# Probability $\sum_{j=n+1}^{2n+1} {M \choose m+1}{M-m-1 \choose j-m-1}/{N \choose j}$

I have a prob. problem:

A school has $N$ students in which $M$ students are leader (of each class in school), and $N>M$. There are $2n+1$ balls in the black box including $n+1$ blue balls and $n$ red balls (where $2n+1<M$).

The balls are randomly distributed to students in the school such that one student get only one (blue or red) ball.

1/ Given value $m$ ($m\le n$), what is the probability $P_1$ that $M$ leaders take $m+1$ blue balls and at most $m$ red ones?

2/ What is the probability $P_2$ that $M$ leaders take the number of blue balls greater than red ones?

I already computed $P_1=\sum_{j=n+1}^{2n+1} {M \choose m+1}{M-m-1 \choose j-m-1}/{N \choose j}$ but not sure it's correct. but not $P2$.

Can anyone help me to find out $P_1$? and also $P_2$? And could you give the simplification and upper bound of them? Thank you.

## 1 Answer

I think P1should be $P1 = \binom{M}{m+1}\sum ^m_{j=0} \binom{M-m-1}{j}$

$P2 = \sum ^n _{i=1} \binom{M}{i} \sum _{j=0} ^{j \lt i} \binom {M-i}{j}$

• in your case $P1$ and $P2$ are $>1$? – Alex Jul 29 '14 at 8:41
• Those aren't probabilities, they are counts of events. Specifically this attempt at $P1$ is only the count of ways to distribute $m+1$ blue balls and up to $m$ red balls among the $M$ leaders. It doesn't include the ways to distribute the remaining balls among the rest of the students. Nor does it divide by the total space. – Graham Kemp Jul 29 '14 at 9:16
• no, the problem is that the balls are distributed to all students, then we have to find the probability M leaders take $m+1$ blue balls and up to $m$ red balls. – Alex Jul 29 '14 at 12:21
• They are not probabilities. My mistake. They are counts. – tpb261 Jul 30 '14 at 8:13