Choose right pdf for lightning strike simplification [Exercice Question] First the problem:
The number of lightning strikes during a year in a large area was counted. This are is divided into equally large 300 sections. It is to assume that if a lightning strike occurs, the probability for it to occur in a specific location is uniformly distributed over the entire area. During the year, a total of 213 strikes were counted.
For a single section, what is the probability that it has been hit by any lightning strikes during this year?
Now, the Idea of this problem is to choose a fitting probability distribution function (pdf) for discrete random variables which is appropriate for this scenario. It doesn't state, that each section was only hit once so this should be applied as well I guess.
I already have thought of two possible pdfs, the normal or the binomial distribution. For the normal I used a mean of $N\cdot p$ with $N=300$ and $p=1/300$ (uniformly distributed strikes) and a standard deviation of $Np\cdot(1-p)$. This gets a probability of approx. $0.24$ for $x=0$.
For the binomial I used a success rate $k=0$, the same $p$ and $N$ to get an approx. of $0.38$.
Typically I would have used the binomial distribution for this, but a later follow up question uses the binomial one and as the question is asked it seems they should be different. But I feel uncomfortable using the normal distribution for this problem.
Am I approaching the problem in a wrong way or do I make a different mistake in solving it?

It seems to me that the binomial distribution is more appropriate. This is how I see it:
A given section has a $p = 1/300$ chance of being hit by a strike. There have been 213 strikes so $N = 213$. The probability of being hit by exactly 1 strike is then (in R code):
dbinom(1, size= 213, prob= 1/300)
0.3498

by exactly two strikes:
dbinom(2, size= 213, prob= 1/300)
0.124

and so on for 3, 4, ..., 213. So the probability of being hit by 1 or more strikes is the sum of the individual probabilities, about 51%:
sum(dbinom(1:213, size= 213, prob= 1/300))
0.5089


As noted by @Wolfmercury in comment, one can simply use $1 - binom(k= 0, N, p)$:
1 - dbinom(0, size= 213, prob= 1/300)
0.5089385

However, I thought the long-winded approach above makes the reasoning more explicit.
