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?

The ratio of two successive terms of the sum is 
$$\frac p{1-p}\frac{3000-k+1}{k}<0.3$$as of $k=2$.
As $0.3^{25}<10^{-13}$, the tail is neglectible and the sum is $1$.
