= Solution
Use the normalized $L_3$ <eigenstates> from part (ii). The <quantum state> found in part (iii) has the decomposition
$$
\psi_x=\sqrt{\frac3{4\pi}}\frac{\sin\theta e^{i\phi}+\sin\theta e^{-i\phi}}2=\frac{\chi_{+1}+\chi_{-1}}{\sqrt2}.
$$
There is no $\chi_0$ component. Equivalently, its <inner product> with $\chi_0$ contains $\int_0^{2\pi}\cos\phi\,d\phi=0$. The <Born rule> therefore gives
$$
\boxed{\Pr(L_3=0)=|\langle\chi_0,\psi_x\rangle|^2=0.}
$$
For completeness the only outcomes are $L_3=\pm\hbar$, each with <probability> $1/2$. Thus $\langle L_3\rangle=0$ does not mean that zero is a possible result in this state: its <quantum uncertainty> in $L_3$ is $\hbar$, even though $L_1$ is exactly zero.
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