There is some freedom in deciding which particle in a particle-antiparticle pair is called "matter" and which is called "antimatter" but the freedom is smaller than you think. A basic problem is that your sentence
Antimatter of course annihilates ordinary matter, but the more precise statement is that antiparticles annihilate the same types of particles.
isn't really true. In fact, the opposite statement, while inaccurate, is much closer to the truth: matter and antimatter often can annihilate even if they belong to different species.
For example, a proton will rapidly annihilate with an antineutron (or an up-quark with anti-down-quark, if we look at the same process at the quark level), leaving some positron and neutrino (whose rest mass is much lower than the rest mass of either proton or antineutron) with lots of energy.
The up-quark and down-quark are different species or flavors but it would make no sense to call one of them "matter" and the other "antimatter" because they can "almost annihilate" to "almost nothing". More generally, all quark flavors are similar and it's better to call all of them "matter", especially because they may be related by symmetries that don't have a reason to include charge conjugation C.
Now, the atoms are composed of protons and electrons – and we call the atomic bound states "matter". That implies that an electron has the same "pro-anti" label as the six quarks. There are no bound states between positrons and protons so there would be no atomic "matter" if you flipped the convention for electrons but not protons.
Grand unified theories actually do link some 2-component spinors to larger representations and these representation contain fields that create both matter and antimatter so the binary label "pro-anti" becomes more subtle in such theories. We must still carefully distinguish a field and its Hermitian conjugate.
The "pro-anti" dichotomy remains meaningless for some particles, anyway. There are totally neutral particles – photons, Z-bosons, gluons, gravitons – that are identical to their antiparticles so here there is no "polarization", of course. There are also charged particles, W-bosons, for which it makes no sense to ask which of them is matter and which of them is antimatter. A W-boson may decay to a quark-antiquark pair so it's equally "far" from matter as it is from antimatter.
Neutrinos seem to be Majorana particles so far so they are identical to their antiparticles, too. However, the helicity (left-handed, right-handed) is correlated with the usual labels "pro-anti" which means that we can distinguish matter from antimatter, after all. There can also be right-handed neutrinos in which case the separation of neutrinos to "matter" and "antimatter" is exactly as possible (in principle) as it is for electrons and positrons.
Yes, to some extent. Once you choose which of the electron or positron is to be considered the normal particle, then that fixes your choice for the other leptons, because of neutrino mixing. Similarly, choosing one quark to be the normal particle fixes the choice for the other flavors and colors of quarks. But I can't think of a reason within the standard model that requires you to make corresponding choices for leptons and quarks.
In particle terms, you can think about it like this: say you start by choosing the electron to be the particle and the positron to be the antiparticle. You can then distinguish electron neutrinos and electron antineutrinos because in weak decay processes, an electron is always produced with an antineutrino and a positron with a normal neutrino. Then, because of neutrino oscillations, you can identify the other two species of neutrinos that oscillate with electron antineutrinos as antineutrinos themselves, and in turn you can identify the muon and tau lepton from production associated with their corresponding antineutrinos.
In terms of QFT, the relevant (almost-)conserved quantity is the "charge parity," the eigenvalue of the combination of operators $\mathcal{CP}$.
Best Answer
Those neutrinos come from the Sun. Fusion converts protons to neutrons, so that must produce neutrinos. One can calculate the number of nuclear reactions necessary for the power output, and get a number for the neutrino flux.
One can also estimated the flux from the cross section of the detector.
The two rates differ by a factor of about three. That was resolved by the neutrino oscillations between the three flavors (electron, muon and tauon neutrinos).