Trees in a forest are distributed according to a Poisson process. Let $X$ be the distance from an arbitrary starting point to the nearest tree. Given that the average number of trees per square metre is $\lambda,$ find the density $f(x)$ of $X.$
The probability that there are $0$ trees within a circle of radius $x$ from the starting point is $e^{-\lambda \pi x^2},$ since $\lambda \pi x^2$ is the average number of circles that would be expected in that region. But I'm not sure how to relate this to $X,$ the distance from an arbitrary starting point to the nearest tree. Fix a starting point $S.$ If the distance from $S$ to the nearest tree is $x,$ then doesn't that mean that there's at least one tree within that distance of $x$? If so, then the probability of the nearest tree being a distance of $x$ from the starting point should be the probability of at least one tree occurring in the circle of radius $x,$ which is $1-P(no \,trees\, occur).$ This is the cumulative distribution function as it's the sum of the probabilities that the nearest tree is a distance of $0\leq y\leq x$ from the start point, and so to find the density, we just take the derivative.
Is this correct?
What you have said looks correct to me but does not seem answer the original question:
The expected number of trees within $x$ of the point is $\lambda \pi x^2$
The probability there is exactly one tree within $x$ of the point is, as you say, $\lambda \pi x^2 e^{-\lambda \pi x^2}$ as this is a Poisson process, but this is not particularly helpful
The probability there are no trees within $x$ of the point is $e^{-\lambda \pi x^2}$ as this is a Poisson process
The probability there is at least one tree within $x$ of the point is $1-e^{-\lambda \pi x^2}$ and this is a cumulative probability
The density for the distance of the nearest tree is then $2 \lambda \pi x e^{-\lambda \pi x^2}$ by taking the derivative