Is there a smart way to compute $\sum^{25}_{k=0}\binom {3000}{k} p^k (1-p)^{3000-k}$?

Question

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?

Answer 1

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 $1$.

Answer 2

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.

Answer 3

I use the approximation results given here:

Let $\mu=3000 \times 0.0001 =0.3$ and $\sigma^2=3000\times 0.0001 \times (1-0.0001)=0.29997.$ Now note that $$\frac{X-\mu}{\sigma}\sim N(0,1)$$ thus $$\Pr(X\leq 25)=\Pr\Big(\frac{X-0.3}{\sqrt{0.29997}}\le\frac{25-0.3}{\sqrt{0.29997}}\Big)\simeq \Pr\Big(Z\le45.0981\Big)\simeq 1, $$where $Z\sim N(0,1).$