What is the probability of at least an elderly person in my neighborhood dying in the next 1 year?

Question

This is a random question that came to my mind when I went for a long run this morning. I tried to solve in it my mind and then on paper, but the number I come up with doesn't feel good intuitively.

First let me formulate the question and then explain my approach,

"What is the probability of at least an elderly person in my neighborhood dying in the next 1 year?"

I assumed,

1. let "Elderly person" be anyone with $age \geq 70$.
2. let Neighborhood mean that there are $50$ houses near my house.
3. there can be only a single elderly person residing in a house and there is 25% chance of an elderly person residing in a house(let it be called $E$ where $P(E)$=.25).
4. let X be a r.v. defining the current age of an elderly person, which is $unif(70, 100)$.
5. let Y be a r.v. defining the age at which an elderly person dies, which is $unif(70,100)$.

My solution,
Let,
$A$ be an event that an elderly person dies(anywhere) in the next 1 year,
$B$ be an event that an elderly person resides in a house and dies in the next 1 year,

$$P(A) = P(X+1>Y|X<Y) = \frac{30^2 - 29^2}{30^2} = 0.065$$
$$P(A^c) = 1-0.065 = 0.9345$$

$$P(B) = P(A| E)P(E) + P(A| E^C)P(E^C) = P(A|E)P(E) = 0.065*.25 = 0.01625 $$

$\textbf{Note}$: The idea is if there are 50 houses in the neighborhood, not all houses have an elderly person residing there. P(A) is in "general" probability of any elderly person dying in next 1 year. P(B) is supposed to mean if an elderly person resides in a house "and" he will die in next year.

$P($at least an elderly person dies in my neighborhood) = $1 - P($none of the elderly person dies in my neighborhood).

$P($none of the elderly person dies in my neighborhood) = $(1-P(B))^{50} = (1-0.01625)^{50} = 0.4408$

Therefore, I concluded that
$P($ at least an elderly person dies in my neighborhood) = $1-0.4407 = 0.5593$,

Does the solution seem fine? Am I incorrect in my assumptions or solution? Please help.