | 
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