How do you interpret fluxes derived from Monte-Carlo simulations, specifically neutrino fluxes coming from the Sun I'm reading the neutrino oscillations paper from the SNO and I'm having a bit of trouble interpreting their results on page 4.  
I understand that they fitted PDFs to the data to get an amount of events and subsequently the fluxes, but how do you interpret these fluxes?
The CC channel is only sensitive to electron-neutrinos, so it makes sense that this is just the electron-neutrino flux coming from the Sun.
The NC channel is equally sensitive to all neutrino flavours, so it also makes sense to interpret this as the total neutrino flux coming from the Sun.
This leaves us with the ES channel. It is more sensitive to electron neutrinos than to muon and tau neutrinos, but it's still sensitive to all flavours, so why doesn't this also give us the total flux coming from the Sun?
To clarify:
A given electron, muon or tau neutrino has a certain chance to be detected by either CC, NC or ES.
Conversely:
If a CC, NC or ES event occurs, there's a certain chance that it's an electron, muon or tau neutrino.
How do you translate these statements into interpretations for the ES flux?
 A: 
Question: "To clarify: A given electron, muon or tau neutrino has a certain chance to be detected by either CC, NC or ES.
Conversely: If a CC, NC or ES event occurs, there's a certain chance that it's an electron, muon or tau neutrino.
How do you translate these statements into interpretations for the ES flux?

and

Comment to YSelf: "I don't see how that leads you to conclude that the amount of elastic scattering events is the amount of electron neutrinos + 1/3 (muon neutrinos + tau neutrinos).".

The paper you refer to, refers us to [2] Q.R. Ahmad et al., Phys. Rev. Lett. 87, 071301 (2001), on page 5 it explains:

"The best fit to $\phi(ν_{µτ})$ is: $\phi(ν_{µτ}) = 3.69 ± 1.13 × 10^6 cm^{−2} s^{−1}$, and $\phi\begin{smallmatrix}{}SK\\ES\end{smallmatrix} = \phi(ν_e) + 0.154  \;\phi(ν_{µτ})$.
Abstract
Solar neutrinos from the decay of $^8\text{B}$ have been detected at the Sudbury Neutrino Observatory (SNO) via the charged current (CC) reaction on deuterium and by the elastic scattering (ES) of electrons. The CC reaction is sensitive exclusively to $ν_e$’s, while the ES reaction also has a small sensitivity to $ν_µ$’s and $ν_τ$’s. The flux of $ν_e$’s from $^8\text{B}$ decay measured by the CC reaction rate is $\phi^{CC}(ν_e) = 1.75 ± 0.07$ (stat.) $\begin{smallmatrix}+0.12 \\ −0.11\end{smallmatrix}$ (sys.) $± 0.05$ (theor.) $× 10^6 cm^{-2} s^{−1}$. Assuming no flavor transformation, the flux inferred from the ES reaction rate is $\phi ^{ES}(ν_x) = 2.39 ± 0.34$ (stat.) $\begin{smallmatrix}+0.16 \\ −0.14\end{smallmatrix}$ (sys.) $× 10^6 cm^{-2} s^{−1}$. Comparison of $\phi ^{CC}(ν_e)$ to the Super-Kamiokande Collaboration’s precision value of $\phi ^{ES}(ν_x)$ yields a $3.3σ$ difference, assuming the systematic uncertainties are normally distributed, providing evidence that there is a non-electron flavor active neutrino component in the solar flux.
The total flux of active $^8\text{B}$ neutrinos is thus determined to be $5.44 ± 0.99 × 10^6 cm^{-2} s^{−1}$, in close agreement with the predictions of solar models.

A: SNO measures the number of events in three categories: $n_{CC}$, $n_{NC}$ and $n_{ES}$, which relate to the number of events per particle type $n_{\nu_e}$, $n_{\nu_\mu}$ and $n_{\nu_\tau}$:
CC works only with $\nu_e$:
$n_{CC} = \nu_e$.
NC works with all three kinds (with the same probability):
$n_{NC} = \nu_e + \nu_\mu + \nu_\tau$.
ES also works with all three kinds, but with different probabilities:
$n_{ES} = \nu_e + \nu_\mu/3 + \nu_\tau/3$.
Now, $\nu_\mu$ and $\nu_\tau$ are indistinguishable, so simplify to $\nu_e$ and $\nu_{\mu,\tau}$ and solve the (overdetermined) equations for both.
This is graphically represented by Fig. 3 on page 5.
The reason for the 1/3 probability is the number of possible channels. Simply try to draw the Feynman diagrams for ES, and you will see that two out of three require an electron neutrino.
Or maybe I can look those diagrams up, but not right now.
