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I have a question about the following problem:

Consider a sample of size 3 drawn in the following manner: We start with an urn containing 5 white and 7 red balls. At each stage, a ball is drawn and its color is noted. The ball is then returned to the urn, along with an additional ball of the same color. Find the following probabilities: 0 white balls, 1 white ball, 3 white balls, and 2 white balls.

I believe the probabilities for white and red are 5/12 and 7/12. If there were 3 balls drawn, then that would mean that three additional balls for the same color are drawn as well, as noted above. So for instance, 0 white balls would mean RRRRRR = 5/12 * 5/12 * 5/12? Is that correct? I guess I'm confused on how to account for the additional ball of the same color being added.

Thanks in advance.

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  • $\begingroup$ No the probability of picking a red changes with each pick. So to get three reds it's $\frac7{12}\cdot\frac8{13}\cdot\frac9{14}$. $\endgroup$ – Gregory Grant Feb 22 '16 at 20:30
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No the probability of picking a red changes with each pick. So to get three reds it's $\frac7{12}\cdot\frac8{13}\cdot\frac9{14}$.

It's harder to compute the probability of two reds, because you have to break it up into the two cases of picking a red or picking a white at each step.

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  • $\begingroup$ So for RRW, would it be 7/12 * 8/13 * 5/14 ? $\endgroup$ – Slae Feb 22 '16 at 20:33
  • $\begingroup$ @Slae Yes but you have to consider all three possibilities RWR and WRR as well as RRW. Do the three separately and then add them up. $\endgroup$ – Gregory Grant Feb 22 '16 at 20:34
  • $\begingroup$ Ah, I was just thinking about the case if the order were to change. Makes sense now. $\endgroup$ – Slae Feb 22 '16 at 20:36
  • $\begingroup$ @Slae excellent! $\endgroup$ – Gregory Grant Feb 22 '16 at 20:37
  • $\begingroup$ and just to verify, the probability would be 5/39 regardless of order, and yet we would still have to add them all together correct? Making it 5/13 in the end? $\endgroup$ – Slae Feb 22 '16 at 20:41
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a)$\frac{7}{12}\cdot\frac{8}{13}\cdot\frac{9}{14}=\frac{3}{13}$

b)$\frac{3\cdot(5\cdot6\cdot7)}{12\cdot13\cdot14}=\frac{5}{13}$

c)$\frac{5\cdot6\cdot7}{12\cdot13\cdot14}=\frac{5}{52}$

d)$\frac{3\cdot(5\cdot6\cdot7)}{12\cdot13\cdot14}=$

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  • $\begingroup$ This is not really very helpful without explaining why you did what you did. $\endgroup$ – The Count Feb 12 '17 at 18:14
  • $\begingroup$ Please write clearer your answer! $\endgroup$ – Kal S. Feb 12 '17 at 18:30

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