An urn containing $r$ red balls and $b$ blue balls. 
Suppose an urn contains $r$ red balls and $b$ blue balls. Suppose $n$ balls are drawn sequentially without replacement. Find the probability that $k$ of the $n$ balls are blue and that the first one is blue.

My try:
Let $A$ be event first ball is blue and $B$ be event $k$ of $n$ balls are blue. I want to find $P(A \cap B)$. In total we have ${r+b \choose n }$ outcomes in sample space. for the number of ways $A \cap B$ occurs we have first one is blue, and so we have $b-1$ blue balls from there we pick $k$ of them so ${b - 1 \choose k}$ and for red balls we have ${r \choose n - k}$. Therefore
$$ P(A \cap B) = \frac{ b {b - 1 \choose k} {r \choose n - k} }{{r+b \choose n } }. $$
Is this correct?
 A: $$P(A\cap B)=P(B)P(A|B)$$
$$P(B)=\frac{\binom bk\binom r{n-k}}{\binom{r+b}n}$$
$$P(A|B)=\frac kn$$
Alternatively
$$P(A\cap B)=P(A)P(B|A)$$
$$P(A)=\frac b{r+b}$$
$$P(B|A)=\frac{\binom{b-1}{k-1}\binom r{n-k}}{\binom{r+b-1}{n-1}}$$
A: A couple of alternative formulas have been given;
as noted in a comment, the formula
$$ P(A \cap B) \stackrel?= \frac{ b {b - 1 \choose k} {r \choose n - k} }{{r+b \choose n } }$$
gives wrong answers for some values of the input.
One obvious flaw in the formula is that the factor $\binom{b-1}{k}$
seems to assume that after choosing the first ball, which is blue,
you still have to choose $k$ balls from among the remaining $b-1$
blue balls, when in fact you really only want $k-1$ of those balls.
So we can try tweaking the formula as follows:
$$ P(A \cap B) \stackrel?= \frac{ b {b - 1 \choose k-1} {r \choose n - k} }{{r+b \choose n } }.$$
Now for the case $r=0, b=3, n=k=1,$ the formula gives the correct answer,
$1,$ where the earlier formula gives $2.$
But let's try $r=0, b=3, n=k=2.$ Now the revised formula evaluates to $2,$
which is clearly wrong.
So what's the flaw in the second formula? The flaw comes from your original approach to the problem. For the first ball you assumed it mattered which of the blue balls came first (which is how you got $b$ possibilities for that event), but for the rest of the balls you assumed it did not matter which balls came in which order (so you just choose some number from $b-1$ and some number from $r$). Meanwhile, in the denominator you assume that it does not matter in what order any of the balls occur, including the first one.
If you take the order of the balls into account in all cases,
then in the denominator you have $b$ ways to choose the first ball,
$\binom{b-1}{k-1}(k-1)!$ ways to choose the other $k-1$ blue balls,
$\binom{r}{n-k}(n-k)!$ ways to choose the red balls,
and $\binom{n-1}{k-1}$ ways to interleave the last $n-1$ red and blue balls.
In the denominator you have $\binom{r+b}{n}n!$ ways to choose the $n$ balls.
The result is
$$ P(A \cap B)
 = \frac{ b {b - 1 \choose k-1} (k-1)!
           {r \choose n - k}(n-k)! \binom{n-1}{k-1}}
        {{r+b \choose n }n! }
 = \frac{ b {b - 1 \choose k-1} {r \choose n - k}}
        {{r+b \choose n }n },$$
which agrees with the other answers.
