= Solution
The first <Stern-Gerlach spin measurement> transmits $x+$ with <probability> $|\langle x+|\uparrow\rangle|^2=1/2$. The selected <quantum state> $|x+\rangle$ then yields $z-$ with conditional <probability> $|\langle\downarrow|x+\rangle|^2=1/2$. Relative to the original beam,
$$
\boxed{p(x+\text{ then }z-)=\frac12\cdot\frac12=\frac14.}
$$
Among particles reaching the second apparatus, the negative-$z$ fraction is $1/2$; the requested fraction uses the original intensity and is $1/4$.
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