= Solution
For the reduced <quantum harmonic oscillator>, $V(q)=q^2/2$ and the ground-state <energy> is $E_0=1/2$. Thus the <classical turning points> satisfy $q^2/2=1/2$, giving $q=\pm1$. Normalizing the <wavefunction> gives $\phi_0(q)=\pi^{-1/4}e^{-q^2/2}$, because $\int_{-\infty}^{\infty}e^{-q^2}\,dq=\sqrt\pi$. The <Born rule> therefore gives the <harmonic-oscillator forbidden-region probability>
$$
P(|q|>1)=\frac{2}{\sqrt\pi}\int_1^\infty e^{-q^2}\,dq
\simeq\frac{2(0.1394)}{\sqrt\pi}=\boxed{0.1573\simeq15.7\%}.
$$
The factor two includes both tails. Using the exact <complementary error function> gives $\operatorname{erfc}(1)=0.1572992$. A stationary real <wavefunction> has zero <probability current>, so this result describes forbidden-region occupancy, not continuous loss of the bound particle.
\Image[/past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2b-oscillator-tails.png]
{title=Normalized ground-state probability with both classically forbidden tails shaded}
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