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There are two urns $U_1,U_2$. First urn contains $2$ white and $8$ black balls. Second urn contains $4$ white and $6$ black balls. If a urn is selected at random and a ball is drawn, its color is noted and replaced. This process is repeated $3$ times and as a result one ball of white color and $2$ balls of black colour are obtained. What is the probability that urn selected was $U_1$.

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closed as off-topic by Parcly Taxel, Davide Giraudo, Shailesh, iadvd, Jack's wasted life Oct 29 '16 at 4:22

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  • $\begingroup$ Thought about it , but got nothing $\endgroup$ – user123733 Oct 26 '16 at 12:27
  • $\begingroup$ If you had had the first urn, what would probability of the draw you observed have been before the draw? If you had had the second urn, what would probability of the draw you observed have been before the draw? $\endgroup$ – Henry Oct 26 '16 at 12:38
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Assuming that your question is We are selecting the Urn only at the start and all 3 picks were done from the same Urn. [Clarify if intent of last line is not same.]

Then it's simply a bayes' theorem problem.

Let events...

  E : 1 white and 2 black ball comes. [irrespective of the order]
  A : U1 is selected
  B : U2 is selected 

P(A) = 1/2.

P(B) = 1/2.

P(E|A) = 2/10 * 8/10 * 8*10. (say P1)

P(E|B) = 4/10* 6/10 * 6/10. (say P2)

P(E) = 1/2 * P1 + 1/2 * P2.

now you can work it out applying the bayes' theorem for P(A|E).

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  • $\begingroup$ How you wrote P(E l B) $\endgroup$ – user123733 Oct 27 '16 at 2:34
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    $\begingroup$ Given that you have selected urn U2, what is the probability of getting one white, it's 4/10. Similarly for a black it's 6/10, since you are replacing the ball in the Urn, next time as well the probability of black ball would be same. $\endgroup$ – Shiv Gupta Oct 27 '16 at 4:06

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