I haven't had a course or any formal training in statistics so I apologize in advance for any butchered explanation/terminology.
Consider a particle at the origin pointing in some direction. It rests in a uniform magnetic field and given a kick. See image:
Unless $\theta = 0,\pi$, it will take a finite amount of time to hit one of the two represented planes. The particle will travel in the familiar helix pattern and as such two angles are unnecessary. We consider $\theta'$, the angle from the plane normals. Then the time of flight is easily found from the transverse velocity,
$$ t = \frac{d}{v}\frac{1}{\cos\theta'} $$
$d$ and $v$ are constant (to simplify things). The only distribution going into this is the emission angle (isotropic). I understand this to be generated as $\theta' = \arccos(1 - 2q)$ with the uniform $q:[0, 1)$.
The question at this point, is what is the PDF for $t$? I.e. what's the probability of a particle reaching a plane within some window $[t_i, \, t_i+dt]$?
Side comment: If anyone's got any good stat book recommendations, that'd be awesome.
Modify $T = \frac{d}{v |\cos \theta'|}$ because walls are symmetric in both directions
$$\theta' = \cos^{-1}(1-2q)$$
So
$$T = \frac{d}{v |1 - 2q|}$$
To avoid annoying absolute values, assume $q' \sim U(0, 0.5)$, because the other half produces exactly the same result, so has the same impact on the time distribution. So
$$T = \frac{d}{v (1 - 2q')}$$
According to Theorem 4.1, given a monotonic relationship between $T$ and $q$, the PDF of a function of a random variable can be written as
$$ \rho(T) = \rho(q') \bigl| \frac{dq'}{dT} \bigr| = \rho(q') \bigl| -\frac{v}{2d}(1-2q')^2 \bigr| $$
Finally, we must plug in explicit expression $q' = \frac{1}{2}\bigl (1 - \frac{d}{Tv} \bigr)$ to get
$$ \rho(T) = \rho (q') \frac{d}{2vT^2} = \frac{d}{vT^2} $$
with the constraint that $q' \in [0, 0.5]$, which translates to the constraint $T > T_{\min} = \frac{d}{v}$. The same theorem prescribes that $\rho(T) = 0$ whenever the constraint is not met, which makes sense, because it is not possible to reach the target faster.