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A bag X  contains 4 white balls and 2 black balls, while another bag Y contains 3 white balls and 3 black balls. Two balls are drawn (without replacement ) at random from one of the bags and were found to be white and one black. Find the probability that the balls were drawn from bag Y.

 

 

 

 
 
 
 
 

Answers (1)

Let A be the events of drawing one white and one black ball from any one of the bag without replacement.

E_1  and  E_2  be the events of drawing bag X and Y resp. Then   P(E_1)=P(E_2)=\frac{1}{2}

P(A|E_1)=\frac{4}{6}\times \frac{2}{5}+\frac{2}{6}\times \frac{4}{5}=\frac{16}{30}    and 

P(A|E_2)=\frac{3}{6}\times \frac{3}{5}+\frac{3}{6}\times \frac{3}{5}=\frac{18}{30}

By Bayes theorem, 

P(E_2|A)=\frac{P(A|E_2)P(E_2)}{P(A|E_1)P(E_1)+P(A|E_2)P(E_2)}

                    =\frac{\frac{18}{30}\times \frac{1}{2}}{\frac{16}{30}\times \frac{1}{2}+\frac{18}{30}\times \frac{1}{2}}=\frac{18}{34}=\frac{9}{17}

 

 

 

Posted by

Ravindra Pindel

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