Neutrino event classification model ($K\sim \mathrm{Bin}(N,q)$)

How can the relative abundance of tracks and showers distinguish hypotheses? Consider an original teaching model, with chosen inputs throughout. Condition on a sample of $N$ selected events, each assigned exactly one of two labels: track or shower. Multitrack events and unclassified events lie outside this model. A shower label need not identify an Electron; background can imitate it. Conversely, a true Electron interaction need not receive a shower label.

Let $b$ be the background fraction among selected events. Among the selected signal events, let $f$ be the fraction whose associated charged lepton is an Electron, with the remainder associated with a muon. These are fractions after event selection, rather than fractions of the incident neutrino particle flux. Define the conditional probabilitiesHere $a$ and $r$ describe classification within the selected sample. They are not measured historical detection efficiencies. The law of total probability gives the overall shower probabilityThis formula is the bridge between an interaction hypothesis and a visible label. It keeps beam contamination, background and mistaken classification conceptually distinct: genuine electron neutrino interactions belong in $f$, while non-signal events belong in $b$.

Assume independent random variables for the event labels, all with this same $q$. If $K$ counts shower labels, every specified pattern with $k$ showers has probability $q^k(1-q{)}^{N-k}$. Counting those patterns with a binomial coefficient yields the binomial distributionThe track count is $N-K$, so its expected value is $N(1-q)$. At fixed $N$, the two counts cannot fluctuate independently. The variance above measures sampling fluctuations under the stipulated model, not uncertainty in $a$, $r$, $b$ or $q_b$.

Choose $N=40$, $a=0.90$, $r=0.05$, $b=0.10$ and $q_b=0.40$. Compare a signal with only muon-associated interactions ($f=0$) against one with equal selected Electron- and muon-associated interactions ($f=1/2$). These give $q=0.085$ and $q=0.4675$, respectively. Their expected shower counts are $3.4$ and $18.7$; their expected track counts are $36.6$ and $21.3$. A background-only model with the specified $q_b$ gives $16$ showers and $24$ tracks on average. These are expectations, so they need not be integers.

For a hypothetical observation $K=4$, the likelihood ratio of the two signal models isThe binomial coefficient cancels because the same hypothetical sample is used in both likelihoods. This ratio is not the probability that either hypothesis is true. A separate original check gives $\mathrm{Pr}(K\le 5)\simeq 0.880$, $4.57\times 10^{-6}$ and $1.44\times 10^{-4}$ for the muon-associated, equal-mixture and specified background-only models, respectively.

Why are the controls essential? If the background-only model may choose any $q_b$, taking $q_b=K/N$ can fit the observed shower fraction. Counts alone then fail to distinguish it from a signal model. Independent background measurements and calibrated particle identification constrain that freedom. Changing the decay path additionally tests where the candidate-producing particles originate. The power of the experiment comes from combining apparatus interventions with event classification; the small binomial distribution model isolates only the latter part of that reasoning.