Probability that a binomial random variable with $n=10.8\cdot 10^6$ and $p=1.1\cdot 10^{-5}$ is larger than $52$ Given $n=10.8\cdot 10^{6}$ independent identically distributed (i.i.d.) random variables $$X_1,\dots, X_n\sim\text{Bernoulli}(p=11\cdot10^{-6}),$$ what is the following probability? $$\mathsf P \left( X_1 + \cdots + X_n \ge 52 \right)$$

Motivation
Warning: the following contains material that may cause discomfort to some readers.

 According to the United Nations Office on Drugs and Crime 2015 crime statistics, the rate of police recorded instances of sexual intercourse without valid consent in Greece in the year 2015 was $1.1$ per $100'000$ people and the population of Greece is around $10.8\cdot 10^6$.


I will just focus on the mathematical aspect of the question and not comment on if and how it applies to the real-world question you mentioned above (there are some important issues raised in the comments to your question).

The formal question is this: Given $n=10.8\cdot 10^{6}$ identically distributed, independent random variables $$X_1,\dots, X_n\sim\text{Bernoulli}(p=11\cdot10^{-6}),$$ what is $$\mathsf P(X_1+\dots+X_n\ge 52)\text{?}$$
Note that
$$\mathsf P(X_1+\dots+X_n\ge 52)=1-\sum_{k=0}^{51}\mathsf P(X_1+\dots+X_n=k).$$
Since $X_1+\dots+X_n$ has a binomial distribution (cf. Proof that a sum of Bernoulli rvs has Binomial distribution), $$\mathsf P(X_1+\dots+X_n=k)=\binom nk p^{k}(1-p)^{n-k}.$$
A numerical computation gives $$\mathsf P(X_1+\dots+X_n\ge 52)\ge 0.9999999999981=1-1.9 \cdot 10^{-12}.$$

If you instead change the formal question to only use half the population, then the computations remained unchanged, except that now $n=5.4\cdot 10^6$, so $$\mathsf P(X_1+\dots+X_n\ge 52)\in[0.847, 0.848].$$
