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.
I think P1should be $P1 = \binom{M}{m+1}\sum ^m_{j=0} \binom{M-m-1}{j}$
$P2 = \sum ^n _{i=1} \binom{M}{i} \sum _{j=0} ^{j \lt i} \binom {M-i}{j}$