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 $U_1$.
closed as off-topic by Parcly Taxel, Davide Giraudo, Shailesh, iadvd, Jack's wasted life Oct 29 '16 at 4:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Parcly Taxel, Davide Giraudo, Shailesh, iadvd, Jack's wasted life
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.
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).