= Solution
For a uniformly distributed pure <qubit> on the <Bloch sphere>, rotational symmetry gives $\langle r_z^2\rangle=1/3$. Thus
$$
\boxed{\overline F=\frac12\left(1+\frac13\right)=\frac23.}
$$
Equivalently, $p=|\alpha|^2$ is uniform on $[0,1]$ under this measure and $\int_0^1[p^2+(1-p)^2]dp=2/3$. The three positive coordinate-axis states mentioned in the source give $F_Z=1$, $F_X=F_Y=1/2$, whose average is the same because this <quantum fidelity>[fidelity] depends only on the quadratic component $r_z^2$. This is a specified uniform-state average, not a statement about every possible prior ensemble.
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