 Solution

Expand the possibly unnormalized second-qubit kets as $|\alpha ⟩=a_0|0⟩+a_1|1⟩$ and $|\beta ⟩=b_0|0⟩+b_1|1⟩$. The joint quantum state isThe 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 givesEquivalently these are the squared norms of $(|0⟩⟨0|\otimes I)|\psi ⟩$ and $(|1⟩⟨1|\otimes I)|\psi ⟩$. Orthogonality of the first qubit's basis gives $P_0+P_1=⟨\psi |\psi ⟩=1$, even if $⟨\alpha |\beta ⟩$ is nonzero. For a nonzero-probability outcome, the conditional second-qubit quantum state is the corresponding ket divided by $\sqrt{P_i}$.