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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?

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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

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

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  • $\begingroup$ 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 $\endgroup$ – Lola Feb 7 '18 at 21:48
  • $\begingroup$ Yep, that exactly! $\endgroup$ – Remy Feb 7 '18 at 22:12
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(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!

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    $\begingroup$ You should write this as a comment then. $\endgroup$ – Aqua Feb 7 '18 at 18:47
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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

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