Lets say I have a coin with P(heads) = 1/2 and P(tails) = 1/2. I toss that coin 100 times. What is then the probability of getting less than 45 heads?
Because I though it is a Bin(100, 0.5) I tried to calculate it like this:
$$\sum_{n=0}^{44} {100\choose n}*0.5^{n}*(1-0.5)^{100-n}$$
Which works fine but the problem is that if I were to do this by hand it would take a long time, so is there a shorter/better way of solving this problem or should I use another distribution for this (e.g. Poisson)?
When the numbers are large, the normal approximation to the binomial works well. If there are $N$ trials with probability $p$, the mean is $Np$ and the standard deviation is about $\sqrt{Np(1-p)}$. It is rather accurate near the mean, less so further out. You can then use the z-score table. In your example, $N=100, p=\frac 12, \mu=50, \sigma\approx\sqrt{100\cdot \frac 12 \cdot \frac 12}=5$, so $45$ heads is $1\ \sigma$ low.