= Solution
Write the proposed <quantum state ensemble> as $\{(|u\rangle,p),(|v\rangle,1-p)\}$ with normalized <pure states>, so that
$$
\frac I2=p|u\rangle\langle u|+(1-p)|v\rangle\langle v|.
$$
Taking the <expectation values> in $|u\rangle$ and $|v\rangle$, and setting $c=|\langle u|v\rangle|^2$, gives
$$
\frac12=p+(1-p)c\geq p,
\qquad
\frac12=pc+1-p\geq1-p.
$$
Consequently $p\leq1/2$ and $p\geq1/2$. Substitution then gives $c=0$. The two <pure states> are therefore <orthogonal>, with equal <probabilities>:
$$
\boxed{p=\frac12,\qquad |v\rangle=|u^\perp\rangle\ \text{up to a global phase}.}
$$
Conversely, any <orthonormal basis> of a <qubit> has projectors summing to $I$, so such a <quantum state ensemble> does give $I/2$. This proves the characterization of <two-state ensembles of a maximally mixed qubit>.
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