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Oct
8
comment Probability that the ball is red
Thanks, Henry! I now understand the question. Yes, the balls are sampled without replacement. That's why I think the probability of there having been X red balls out of N initially and R were found when P were drawn would not be $$\frac{1}{N+1} {P \choose R}\left(\frac{X}{N}\right)^R\left(\frac{N-X}{N}\right)^{P-R}$$ It should be $$\frac{\binom{X}{R} \binom{N - X}{P-R}}{\binom{N}{P}}\cdot \frac{1}{N+1}$$ as @Alex said.