= Solution
Measure the first <qutrit> in the <computational basis> and the second in the <orthonormal basis> $\{|\phi_0\rangle,|\phi_1\rangle,|\phi_2\rangle\}$ from Question 1(a), with those labels as outcomes. For first-party outcome 0, the unnormalized conditional <vector> of the second party is $|0\rangle/\sqrt3$, so this outcome occurs with <probability> $1/3$ and leaves the normalized state $|0\rangle$.
The second-party <Born rule> <probabilities> are then $|\langle\phi_0|0\rangle|^2=1/2$, $|\langle\phi_1|0\rangle|^2=0$, and $|\langle\phi_2|0\rangle|^2=1/2$. Hence
$$
\boxed{\Pr(B=0\mid A=0)=\Pr(B=2\mid A=0)=\tfrac12,\qquad\Pr(B=1\mid A=0)=0.}
$$
The relevant joint <probabilities> are $1/6,0,1/6$, each divided by $\Pr(A=0)=1/3$. This distinguishes the conditional statement from an unconditional outcome <probability>. More generally, <conjugate-state steering from a maximally entangled pair> explains how the first local measurement determines the second party's conditional <vector>; the <basis> choice here makes that <vector> especially simple.
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