Please help me with comprehension 1

Dear Student,

Please find below the solution to the asked query:
$$\begin{gathered} \left(1\right) \\ \mathrm{Let} \mathrm{the} \mathrm{side} \mathrm{of} \mathrm{cube} \mathrm{be} \mathrm{a} \\ \mathrm{Then} \mathrm{radius} \mathrm{of} \mathrm{sphere} \mathrm{is} \frac{\sqrt{3}}{2}\mathrm{a} \\ \mathrm{Now} \mathrm{Probabolity}=\frac{\mathrm{V}\left(\mathrm{sphere}\right)-\mathrm{V}\left(\mathrm{cube}\right) \\ }{\mathrm{V}\left(\mathrm{sphere}\right)} \\ \mathrm{P}=1-\frac{\mathrm{V}\left(\mathrm{cube}\right)}{\mathrm{V}\left(\mathrm{sphere}\right)}=1-\frac{{\mathrm{a}}^3}{\left({\displaystyle \frac{4}{3}{\mathrm{\pi r}}^3}\right)}=1-\frac{3{\mathrm{a}}^3}{4\mathrm{\pi }{\left({\displaystyle \frac{\sqrt{3}\mathrm{a}}{2}}\right)}^3}=1-\frac{3{\mathrm{a}}^3}{4\mathrm{\pi }{\displaystyle \frac{3\sqrt{3}{\mathrm{a}}^3}{8}}} \\ \Rightarrow \mathrm{P}=1-\frac{2}{\mathrm{\pi }\sqrt{3}} \mathrm{Answer} \mathrm{OptionA} \end{gathered}$$

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