Probability that the ball is red There is a bag with N balls. Some of them are black and some are red. In the bag there can be 0 red balls or 1 red ball or 2 red balls or N red balls. It is given that each of these N+1 options is equally likely.
Now, I pick up P balls from the bag and found that R of them are red.
After this I pick up another ball from the bag, what is the probability that this ball is red?

Like Alex, I will assume sampling without replacement.  
The probability of there having been X red balls out of N initially and R red were found when P were drawn is $$\frac{1}{N+1} \frac{\binom{X}{R} \binom{N - X}{P-R}}{\binom{N}{P}}$$  and, given that, the probability the next ball is red is $\frac{X-R}{N-P}$, which only makes sense (and is between $0$ and $1$) if $R \le X \le R+N-P$ so the result is: 
$$\frac{\sum \left(\frac{X-R}{N-P}\right)\frac{1}{N+1} \frac{\binom{X}{R} \binom{N - X}{P-R}}{\binom{N}{P}} }{\sum \frac{1}{N+1} \frac{\binom{X}{R} \binom{N - X}{P-R}}{\binom{N}{P}}}$$ 
$$= \frac{\sum (X-R)  \binom{X}{R} \binom{N - X}{P-R} }{(N-P)\sum  \binom{X}{R} \binom{N - X}{P-R} }$$
where the sums are over the possible values of $X$, i.e. $\sum_{X=R}^{R+N-P}$.
