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.
Best Answer
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.