InterviewSolution
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InterviewSolution
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There are three urns containing 2 white and 3 black balls, 3 white and 2 black balls and 4 white and 1 black balls, respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn and it is found to be white. Find the probability that the ball drawn was from the second urn. |
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Answer» Let `U_(1)`={2 white, 3 black balls} `U_(2)`={3 white , 2 black balls} and `U_(3)`={4 white ,1 black balls} `thereforeP(U_(1))=P(U_(2))=P(U_(3))=1/3` Let `E_(1)` be the event that a ball is chosen from urn U, E be the event that a ball is chosen from urn `U_(1)` and `E_(2)` be the event that a ball is chosen from urn `U_(2)` and `E_(3)` be the event that a ball is chosen from urn `U_(3)`. Also, `P(E_(1))=P(E_(2))=P(E_(3))=1//3` Let E be the event that white ball is drawn `therefore P(E//E_(1))=2/5,P(E//E_(2))=3/5,P(E//E_(3))=4/5` Now, `P(E_(2)//E)=(P(E_(2))cdotP(E//E_(2)))/(P(E_(1))cdotP(E//E_(1))+P(E_(2))cdotP(E//E_(2))+P(E_(3))cdotP(E//E_(3)))` `=(1/3cdot3/5)/(1/3cdot2/5+1/3cdot3/5+1/3cdot4/5)` `=(3/15)/(2/15+3/15+4/15)=3/9=1/3` |
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