= Solution
{c}
Expand the possibly unnormalized second-<qubit> kets as $|\alpha\rangle=a_0|0\rangle+a_1|1\rangle$ and $|\beta\rangle=b_0|0\rangle+b_1|1\rangle$. The joint <quantum state> is
$$
|\psi\rangle=a_0|00\rangle+a_1|01\rangle+b_0|10\rangle+b_1|11\rangle.
$$
The <Born rule> in the orthonormal <computational basis> gives $P_{00}=|a_0|^2$, $P_{01}=|a_1|^2$, and similarly for the two outcomes beginning with one. Ignoring the second outcome therefore gives
$$
\boxed{P_0=|a_0|^2+|a_1|^2=\langle\alpha|\alpha\rangle,\qquad
P_1=|b_0|^2+|b_1|^2=\langle\beta|\beta\rangle.}
$$
Equivalently these are the squared norms of $(|0\rangle\langle0|\otimes I)|\psi\rangle$ and $(|1\rangle\langle1|\otimes I)|\psi\rangle$. Orthogonality of the first <qubit>'s basis gives $P_0+P_1=\langle\psi|\psi\rangle=1$, even if $\langle\alpha|\beta\rangle$ is nonzero. For a nonzero-probability outcome, the conditional second-<qubit> <quantum state> is the corresponding ket divided by $\sqrt{P_i}$.
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