 Solution

The basic Feynman diagrams have ${\tau }^{-}\to {\nu }_{\tau }{W}^{-\ast }$ followed by $W^{-\ast }\to e^{-}{\bar{\nu }}_e$, ${\mu }^{-}{\bar{\nu }}_{\mu }$, or $d\bar{u},s\bar{u}$, with the latter pairs undergoing hadronization. See the charged-current diagrams in a. Neglect all daughter masses, QCD corrections and inaccessible charm final states, and assume lepton universality. Each leptonic mode then has one unit of decay width; the two accessible quark combinations have three color charge states and total CKM weight $|V_{ud}{|}^2+|V_{us}{|}^2\simeq 1$. ThereforeThe result is an inclusive parton model estimate. Perturbative QCD corrections increase the hadronic contribution, while muon mass, nonperturbative dynamics and radiative corrections also matter; the physical fraction is roughly $0.65$.

Every ordinary hadronic mode already contains a tau neutrino, so it contains a lepton as well as hadrons if invisible particles are counted. For an additional visible charged lepton, a virtual photon can internally convert in a hadronic mode, for example ${\tau }^{-}\to {\nu }_{\tau }{\pi }^{-}{\gamma }^{\ast }\to {\nu }_{\tau }{\pi }^{-}{e}^{+}{e}^{-}$. The extra electromagnetic vertices suppress such modes; they are not a large fraction of the inclusive width. Alternatively a daughter pion or kaon can decay semileptonically or leptonically in flight, for example ${\tau }^{-}\to {\nu }_{\tau }{\pi }^{-}{K}^0$ followed by a semileptonic neutral-kaon decay. Those are secondary decay chains, not independent primary tau lepton branching fractions. Lepton-flavour-violating neutrinoless modes such as ${\tau }^{-}\to {\mu }^{-}{\pi }^{+}{\pi }^{-}$ do not contribute appreciably in the Standard Model.