The first orbital of H is represented by $\psi =\frac{1}{\sqrt{\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}{e}^{-1/a_0},where{a}_0=\text{\hspace{0.17em}Bohr's\hspace{0.17em}orbit}$ The probability of  finding the electron at a distance r from the nucleus is 

The probability of the electron being in an infinitesimal region dv at distance r is given as probability ${\text{=\hspace{0.17em}ψ}}^{\text{2}}\text{\hspace{0.17em}dv}$$\because$1^st orbitals is spherically symmetrical

$\therefore \text{\hspace{0.17em}V\hspace{0.17em}=\hspace{0.17em}}\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}{\text{\hspace{0.17em}or\hspace{0.17em}dV\hspace{0.17em}=\hspace{0.17em}4πr}}^{\text{2}}\text{dr}$

$\therefore {\text{\hspace{0.17em}Probability\hspace{0.17em}=\hspace{0.17em}ψ}}^{\text{2}}{\text{.\hspace{0.17em}4πr}}^{\text{2}}\text{dr}$