Relationships between optical depth, $\tau$, and probability of being scattered The optical depth is given by many as:
$$
\tau=n_e\sigma_{cs}L
$$
Where $n_e$ is the number density of the medium, $\sigma_{cs}$ is the scattering cross section and L is the linear extent of the medium. 
I was wondering, what is the relationship between the optial depth and the probability of scattering? I can't seem to find it amongst online lecture notes and other sources. A really concise answer would be appeciated. 
 A: The probability of light getting to an optical depth $\tau$ is $\exp(-\tau)$. So the probability of it being (singly) scattered would $1 - \exp(-\tau)$.
A: The quantity $$\ell=\frac{1}{n_{\mathrm e}\sigma_{sc}}.\tag1$$
is called the mean free path. It is the average distance a photon travels between two scattering events. To understand this definition, consider a  photon travelling in the medium. When the photon travels a distance $x$, it will interact with a scatterer of cross-section $\sigma_{sc}$ if this scatterer is in the volume $\sigma_{sc} x$. The average number of scatterers in this volume is $N(x)=n_{\mathrm{e}}\sigma_{sc}x$. Therefore, the distance that the photon has covered to meet one scatterer is $\ell$ given by $N(\ell)=1$, which demonstrates formula (1).
When a photon travels a long distance $L$, the scatterieng events are independent. The average number of scattering event is given by our preceding law: $N(L)=n_{\mathrm e}\sigma_{sc}L=L/\ell=\tau$. The number of scattering events is therefore is distributed according a Poisson law
$$\mathbb P(N)=\frac{\tau^N\,\mathrm e^{-\tau}}{N!}.$$
The probability of not being scattered is $\mathbb P(0)=\mathrm e^{-\tau}$, which decreases exponentially as the depth increases (intuitive). 
