What is the probability of getting 2 new balls out of 3 at the third time of picking 3 out of 12? Here is the question: (don't bother with the title if you don't get the question)
You are going to pick 3 balls out of 12 and put it back every time. The ball that gets picked is considered an old one. So what is the probability of getting 2 new balls when picking 3 balls at the third time?
My idea is that at the third time, there could be 3,4,5,6 balls picked before. Under each circumstances, it is a hypergeometric example. And I sum them up and get,
$$P(X=2 ;3|4|5|6)= \sum _{n=3}^6  \frac{ \binom{12-n}{2}\binom{n}{3-2}}{\binom{12}{3}} = \frac{83}{44} = 1.88636$$
which is definitely not the answer since probability can't be more than 1.
I think this is because I did not take conditional probability into account. How to figure that out? Also, is there another simpler way to solve this question? (I don't know if my solution is on the right track, either) Extension of this problem can be what is the probability of getting 2 new balls when picking 3 balls at the nth time, and would that be much more terrible and difficult?
 A: Your suspicion is right in that you didn't take the conditional probabilities into account. The numbers of new/old balls in each draw of $3$ balls do have hypergeometric conditional distributions. As you seem to have done, I'll define "success" in this distribution as getting an old ball. If $X_k$ is the number of old balls in the $k^{th}$ draw, then the probability we require is
\begin{eqnarray*}
P(X_3=1) &=& \sum_{i=0}^{3}{P(X_2=i) P(X_3=1\mid X_2=i)}\qquad\qquad\text{($i=$ #old balls in $2^{nd}$ draw)} \\
&& \\
&=& \sum_{i=0}^{3}{\dfrac{\binom{3}{i} \binom{9}{3-i}}{\binom{12}{3}} \dfrac{\binom{6-i}{1} \binom{6+i}{2}}{\binom{12}{3}}} \\
&& \\
&=& \dfrac{1377}{3025} \approx 0.455.
\end{eqnarray*}
A: Well, from your equation, I notice that it is only increasing, which can't be right, since the probability should keep going down. So, instead of having a constant denominator, there needs to be some variable there. Also, if I checked this right, there should be a 0 probability of getting a "new" ball on your 6th try.
I would say that you actually have 10 balls to choose from after you draw your first 3, since you're allowed to pick one of your old balls. Then, after that, you've added 2 more "old" balls. So, you only have 8 choices. Keep going, you've got 6, 4, 2, and finally 0.  So, yeah, it is much simpler than you thought originally. You should be taking the probability of getting 2 new balls on the first pulls and then multiplying it by the probability of getting 2 more new balls on the second, and so on.
I got $\frac{84}{6655}$ as the probability. Do you know what the correct answer is?
