SOLUTION: Urn A contains five white balls and three black balls. Urn B contains seven white balls and six black balls. A ball is drawn from Urn A and then transferred to Urn B. A ball is the

Question 1152683: Urn A contains five white balls and three black balls. Urn B contains seven white balls and six black balls. A ball is drawn from Urn A and then transferred to Urn B. A ball is then drawn from Urn B. What is the probability that the transferred ball was black given that the second ball drawn was black?
Answer by ikleyn(52788) (Show Source): You can put this solution on YOUR website!
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The conditional probability under the question, by the definition, is the ratio of two probabilities

                                   
                                                                                    P(the transfered ball was black)
    P(the transfered ball was black given that the second drawn ball was black) = -------------------------------------   (1)
                                                                                    P(the second drawn ball was black)


The numerator is  P(the transfered ball was black)) =  = .


The denominator is the sum of probabilities of two disjoint events:  

    Event A = (the transfered ball was white and the second drawn ball was black)  and 

    Event B = (the transfered ball was black and the second drawn ball was black.)


In case of event A, urn B after transfer contains 8 white and 6 black balls; therefore

    P(Event A) =  =  = .


In case of event B, urn B after transfer contains 7 white and 7 black balls; therefore

    P(Event B) =  =  = .


Thus the denominator in (1) is  P(the second drawn ball was black) =  = .


Then according formula (1)

    P(the transfered ball was black given that the second drawn ball was black) =  =  = .


ANSWER.  The probability under the question is  .

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