= Solution
Taking the <outer product> for A and averaging the two <orthogonal projections> for B gives
$$
\boxed{\rho_A=\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}=\frac{I+\sigma_x}{2},\qquad
\rho_B=\frac12\begin{pmatrix}1&0\\0&1\end{pmatrix}=\frac I2.}
$$
The off-diagonal entries in $\rho_A$ retain <quantum coherence> between the $z$-basis states. They are absent from B's classical <quantum state ensemble>. In particular, $\operatorname{Tr}\rho_A^2=1$ and $\operatorname{Tr}\rho_B^2=1/2$: A is a <pure state>, B a <mixed state>.
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