Smallest $n$ where $2 \cdot \sum_{k = 1}^{n} \frac{\left(\frac{n}{100}\right)^k}{k!} \geq e^{\frac{n}{100}}$ I'm having a problem with solving the following relation for $n$:
$$2 \cdot \sum_{k = 1}^{n} \dfrac{\left(\frac{n}{100}\right)^k}{k!} \geq e^{\frac{n}{100}}$$
By trial-and-error I was already able to confirm, that $n \geq 70$ but I have yet to show it mathematically. I somehow cannot find a way to get rid of the summation, which looks a lot like the exponential function for $e^{\frac{n}{100}}$.
I transformed this relation from the following task:
A randomly chosen person has blood type AB- with a probability of about 1%. How many persons have to be chosen at least, so that there is at least one person with blood type AB- with a probability of at least 50%. (You may use a suitable approximation.)
Choosing poisson distribution, I came up with:
$$\sum_{k = 1}^{n} \mathcal{P}_{\lambda = \left(n \cdot \frac{1}{100}\right)} \left(\left\lbrace k \right\rbrace\right)= \sum_{k = 1}^{n} e^{-\left(n\cdot\frac{1}{100}\right)} \cdot \dfrac{\left(n\cdot\frac{1}{100}\right)^k}{k!} \geq \dfrac{1}{2}$$
 A: for your equation i have found
$$2\,{\frac {1}{\Gamma  \left( n+1 \right) }{{\rm e}^{{\frac {n}{100}}}}
\Gamma  \left( n+1,{\frac {n}{100}} \right) }-2-{{\rm e}^{{\frac {n}{
100}}}}=0
$$
and a solution is $\approx 69,31471806$
A: So, I was able to come up with a fairly simple answer.
Instead of finding an $n$, so that $1, 2, \ldots, n$ people have have blood type AB- and are amongst the chosen group of people with a probability of at least 50%, it is easier to find an $n$, so that the probability of $0$ people having blood type AB- is at most 50%:
\begin{alignat*}{2}
 \mathcal{P}_{\lambda = \left(n \cdot \frac{1}{100}\right)} \left(\left\lbrace 0 \right\rbrace\right) = \ & e^{-\frac{n}{100}} \cdot \underbrace{\dfrac{\left(\frac{n}{100}\right)^0}{0!}}_{= 1} &&\leq \dfrac{1}{2}\\
 \Leftrightarrow \ & e^{-\frac{n}{100}} &&\leq \dfrac{1}{2}\\
 \Leftrightarrow \ & -\frac{n}{100} &&\leq \ln\left(\dfrac{1}{2}\right)\\
 \Leftrightarrow \ & n &&\geq -100 \cdot \ln\left(\dfrac{1}{2}\right)\\
 \Leftrightarrow \ & n &&\geq 100 \cdot \ln\left(2\right)\\
 \Rightarrow \ & n &&= \lceil 100 \cdot \ln\left(2\right) \rceil = 70
\end{alignat*}
As @Travis has already mentioned in a comment, $n$ is indeed $100 \cdot \ln\left(2\right)$.
@Alex M. raised the question, whether the obviously binomial distributed problem could even be approximated by Poisson distribution:
\begin{alignat*}{3}
& \mathcal{B}_{\left(70, \frac{1}{100}\right)}\left(\left\lbrace 0 \right\rbrace \right)  &&= \begin{pmatrix} 70 \\ 0 \end{pmatrix} \cdot \left(\dfrac{1}{100}\right)^0 \cdot \left(\dfrac{99}{100}\right)^{70} &&\approx 0.4948\\
& \mathcal{P}_{\lambda = \left(\frac{70}{100}\right)} \left(\left\lbrace 0 \right\rbrace\right) &&= e^{-\frac{70}{100}} \cdot \dfrac{\left(\frac{70}{100}\right)^0}{0!} &&\approx 0.4966\\
& \sum_{k = 1}^{70} \mathcal{B}_{\left(70, \frac{1}{100}\right)}\left(\left\lbrace k \right\rbrace \right) &&= \sum_{k = 1}^{70} \begin{pmatrix} 70 \\ 0 \end{pmatrix} \cdot \left(\dfrac{1}{100}\right)^k \cdot \left(\dfrac{99}{100}\right)^{70-k} &&\approx 0.5052\\
& \sum_{k = 1}^{70} \mathcal{P}_{\lambda = \left(\frac{70}{100}\right)} \left(\left\lbrace k \right\rbrace\right) &&= \sum_{k = 1}^{70} e^{-\frac{70}{100}} \cdot \dfrac{\left(\frac{70}{100}\right)^k}{k!} &&\approx 0.5034\\
\end{alignat*}
One could indeed find, that $n \geq 69$ is a little more exact, when using the binomial distribution. But Poisson distribution is still a very precise approximation.
