# what is the probability that after 220 coin tosses one gets between 120 and 133 tails?

Attempt: the straightforward solution is according to this formula: $\sum_{k=120}^{133}\binom{220}{k} (\frac{1}{2})^{220-k}(\frac{1}{2})^{k}$. My problem is that term $\binom{220}{k}$ is a very large number. I can calculate it but I wonder if there any other simpler way to find the answer.

• Try a normal approximation – carmichael561 Mar 22 '16 at 21:14
• Related: math.stackexchange.com/questions/1707355/… the only difference is the number of trials and the probability for success. – JMoravitz Mar 22 '16 at 21:18
• @Kirill I'll write up a longer one with a bit more detail. – JMoravitz Mar 22 '16 at 21:21

Indeed, $\sum\limits_{k=120}^{133}\binom{220}{k}(\frac{1}{2})^{220}$ will be the exact probability, but as in this question, it will be difficult to calculate manually.
$n=220, p=\frac{1}{2}$ so $\mu=np=220\cdot \frac{1}{2}=110$ and $\sigma = \sqrt{npq}=\sqrt{220\cdot \frac{1}{2}\cdot \frac{1}{2}} = \sqrt{55}\approx 7.416$
We have $Pr(120\leq X\leq 133)\approx Pr(119.5\leq Y\leq 133.5)\approx Pr(\frac{119.5-110}{7.416}\leq Z\leq \frac{133-110}{7.416})\\ \approx A(3.17)-A(1.28)\approx 0.9992 - 0.8997=0.0995$
where $X$ is the binomially distributed variable for number of tails, $Y$ is the normal distribution with the same mean and variance as $X$, and $Z$ is the standard normal distribution and $A(z)$ denotes the area under the standard normal curve to the left of $z$.
The normal approximation will not be perfectly accurate, but it will provide a very close estimate. As it so happens, the exact value is closer to $0.099309159$, but the estimation got us within $0.0002$ of that value.