= Solution
{c}
Expand the given <pure state> in the ordered <computational basis> $00,01,10,11$:
$$
|\psi\rangle=\frac1{\sqrt3}|00\rangle-\frac1{\sqrt6}|01\rangle
+\frac1{\sqrt3}|10\rangle+\frac1{\sqrt6}|11\rangle,
\qquad
v=\frac1{\sqrt6}\begin{pmatrix}\sqrt2\\-1\\\sqrt2\\1\end{pmatrix}.
$$
The coefficients have squared magnitudes $1/3,1/6,1/3,1/6$, summing to one. The <density matrix> is the <outer product> $vv^\dagger$, not the diagonal matrix of these four <probabilities>: its off-diagonal entries retain <quantum coherence>. Therefore
$$
\boxed{\rho=|\psi\rangle\langle\psi|
=\frac16\begin{pmatrix}
2&-\sqrt2&2&\sqrt2\\
-\sqrt2&1&-\sqrt2&-1\\
2&-\sqrt2&2&\sqrt2\\
\sqrt2&-1&\sqrt2&1
\end{pmatrix}.}
$$
The <outer product> is a <Hermitian matrix> and is <positive semidefinite>: $u^\dagger\rho u=|v^\dagger u|^2\geq0$. Its <trace> is one, and $\rho^2=\rho$, as required for a <pure state>.
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