Extract balls from urn probability We have two urns, the first with 6 white balls and 7 black balls and the second with 10 white balls and 5 black balls.
We extract a ball from the first urn and introduce it into the second one, then we extract from the second urn 5 balls,reintroducing them back after each extraction.Whats the probability all the 5 balls are white? What scheme could be used here? Is it Poisson and if yes how to use it given the fact that theres and extraction with replacement?
 A: Case 1) We initially extracted a white ball with probability $\frac{6}{13}$. Then the second urn has $11$ white balls and $5$ black balls. The probability that all $5$ selected are white is then $$\left(\frac{11}{16}\right)^5$$
Case 2) We initially extracted a black ball with probability $\frac{7}{13}$. Then the second urn has $10$ white balls and $6$ black balls. The probability that all $5$ selected are white is then $$\left(\frac{10}{16}\right)^5$$
All together we get 
$$\left(\frac{6}{13}\cdot\left(\frac{11}{16}\right)^5\right)+\left(\frac{7}{13}\cdot\left(\frac{10}{16}\right)^5\right)\approx0.1222$$
A: (not an answer, just a comment)
I took probability last fall and got a B, so take this with a grain of salt, but when you see a question about picking balls from an urn without replacement, you likely want to think hypergeometric.
Good luck!
A: Lets consider: 
A : choosing a white ball from the first urn
B : choosing a black ball from the first urn
C : the probability you're looking
Then: p(A) = 6/13 and p(B) = 7/13
Then: p(C) = p(choose a white ball and extract 5 whites after) + p(choose a black ball and extract 5 whites after)= 
(6/13)* (11/16)^5 + (7/13)*(10/16)^5
