Plz explain in detail with basics

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{toal} \mathrm{subset} \mathrm{of} \mathrm{A}=2^{\mathrm{n}} \\ \mathrm{Now} \\ \mathrm{We} \mathrm{need} \mathrm{to} \mathrm{choose} \mathrm{two} \mathrm{subsets} \mathrm{out} \mathrm{of} 2^{\mathrm{n}}{=}^{2^{\mathrm{n}}}{\mathrm{C}}_2 \\ =\frac{2^{\mathrm{n}}\left(2^{\mathrm{n}}-1\right)}{2\times 1} \\ =2^{\mathrm{n}-1}\left(2^{\mathrm{n}}-1\right) \\ \mathrm{We} \mathrm{need} \\ \mathrm{P}=\mathrm{Q}\text{'} \\ \mathrm{Q}\text{'}=\mathrm{All} \mathrm{elements} \mathrm{of} \mathrm{A} \mathrm{which} \mathrm{are} \mathrm{not} \mathrm{in} \mathrm{Q} \\ \mathrm{So} \mathrm{we} \mathrm{need} \mathrm{to} \mathrm{choose} \mathrm{Q} \mathrm{and} \mathrm{put} \mathrm{remaining} \mathrm{elements} \mathrm{of} \mathrm{A} \mathrm{in} \mathrm{P}. \\ \mathrm{Number} \mathrm{of} \mathrm{ways} \mathrm{of} \mathrm{choosing} \mathrm{P} \mathrm{and} \mathrm{Q} \left(\mathrm{subsets}\right) \mathrm{is} 2^{\mathrm{n}} \\ \mathrm{But} \mathrm{every} \mathrm{selection} \mathrm{of} \mathrm{P}\hspace{0.17em}\mathrm{and} \mathrm{Q} \mathrm{appears} \mathrm{twice}, \mathrm{hence} \mathrm{total} \mathrm{number} \mathrm{o} \\ \mathrm{ways} \mathrm{of} \mathrm{choosing} \mathrm{P} \mathrm{and} \mathrm{Q}=\frac{2^{\mathrm{n}}}{2}=2^{\mathrm{n}-1} \\ \mathrm{Required} \mathrm{Probabilty}=\frac{2^{\mathrm{n}-1}}{2^{\mathrm{n}-1}\left(2^{\mathrm{n}}-1\right)}=\frac{1}{2^{\mathrm{n}}-1} \left[\mathrm{Answer}\right] \end{gathered}$$

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