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Urn A contain 1 white, 2 black, 3 red balls. Urn B contain 2 white, 1 black, 1 red balls. Urn C contains 4 white, 5 black, 3 red balls. Two balls are drawn from one of the Urn and found to be one white and one red. Find the probabilities that they come from Urns A, B or C.

Urn A contain 1 white, 2 black, 3 red balls. Urn B contain 2 white, 1 black, 1 red balls. Urn C
contains 4 white, 5 black, 3 red balls. Two balls are drawn from one of the Urn and found to be one
white and one red. Find the probabilities that they come from Urns A, B or C.

Grade:Upto college level

2 Answers

Saurabh Koranglekar
askIITians Faculty 10335 Points
4 years ago
Dear student

Since all the balls are drawn from the given sample space

the probability is 1

Regards
satyam suman
15 Points
3 years ago
Let E1,E2,E3 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(E1)=P(E2)=P(E3)=13.
P(E/E1)= probability that the balls drawn are one white and one red, given that the balls are from urn A
=.1C1×.3C1.6C2=315=15.
P(E/E2)= probability that the drawn are one white and one red,given that the balls are from urn B
=.2C1×.1C1.4C2=26=13
P(E/E3)= probability that the balls drawn are one white and one red, given that the balls are from urn C
=.4C1×.3C1.12C2=1266=211
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/E1).P(E1)P(E/E1).P(E1)+P(E/E2).P(E2)+P(E/E3).P(E3)[by Bayes's theorem]
(15×13)(15×13)+(13×13)+(211×13)
=(115×495118)=33/118.
Hence, the required probability is 33/118.

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