Background
In physics, there is the concept of "mean free path," which is the expected value for the distance a molecule (for example) can travel before it hits another one.
If they're all the same size, and the neighbor molecules are motionless, this can be calculated as $\frac{1}{\pi d^2 n_v}$ (where $d$ is diameter of a molecule, $n_v$ is number of molecules per-volume).
That all makes sense to me.
What I want to determine is: what is the probability of collision, for a given molecule which travels some distance $D$ ?
Initial Stumblings
I would expect the probability graph to look something like this:
That is: as the particle travels further and further, the odds of it hitting a neighbor approach 1 (assuming infinitely large surroundings of of uniform density).
I'm trying to find a formula for this, but am a little stuck.
Also, I'm trying to find the relationship between this and the mean free distance formula, which also seems elusive.
Could you help me?
Update:
I made some more progress: I think I was looking for the Poisson Distribution, and in particular the graph I drew is the CDF for that distribution.
The general formula for Poisson is:
$$ \frac{\lambda^k e^{-\lambda}}{k!} $$
And in my case, $k = 1$ (one collision).
I did find this source, which explicitly states the formula for the answer to my exact question (but does not explain it):
$$ \frac{1}{ l_{free} } exp(\frac{-r}{l_{free}}) $$
i.e.
$$ \frac{1}{ l_{free} } e^{\frac{-r}{l_{free}}} $$
where $r$ is distance traveled and $l_{free}$ is the length of the mean free path.
That looks very close to a Poisson distribution, except for that $r$ term in the exponent.
There is also a formula buried in the Wikipedia page for mean free path:
... the probability that a particle is absorbed between x and x + dx is given by $$ \frac{1}{L}e^{-x/L}{dx} $$
which again is very close to the stated "correct answer," but if I try to plug in "$x=0, dx=D$" to see the probability of absorption after traveling distance D, it is wrong. Presumably it is actually an infintesimal formula (maybe the wikipedia description is not entirely accurate?).
Let's assume that other particles are motionless and "located independently of each other".1) Under this assumption, the probability of a molecule freely traveling an infinitesimally small distance $\mathrm{d}l$ is $1 - \pi d^2 n_v \mathrm{d}l$, and this event is independent of what will happen in the remaining travelling path of that molecule.
So, if $L$ denotes the distance the molecule travels before hitting another one, then $\mathbb{P}(L > l)$, the probability of traveling a distance of $l > 0$ freely, is given by the following product integral:
$$ \mathbb{P}(L > l) = \prod_{0}^{l}(1 - \pi d^2 n_v \, \mathrm{d}l) = \exp\left(-\int_{0}^{l} \pi d^2 n_v \, \mathrm{d}l\right) = e^{-\pi d^2 n_v l} = e^{-l / l_{\text{MFP}}} $$
where $l_{\text{MFP}} = \frac{1}{\pi d^2 n_v}$ stands for the mean free path. Hence $L$ is distributed according to the exponential distribution with mean $l_{\text{MFP}}$:
$$ \mathbb{E}[L] = \int_{0}^{\infty} \mathbb{P}(L > l) \, \mathrm{d}l = l_{\text{MFP}} $$
Then the probability density function $f_L(\cdot)$ of $L$ is given by
$$ f_L(l) = \frac{\mathbb{P}(\text{$L$ is between $l$ and $l+\mathrm{d}l$})}{\mathrm{d}l} = \frac{\mathrm{d}}{\mathrm{d}l} \mathbb{P}(L \leq l) = \frac{1}{l_{\text{MFP}}} e^{-l/l_{\text{MFP}}} $$
for $l > 0$. This coincides with what is explained in OP's link. Also, the the probability of collision,
$$ \mathbb{P}(L \leq l) = 1 - e^{-l/l_{\text{MFP}}}, $$
as a function of $l / l_{\text{MFP}}$ then looks like:
1) Mathematically, this is the same as assuming that other particles are located according to the Poisson point process in $\mathbb{R}^3$ with intensity $n_v$.