# Fast(er) way of computing the cumulative binomial probability? You flip a fair coin 100 times. What is the probability that you have less than 45 heads?

Fast(er) way of computing the cumulative binomial probability?
While revising for my probability test, I saw this question from one of the previous exams:
You flip a fair coin 100 times. What is the probability that you have less than 45 heads?
My question is a simple one. I know that to calculate the probability of a specific amount of n heads (or tails), we can use the binomial distribution with formula $P\left(X=k\right)=\left(\genfrac{}{}{0}{}{n}{k}\right){p}^{k}\left(1-p{\right)}^{n-k}$, with $n=100$ and $p=0.5$ in this case. I also know that we can compute the chance of $P\left(X as either $P\left(X=0\right)+P\left(X=1\right)+...+P\left(X=k-10\right)$ or $1-\left(P\left(X=k\right)+P\left(X=k+1\right)+...+P\left(X=n\right)\right)$
However this is a simple question only worth two points out of over 60 total points. I cannot imagine that you need to compute $P\left(X=n\right)$ 44 separate times and sum them together for such a small amount of points.
Is there any way to rewrite the formula or apply a different trick to drastically lower the amount of computations you need to do?
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Denier5h
Step 1
The trick is to indeed use the central limit theorem since we already have a large n.
Since our distribution is a binomial one, we have $E\left[X\right]=\mu =0.5\cdot 100=50$ and $Var\left(X\right)={\sigma }^{2}=0.5\cdot 100\left(1-0.5\right)=25$
Step 2
Then, we can compute $\frac{45-\mu }{\sqrt{{\sigma }^{2}}}=\frac{45-50}{\sqrt{25}}=-1$. Looking this up in the Z-table gives 0.1587. This is close to the exact answer of 0.14 (and indeed, both 0.14 and 0.16 are marked as correct in the answer sheet).