For s-orbital, since ($\mathrm{\psi }$ orbital wave function) is independent of angles, the probability density $\left({\mathrm{\psi }}^2\right)\mathrm{is}$

For spherical orbital

$\mathrm{\psi }=\mathrm{f}\left(\mathrm{r}\right)$ (r = distance from nucleus)

since $\mathrm{\psi }$ (orbital) is independent of angles therefore probability $\left({\mathrm{\psi }}^2\right)$ also independent of angles and spherically symmetric.