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 $10.8\cdot 10^6$.


The expected number of occurrences would be $10800000\cdot1.1/100000\approx119$. So having at least 52 is very very likey. In fact, R tells me:
> binom.test(52,10800000,1.1/100000, alternative="greater")

        Exact binomial test

data:  52 and 10800000
number of successes = 52, number of trials = 10800000, p-value = 1
alternative hypothesis: true probability of success is greater than 1.1e-05
…

Somehow, this is confusing, since you said (before editing your question if I remember correctly) that there has been an increases in the number of occurrences. This scenario suggests that there has actually been a very significant drop:
> binom.test(52,10800000,1.1/100000)

        Exact binomial test

data:  52 and 10800000
number of successes = 52, number of trials = 10800000, p-value = 8.528e-12
alternative hypothesis: true probability of success is not equal to 1.1e-05
…

