Probability Question, Possible Inclusion/Exclusion?

Question

The question is as follows:

Given $x + y$ students in a class, and $r + s$ girls in the class, $x \geq r$. Randomly selecting $x$ students, what is the probability that exactly $r$ of the students are girls?

What I have so far:

First we have to choose the girls who will be in the group of $x$ students, and this is done by:

$\displaystyle ^{r + s} \mathbf{C}_{r}$

Then we multiply that value by the number of ways there are $r$ girls in a set of $x$ students and this is given by:

$\displaystyle ^x \mathbf{C}_r$

Then this is all divided by the total number of ways of choosing a set of $x$ students which is:

$\displaystyle ^{x + y} \mathbf{C}_x$

So in the end I get:

$\displaystyle \frac{^{r + s} \mathbf{C}_r \, ^x \mathbf{C}_r }{ (^{x + y} \mathbf{C}_x)}$

Is this right?

Answer 1

You multiply the ways of choosing girls by the number of ways of choosing students who are not girls, which is $\displaystyle ^{x+y-r-s} \mathbf{C}_{x-r}$. So your final answer should look something like: $$ \frac{^{r+s} \mathbf{C}_r \,^{x+y-r-s} \mathbf{C}_{x-r}}{^{x+y} \mathbf{C}_x} $$ Otherwise the reasoning is correct.

Answer 2

There doesn't seem to be any inclusion/exclusion involved.

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The total number of ways to choose $x$ out of $x+y$ students is:

$$\binom{x+y}{x}=\frac{(x+y)!}{{x!}\cdot{y!}}$$

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The number of ways to choose $x$ out of $x+y$ students with $r$ out of $r+s$ girls is:

$$\binom{r+s}{r}\cdot\binom{x-r+y}{x-r}=\frac{(r+s)!}{{r!}\cdot{s!}}\cdot\frac{(x-r+y)!}{{(x-r)!}\cdot{y!}}$$

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Hence the probability to choose $x$ out of $x+y$ students with $r$ out of $r+s$ girls is:

$$\frac{\frac{(r+s)!}{{r!}\cdot{s!}}\cdot\frac{(x-r+y)!}{{(x-r)!}\cdot{y!}}}{\frac{(x+y)!}{{x!}\cdot{y!}}}=\frac{{(r+s)!}\cdot{(x-r+y)!}\cdot{x!}}{{r!}\cdot{s!}\cdot{(x-r)!}\cdot{(x+y)!}}$$