= Solution
The <Mott two-current model> treats the two <spin> populations as parallel conduction channels when <spin-flip scattering> is weak. Thus $\sigma=\sigma_\uparrow+\sigma_\downarrow$ and $\rho^{-1}=\rho_\uparrow^{-1}+\rho_\downarrow^{-1}$. Light, broad-band carriers conduct efficiently; scattering into narrow $d$-like states relaxes their current. In a simple spin-conserving $s$–$d$ picture,
$$
\tau_\sigma^{-1}\propto|V_{sd,\sigma}|^2D_{d,\sigma}(E_F),\qquad
\sigma_\sigma\simeq\frac{e^2}{3}D_\sigma(E_F)\langle v_\sigma^2\tau_\sigma\rangle.
$$
In <copper>, the filled $d$ bands supply little final-state phase space at $E_F$, so both spin channels have low resistivity. In <cobalt>, the nearly filled majority $d$ band suppresses majority-channel scattering, while minority $d$ states remain available. The stated <iron> reversal corresponds to stronger scattering-weighted majority-channel phase space: its majority $d$ band is not closed as in cobalt. Band filling, exchange splitting, velocities and scattering matrix elements set this <scattering-weighted spin-channel asymmetry>; majority population alone does not select the better conductor. \b[The given iron ordering is a material/scattering input, not a universal consequence of the two-current postulate.] Different impurities and interfaces can change the asymmetry.<Spin dependent transport: GMR & TMR by Alain Schuhl and Daniel Lacour>[§2.1, printed p.947: intrinsic transport weights and impurity-dependent asymmetry]{ref}
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