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.