Is there a smart way to compute $\sum^{25}_{k=0}\binom {3000}{k} p^k (1-p)^{3000-k}$? I am trying to compute probability, and I give a:
$$\sum^{25}_{k=0}\binom {3000}{k} p^k (1-p)^{3000-k}$$
$$p=0,0001$$$
Is there a smart way to compute it?
Another else, I am thinking about Poisson. What do you think?

You're on the right track. If $p=c/n$, then $\binom{n}{k}p^k(1-p)^{n-k}\rightarrow e^{-c}\frac{c^k}{k!}$. In other words, the binomial distribution goes to a Poisson for rare events. You can use the above approximation to get a quick estimate, which is equivalent to asking $P(X\le25)$ for a Poisson with $n=3000$, giving $c=0.3$. Only the first few terms will matter. 
