# Thread: Solution to a Modified Binomial Probability Distribution

1. ## Solution to a Modified Binomial Probability Distribution

I could use help determining whether the following can be solved analytically.

I have an equation $\displaystyle$\sum_{s=c}^{Q}\frac{s-c+1}{s+1}{Q\choose s}p^{s}(1-p)^{Q-s}=\frac{a-b}{a-d}$$. Notice this equation is the binomial probability distribution from c to Q (as opposed to going from 0 to Q), and it is multiplied by \displaystyle \frac{s-c+1}{s+1}$$.

I understand that $\displaystyle$\sum_{s=0}^{Q}s{Q\choose s}p^{s}(1-p)^{Q-s}$$simplifies to \displaystyle pn$$. Can a similar simplification be made to the equation I've presented above? Any thoughts on how I might go about it?

$\displaystyle \sum _{s=c}^Q \frac{(-c+s+1) p^s \binom{Q}{s} (1-p)^{Q-s}}{s+1} =$
$\displaystyle =p^c \Gamma (c+2) (1-p)^{Q-c} \left(\frac{\binom{Q}{c} \, _2\tilde{F}_1\left(1,c-Q;c+2;\frac{p}{p-1}\right)}{c+1}-\frac{p \binom{Q}{c+1} \, _2\tilde{F}_1\left(2,c-Q+1;c+3;\frac{p}{p-1}\right)}{p-1}\right)$
2F1=$\displaystyle Hypergeometric function \left._2F_1(a,b;c;z)\right/\Gamma (c)$ ref