If 0.01% of a population has a disease, what sample size k is needed so that there is a 95% chance that one person with this disease falls into k? I read that this can be done using the Poisson Approximation, which I understand how to do when you are already given a sample size. How would one find the solution to this problem to find the sample size, using a Poisson approximation?
 A: In this case, you can use the "exact" Binomial distribution. 
We want $k$ such that $\mathbb{P}(j \geq 1) \geq 95\%$, where $j$ is the number of persons with the disease.
This is equivalent to say that $\mathbb{P}(j < 1) = \mathbb{P}(j = 0)\leq 5\%$
The probability is:
$$
\mathbb{P}(j = 0) = \binom{k}{0} p^0 (1-p)^{k-0} = (1-p)^k
$$
where $p = 0.0001$ is the probability of getting the disease.
Therefore, we have to solve for k
$$
0.9999^k = 0.05
$$
I.e., 
$$
k \log 0.9999 = \log 0.05 
$$
Which gives
$$
k \approx 29\ 955.8\ldots
$$
So, a sample size of $29\ 956$ individuals.

If you really want to use the Poisson approximation, then you need the same reasoning, but now the probability is:
$$
\mathbb{P}(j = 0) = e^{-pk} \frac{(pk)^0}{0!} = e^{-pk}
$$
where $p = 0.0001$ is the probability of getting the disease and $pk$ is the expected (average) number of people with the disease in a sample of size $k$.
Can you take it from here and get the value of $k$ in this case? The result will not be the same as before.
Edit:
We just need to plug in the value of $p$ and solve:
$$
e^{-0.0001k} = 0.05
$$
So, applying natural logarithms: 
$$
-0.0001k = \log 0.05
$$
Which gives
$$
k \approx 29\ 957.3
$$
So, a sample size of $29\ 957$ will have an approximate probability of $95\%$ of having at least one disease person.
In case you need at least $95\%$ probability, $k$ will need to be at least $29\ 958$.
