Calculating the probability of a fire incident I am currently working on a solution to a dataset that I have been given based on probabilities though I require some further guidance on the calculation.
For example, I have a dataset with the number of fire incidents for a district. I want to calculate the probability of a fire incident occurring within a 2km radius of a fire station given the number of buildings within that radius.
Let's say the total number of fires in this district is 80 in a month. The total number of buildings within the fire-station radius is 200, and the total number of buildings within the entire district is 15000.
What is the probability that one or more of those fires belong to the buildings within the fire-station radius?
It is not necessarily like rolling dice however I thought this approach could be taken: $n = 15000$, then $k = 80$ however given that we are looking at those within the 200 buildings, I was thinking something like:
$\frac{15000!}{(15000-80)\cdot200^{80}}$?
My reasoning for the equation:
I presumed that given a total population size $=15000$ ,then selecting the difference of population size and $k$ occurrences of incidents and finally $\frac{1}{200^{80}}$
$80$ represents the selection and $200$ the number of samples of interest. Though this may be a better equation when calculating for fires occurring in the same building? i.e. sampling with replacement.
Secondary Approach:
Perhaps calculating the daily probability relative to a month helps. Hence, $80/30$ ~ $2.6$ which is equivalent to $\frac{1}{30}$. We know that the sample space is $n = 15000$, and $m = 200$ is a subset of the sample space. What is the probability that in any given day, more than 1 incidences have occurred within $m$.
Given the potential of more than one station in a district, then perhaps a hypergeoemetric distribution works well in this case?
$\frac{\binom{z}{k}\binom{m}{y}\binom{15000-z-m}{80-k-y}}{\binom{15000}{80}}$
Where $z$ represents another fire-station in the district independent to the previous fire-station, and $y$ is the proportion of incidences.
Extra information:

*

*We assume each incident is independent and happening to unique buildings. 2. Again we assume each logged incident is unique. 3. We do not know if they have caught on fire more than once in the month, again we assume uniqueness here also. 4. Presumingly those closer to the station probably have a lower rate of an incident. On this assumption, we could add a weight to those closer to the station by every 100m, the closer then the lower the probability.

Data collection:
Fire stations collected from here:
https://www.datadaptive.com/?pg=5
Fire incidents collected from here:
https://www.gov.uk/government/statistics/fire-statistics-incident-level-datasets
UK administrative boundaries here:
https://www.ordnancesurvey.co.uk/business-government/products/boundaryline
Buildings in UK:
https://www.ordnancesurvey.co.uk/business-government/products/open-zoomstack
Methodology:
Plotting the number of station across the UK on QGIS while overlaying the boundaries data to gather the administrative districts. Secondly, uploading the buildings from the open-zoom stack. Lastly, creating a buffer around each station for a 2km radius and counting the number of buildings within the buffer, and counting the number of buildings within each district.
Given that the fire incidence dataset does not have coordinates, we just take the total count of fires in a district.
More than one fire-station may be within the same district, is this something to account for - if so, how could this be interpreted?

Goal:  Probability of a fire incident within a 2 km radius of a fire station
Data:  $x$ fires in $N$ homes per month.  There are $m$ homes within a 2 km radius
$P$(one or more fires within a 2 km radius in a month)$ = 1 - P$(exactly zero files within a 2 km radius within a month)
Assume the occurrence of a fire in a home is distributed as a Poisson random variable
$P$(k events in interval t)$ = \frac{(rt)^k}{k!} e^{-(rt)}$
$P$(0 events in a month) = $\frac{(\frac{x}{N}m)^0}{0!} e^{-(\frac{x}{N}m)}$
R Code:
# exact calc
1 - exp(-80/15000*200) = 0.6558462

# simulation
set.seed(103400)
x <- sapply(1:10000, function(z) rbinom(n = 200, size = 1, prob = 80/15000))
y <- apply(x, 2, function(x) all(x == 0))
1 - mean(y)
# 0.6557

