Drawing balls from a single urn (conditional probability) I've a urn with $b$ blue and $r$ red balls. I randomly draw $n$ balls from the urn sequentially and without replacement. I'm asked the probability of first ball chosen is blue given that $k$ out of $n$ balls are blue.
My attempt is as follows.
Let,
$B:$ $1^{st}$ chosen ball is blue
$B_k:$ $k$ balls out of $n$ is blue.
Then, what I'm asked is simply $$ P(B|B_k) = \frac{P(BB_k)}{P(B_k)}. $$
We have $$P(B_k) = \frac{\binom{b}{k}\cdot\binom{r}{n-k}}{\binom{b+r}{n}}$$
Also, $P(BB_k)$ is the case where the first ball is blue and we have $k$ blue balls out of $n$ balls. So if we fix the first ball as blue, it should be $$P(BB_k) = \frac{\binom{b-1}{k-1}\cdot\binom{r}{n-k}}{\binom{b+r}{n}}$$
Doing the division I obtain $$\frac{\binom{b-1}{k-1}}{\binom{b}{k}}$$ however the answer is $\frac{k}{n}$.
What's wrong with my reasoning? In particular, why is my calculation of $P(BB_k)$ is incorrect?
 A: $$P(BB_k)=P(B_k\mid B)P(B)=\frac{\binom{b-1}{k-1}\binom{r}{n-k}}{\binom{b+r-1}{n-1}}\frac{b}{b+r}$$
Dividing this by: $$P(B_k)=\frac{\binom{b}{k}\binom{r}{n-k}}{\binom{b+r}{n}}$$gives $$P(B\mid B_k)=\frac{k}{n}$$

There is a much easyer route to  this result: there are $n$ drawn balls and $k$ of them are blue. So what is the probability that one of them (e.g. the first drawn) is blue??...
A: Conditioning on the first ball being blue, your calculation for $P(BB_k)$ is on the right track (although you're really discussing $P(BB_k|B)$). Take your formula for $B_k$ and replace all instances of $b$, $k$, and $n$ with $b-1$, $k-1$, and $n-1$. So you get
$$\frac{{{b-1}\choose {k-1}}\cdot {r\choose {n-k}}}{{b+r-1}\choose{n-1}}$$
A: I don't know if this approach is right or wrong.

*

*We are given that k out of n balls are blue. which is a condition or ground truth.

*We need to calculate the probability for given condition that first ball is blue.

As we have k blue balls we can choose the first blue balls in k ways and rest n-1 balls in (n-1)! ways. So favourable events are $k\cdot(n-1)!$
Now a total number of ways we can arrange n balls is n!
Prob(first ball is blue | k out of n balls are blue) = $\frac{k\cdot(n-1)!}{n!} = \frac{k}{n}$
