Probability Theory - random number hit from a pool With the given data:
After picking 30 random natural integers in a pool of X-natural-numbers (numbers do not disappear from the pool after picking). The probability of NOT picking 1 pre-defined specific number should be around a probability of 0,3%.
This is what I'm trying to calculate:
The size of the pool
What I want to know:
The formula
 A: It is not entirely clear what the problem is, so I will make an interpretation that I hope is the one you intend. Out of habit, I will use $n$ instead of $X$.
We have a pool of $n$ distinct natural numbers, or indeed any $n$ distinct objects. We suppose that $A$ is one of these objects.
We pick from this pool $30$ times, each time replacing the object that we picked, so that the composition of the pool does not change. The probability that in $30$ trials, we never pick $A$, is $0.003$.  We want to know the number $n$ of objects in the pool.
The probability that, on any individual  pick, we get the object $A$, is $\dfrac{1}{n}$.  So the probability that we don't get $A$ is $1-\dfrac{1}{n}$.
The probability that this happens $30$ times in a row is
$$\left(1-\frac{1}{n}\right)^{30}.\qquad\qquad (\ast)$$
So we want to solve the equation
$$\left(1-\frac{1}{n}\right)^{30}=0.003.$$
This equation can be solved in various ways, including "trial and error."  In our particular situation, trial and error is a very good way. If we play with the calculator a bit, using the formula $(\ast)$, we find that if $n=5$, the probability of never getting $A$ is about $0.0012379$, while if $n=6$, the probability is about $0.0042127$.  There is no integer $n$ such that the probability is exactly $0.003$. We get closest with $n=6$.
We now describe a more systematic way of solving our equation.
Take the logarithm of both sides. I will use logarithm to the base $10$, though I would prefer the natural logarithm (base $e$).
We obtain
$$30\log\left(1-\frac{1}{n}\right)=\log(0.003).$$
Calculate. We get
$$\log\left(1-\frac{1}{n}\right)\approx -0.084096.$$
Recall that if $y=\log x$ then $x=10^y$. We conclude that
$$\left(1-\frac{1}{n}\right)\approx 0.823956,$$
which gives $n=5.6803994$.  Of course, that is not right, $n$ must be an integer. If we let $n=5$, the probability we never get $A$ is quite a bit less than $0.003$, while if $n=6$, the probability we never get $A$ is greater than $0.003$.
Remark: You might be interested in numbers other than your special $30$ and $0.003$. More generally, suppose that we pick $k$ times, and we want the probability of never getting $A$ to be $p$. Then we need to solve the equation
$$\left(1-\frac{1}{n}\right)^{k}=p.$$
Like in our concrete case, we can find the appropriate value of $n$ by using logarithms. In general, like in our concrete case, there will not be an integer $n$ that gives probability exactly $p$.
Again, we use logarithms to the base $10$, though any base will do.
We get
$$k\log\left(1-\frac{1}{n}\right)=\log p,$$
and therefore 
$$\log\left(1-\frac{1}{n}\right)=\frac{\log p}{k},$$
and therefore 
$$1-\frac{1}{n}=10^{\frac{\log p}{k}}.$$
Solving for $n$, we obtain
$$n=\frac{1}{1-10^{\frac{\log p}{k}}}.$$
Suppose that instead of your $0.003$, we let $p=0.95$. Let $k=30$. Using the above formula, we get  $n\approx 585.4$.  This could have taken some time to reach by trial and error. 
A: The chance of not picking the predefined integer in our thirty picks $= \left( \dfrac {x-1} {x}\right) ^{30} = 0.3$.
Now solve for x.
