 Solution

Squaring the states gives $\boxed{p_0=e^{-x^2/{\sigma }^2}/(\sqrt{\pi }\sigma ),p_1=2x^2{e}^{-x^2/{\sigma }^2}/(\sqrt{\pi }{\sigma }^3)}$. The first peaks at zero; the second has a node there and peaks at $x=\pm \sigma$. For a centered interval of width $\delta x\ll \sigma$, $P_0\simeq \delta x/(\sqrt{\pi }\sigma )$, whereas $P_1\simeq (\delta x{)}^3/(6\sqrt{\pi }{\sigma }^3)$. Its leading linear term vanishes, but the probability in a finite interval is not exactly zero.