Lets say a production machine makes one fault unity per every $500$ units. Assume that it is making $50$ units per package. Let $S$ be a random variable corresponding to the number of faulty units that the machine makes in a single package.
I calculated the expected number of faulty units per page is $E(S) = \frac{50}{500}$
However, I am not able to wrap my mind around calculating the probability of $S = someNumber$
Also, how exactly can this normal distribution be modeled as a poisson distribution?
The Poisson distribution takes a an average rate, which you've found, and tells you the probability that a number of occurrences will happen. For a Poisson distribution, the density function is given by $$\mathbb P(S=s)=\frac{\lambda^se^{-\lambda}}{s!}$$ where $s$ is a non-negative integer. In this case, we are concerned with the number of faults per $50$ units so $\lambda=\frac{1}{500}\frac{\text{faults}}{1\text{unit}}= 0.1 \frac{\text{faults}}{50\text{units}} \implies \lambda= 0.1$. You can then calculate the probability of $s$ faults happening with the formula above! Note that the maximum number of faults that can occur is $50$, so theoretically $s$ cannot be any positive integer and to be precise you should be using the binomial distribution with $n=50$ and the probability of a fault $p=1/500$, however, the Poisson distribution is derived by taking $n \to \infty$ so for large samples the Poisson distribution is a good approximation.