# 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?

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

$$\left(\frac{6}{13}\cdot\left(\frac{11}{16}\right)^5\right)+\left(\frac{7}{13}\cdot\left(\frac{10}{16}\right)^5\right)\approx0.1222$$

• what if they are without replacement?would that mean it would be something like ((6/13)*combinations of 11 taken 5)/(combinations of 16 taken 5) ? for first case
– Lola
Feb 7 '18 at 21:48
• Yep, that exactly!
– Remy
Feb 7 '18 at 22:12

(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!

• You should write this as a comment then.
– Aqua
Feb 7 '18 at 18:47

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