What's fastest way for calculate this Sigma?

Sorry for bad English.
What's the fastest way to calculate this?

$$ \sum_{i=11}^{35}\left(\begin{array}{c} 35 \\ i \end{array}\right)\left(\frac{1}{10}\right)^{i}\left(\frac{9}{10}\right)^{35-i}=0.0004 $$

Is the solution for this related to the Central limit theorem? If is it, how to solve with this theorem ?
Thank You.

Solution 1:

I would say that the quickest way was with a computer.

For example in R

m <- 35; n <- 11:m; sum(choose(m,n) * 0.1^n * 0.9^(m-n))

gives 0.0004242976.

If you wanted to do this approximately by hand, it seems likely that most of the sum is in the $i=11$ term, which is about $0.0003328$. The $n=12$ term is about $0.000073957$ and the others are rather smaller.

If you were to use the Central Limit Theorem then you have a binomial distribution with mean $35\times0.1 = 3.5$ and variance $35\times0.1 \times0.9 = 3.15$. So with a continuity correction, you want the probability that a standard normal distribution is greater than $\frac{10.5-3.5}{\sqrt{3.15}} \approx 3.944053$ giving $0.000040058$, which is wrong by more than a factor of $10$. You should not expect the Central Limit Theorem to give accurate answers in the tail.