= Solution
In the simple transport model, $\mu_e=e\tau_e/m_e^*$ and $\mu_h=e\tau_h/m_h^*$. Conduction bands in many common semiconductors have lighter transport masses than the dominant valence hole bands. For comparable momentum-relaxation times, this gives larger <electron mobility>. It is a common trend, not a theorem: band anisotropy, light and heavy hole populations and different scattering times can alter the comparison. The hole's <mass> is not the <mass> of an atom, and crystal translational symmetry does not make its mobility infinite.
For the intrinsic <germanium> sample, <electrons> and holes both contribute positive conventional conductivity:
$$
\sigma=e(n\mu_e+p\mu_h)=en_i(\mu_e+\mu_h),\qquad\sigma=1/\rho=2\,\mathrm{S\,m^{-1}}.
$$
Therefore
$$
\boxed{n=p=n_i=\frac1{(0.5)e(0.36+0.18)}
=2.31\times10^{19}\,\mathrm{m^{-3}}.}
$$
This is the density of each species; the combined carrier <number density> is $2n_i$. Assigning separate densities from the two mobilities would violate intrinsic charge neutrality.
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