Probability of remaining balls in a box 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?
 A: 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 $\binom{n}{r} = \binom{n}{n-r}$:

*

*$\binom{25}{22} = \binom{25}{3}$

*$\binom{10}{8} = \binom{10}{2}$

*$\binom{15}{14} = \binom{15}{1}$
and the probability is, once again,
$$\frac{\binom{10}{8} * \binom{15}{14}}{\binom{25}{22}} = \frac{\binom{10}{2} * \binom{15}{1}}{\binom{25}{3}} = \frac{27}{92}.$$
A: The answer given by Rivers McForge is very good, but does presume a familiarity with "n-choose-k" notation. You can arrive at $P\,=\,\frac{27}{92}$ using rules of probability. For this it is strongly suggested to draw a tree diagram (e.g. http://www.bbc.co.uk/schools/gcsebitesize/maths/images/figure_89.gif).
There are three different (and mutually exclusive) ways to win this game: WWB, WBW, and BWW. We calculate the probability of each, and then add them up.
The first one you already did yourself, in your original post. You found that
$$P(WWB)\,\,=\,\,\frac{10}{25}\frac{9}{24}\frac{15}{23}\,\,=\,\,\frac{9}{92}.$$
But then the remaining two turn out to be the same, since they amount to shuffling the factors in the numerator of your original calculation:
$$P(WBW)\,\,=\,\,\frac{10}{25}\frac{15}{24}\frac{9}{23}\,\,=\,\,\frac{9}{92},$$
$$P(BWW)\,\,=\,\,\frac{15}{25}\frac{10}{24}\frac{9}{23}\,\,=\,\,\frac{9}{92}.$$
Then our final result is $P(WWB)\,+\,P(WBW)\,+\,P(BWW)\,=\,\frac{27}{92}$.
If you want to use McForge's way, you can check out:

*

*https://en.wikipedia.org/wiki/Combination


*https://en.wikipedia.org/wiki/Hypergeometric_distribution
A: in those problems we have to be aware about the repetition of cases, in that case , the probability you said is the probaility of getting 1 white,1 white and 1 black. but if you get 1 black, 1 white and 1 white te probability it's (15/25)(10/24)(9/23), and (10/25)(15/24)(14/25) the case with the black in the middle.
So the total probability it's just add up the probability of the above 3 cases.
b) it's the same probability as a) because it's just anothe way of picking 3 balls randomly.
(insted of getting the balls,quit the others you don't need)
A: $\dfrac {10}{25}\dfrac {9}{24}\dfrac {15}{23}$ would be the correct answer if a) read: What is the probability that we first extract 2 white balls and then 1 black ball from the box? To get the answer for the question the way you phrased it - where the order of the balls chosen doesn't matter - the answer is
$$\underbrace{2\cdot\dfrac {10}{25}\dfrac {9}{24}\dfrac {15}{23}}_{W_1W_2B\text{, or }W_2W_1B}+\underbrace{2\cdot\dfrac {10}{25}\dfrac {15}{24}\dfrac {9}{23}}_{W_1BW_2\text{, or }W_2BW_1}+\underbrace{2\cdot\dfrac {15}{25}\dfrac {10}{24}\dfrac {9}{23}}_{BW_1W_2\text{, or }BW_2W_1}=6\cdot\dfrac {10}{25}\dfrac {9}{24}\dfrac {15}{23}$$
As you can see, we have to differentiate three possible orders: Either we first pick two white balls, then a black one ($W_1W_2B\text{, or }W_2W_1B$), or a withe one, a black one, then a white one again ($W_1BW_2\text{, or }W_2BW_1$), or we draw a black one, then two white ones ($BW_1W_2\text{, or }BW_2W_1$), and furthermore we cannot treat the two white balls as identical. Notice that $6=3!$.
We can rephrase b) to make it more approachable: When extracting 22 balls from the box, what is the probability $P$ of extracting 8 white and 14 black balls? From this perspective, b) is almost identical to a):
$$P=22!\cdot\dfrac {10}{25}\cdot\dfrac {9}{24}\cdot...\cdot\dfrac {4}{19}\cdot\dfrac {3}{18}\cdot\dfrac {15}{17}\cdot\dfrac {14}{16}\cdot...\cdot\dfrac {3}{5}\cdot\dfrac {2}{4}$$
