Calculating probabilities over longer period of time

In all calculations, we will assume independence. That may not be reasonable in the case of forest fires.

Suppose that the probability of a fire in the course of a month is $0.05$, that is, $5\%$, which is very high for any individual structure.

Then the probability of no fire in the month is $0.95$.

The probability of no fire for $12$ months in a row is then $(0.95)^{12}$.

It follows that the probability of at least one fire in a year is $1-(0.95)^{12}$.

This is about $0.45964$.

For $30$ years, the same reasoning gives $1-(0.95)^{360}$. This is very close to $1$. That may feel counterintuitive. However, as mentioned earlier, the probability that a house has a fire in a given month is very much smaller than $0.05$.

Now we look at the problem where we have probability $0.05$ each month for $5$ months, and $0.10$ each month for $7$ months. Then the probability of no fire in the $5$ months is $(0.95)^5$, and the probability of no fire in the other $7$ months is $(0.90)^7$. So the probability of no fire in a year is $(0.95)^5(0.90)^7$. It follows that the probability of at least one fire in the year is $1-(0.95)^5(0.90)^7$.
Over $30$ years, in the $5$-$7$ scenario, the probability of no fire is $((0.95)^5(0.90)^7)^{30}=(0.95)^{150}(0.90)^{210}$. So the probability of at least one fire is $1- (0.95)^{150}(0.90)^{210}$. This is nearly $1$.