Compute the probability that the first $k$ draws are red and the next $n-k$ are green
Solution 1:
Maybe I misunderstand the question but I get the answer to the first question to be like this instead.
$\left( \frac{1}{2}\right) \left( \frac{2}{3}\right) ...\left( \frac{k}{1+k}% \right) \left( \frac{1}{2+k}\right) \left( \frac{2}{3+k}\right) ...\left( \frac{n-k}{n+1}\right) =\frac{k!(n-k)!}{(n+1)!}$