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Urn A contains 1 white, 2 blackand 3 red balls, urn B contains 2 white, 1 black and 1 red balls, and urn C contains 4 white , 5 blackand 3 red balls. One urn is chosen at random and two balls are drawn. These happen to be one white and one red. What is the probability that they come from urn A? |
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Answer» SOLUTION :Let `E_1,E_2,E_3` be the EVENTS that the balls are drawn from urn A, urn B and urn C respectively, and let E be the event that the balls drawn are one WHITE and one RED. Then, `P(E_1)=P(E_2)=P(E_3)=1/3`. `P(E//E_1)`= probability that the balls drawn are one white and one red, giventhat the balls are from urn A `=(.^1C_1xx.^3 C_1)/(.^6C_2)=3/15=1/5`. `P(E//E_2)`= probability that the drawn are one white and one red,given that the balls are from urn B `=(.^2C_1xx.^1C_1)/(.^4C_2)=2/6=1/3` `P(E//E_3)`= probability that the balls drawn are one white and one red, given that the balls are from urn C `=(.^4C_1xx.^3C_1)/(.^12C_2)=12/66=2/11` Probability that the balls drawn are from urn A, it being given that the balls drawn are one white and one red `=P(E-1//E)` `(P(E//E_1).P(E_1))/(P(E//E_1).P(E_1)+P(E//E_2).P(E_2)+P(E//E_3).P(E_3))`[by Bayes's theorem] `((1/5xx1/3))/((1/5xx1/3)+(1/3xx1/3)+(2/11xx1/3))` `=(1/15xx495/118)=33/118`. Hence, the required probability is `33/118`. |
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