So I'm just approaching probability theory and I have the following excercise:
We have a box with 10 white balls and 15 black balls. We extract the balls without reintroductions.
a) what is the probability that we extract 2 white balls and 1 black ball?
b) we extract 22 balls from the box. What is the probability that the remaining balls in the box are 2 white and 1 black?
For a) I think the answer is $\dfrac {10}{25}\dfrac {9}{24}\dfrac {15}{23}$ but I really have no idea how to do b). Can anyone help me out?
Best Answer
As other people have pointed out, choosing 3 balls to remove from the box, or choosing 3 balls to leave behind in the box, can be done in the same number of ways, so the answers to (a) and (b) should be equal. But let's check directly that they are the same. For convenience's sake, assume the balls are distinguishable (eg: numbered 1-10 on the white, and 11-25 on the black).
Part (a): There are $\binom{25}{3}$ ways to choose 3 balls out of 25. The number of ways to pick exactly 2 white and 1 black is $\binom{10}{2} * \binom{15}{1}$, so our overall answer is
$$\frac{\binom{10}{2} * \binom{15}{1}}{\binom{25}{3}} = \frac{45*15}{2300} = \frac{27}{92}.$$
Part (b): There are $\binom{25}{22}$ ways to choose 22 balls out of 25. If 2 balls left are white and 1 is black, that means we chose 8 white balls and 14 black balls in our set of 22, and there are $\binom{10}{8}*\binom{15}{14}$ ways to make those picks. But since $\binom{n}{r} = \binom{n}{n-r}$:
and the probability is, once again,
$$\frac{\binom{10}{8} * \binom{15}{14}}{\binom{25}{22}} = \frac{\binom{10}{2} * \binom{15}{1}}{\binom{25}{3}} = \frac{27}{92}.$$