A is a set containing n-elements A subset P₁ of A is choosen. The set A is reconstructed by replacing the elements of P₁. Next a subset P₂ is constructed in the similar way. In this similar way subsets P₃, P₄.....Pm are constructed then the probability that the sets P₁, P₂.....Pm are constructed such that

If ${\mathrm{a}}_1\in \mathrm{A}$ Then${\mathrm{a}}_1\in {\mathrm{P}}_1,{\mathrm{a}}_1\notin {\mathrm{P}}_1\to 2 \mathrm{ways}$
$\begin{array}{l}{\mathrm{a}}_1\in {\mathrm{P}}_2,{\mathrm{a}}_1\notin {\mathrm{P}}_2\to 2\text{ ways }\\ {\mathrm{a}}_1\in {\mathrm{P}}_{\mathrm{m}},{\mathrm{a}}_1\notin {\mathrm{P}}_{\mathrm{m}}\to 2\text{ ways }\end{array}$
$\therefore {\mathrm{P}}_1{\mathrm{P}}_2\dots .{\mathrm{P}}_{\mathrm{m}}$ can be constructed in 2m ways When A–contains n-elements P₁, P₂,...P_m can be constructed in (2^m)ⁿ = 2^mn = n(s)
A) ${\mathrm{P}}_1\cap {\mathrm{P}}_2\cap \dots \cap {\mathrm{P}}_{\mathrm{m}}=\mathrm{\varphi }$
$\begin{array}{l}{\mathrm{a}}_1\notin {\mathrm{P}}_1\cap {\mathrm{P}}_2\cap \dots \cap {\mathrm{P}}_{\mathrm{m}}\text{ in }\left(2^{\mathrm{m}}-1\right)\text{ ways }\\ {\mathrm{a}}_2\notin {\mathrm{P}}_1\cap {\mathrm{P}}_2\cap \dots \cap {\mathrm{P}}_{\mathrm{m}}\text{ in }\left(2^{\mathrm{m}}-1\right)\text{ ways }\\ {\mathrm{a}}_{\mathrm{n}}\notin {\mathrm{P}}_1\cap {\mathrm{P}}_2\cap \dots \cap {\mathrm{P}}_{\mathrm{m}}\text{ in }\left(2^{\mathrm{m}}-1\right)\text{ ways }\\ \therefore \mathrm{n}\left(\mathrm{E}\right)={\left(2^{\mathrm{m}}-1\right)}^{\mathrm{n}}\therefore \mathrm{P}\left(\mathrm{E}\right)=\frac{{\left(2^{\mathrm{m}}-1\right)}^{\mathrm{n}}}{2^{\mathrm{mn}}}\end{array}$
B) ${\mathrm{a}}_1\in {\mathrm{P}}_1\cup {\mathrm{P}}_2\cup \dots ..\cup {\mathrm{P}}_{\mathrm{m}} \mathrm{in} \left(2^{\mathrm{m}}-1\right)$ways
For i = 1 to n→P₁,P₂ .....Pm can be constructed such
that ${\mathrm{a}}_{\mathrm{i}}\in {\mathrm{P}}_1\cup {\mathrm{P}}_2\dots ..\cup {\mathrm{P}}_{\mathrm{m}} \mathrm{in} {\left(2^{\mathrm{m}}-1\right)}^{\mathrm{n}}$ways
$\mathrm{P}\left(\mathrm{E}\right)=\frac{{\left(2^{\mathrm{m}}-1\right)}^{\mathrm{n}}}{2^{\mathrm{mn}}}$
C) For $\mathrm{i}\ne \mathrm{j}$each element A either it does not belongs to any P₁, P₂....Pm or it belongs to at most one set  $\therefore$(m +1) possibilities for each element for
 n-elements →$(\mathrm{m}-1{)}^{\mathrm{n}}$
$\mathrm{P}\left(\mathrm{E}\right)=\frac{(\mathrm{m}+1{)}^{\mathrm{n}}}{2^{\mathrm{mn}}}$
$\text{ D) }{\mathrm{P}}_1\cup {\mathrm{P}}_2\cup \dots .{\mathrm{P}}_{\mathrm{m}}\text{ contain }\mathrm{r}{\text{ elements in }}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{\left(2^{\mathrm{m}}-1\right)}^{\mathrm{r}}$
$\mathrm{P}\left(\mathrm{E}\right)=\frac{{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{\left(2^{\mathrm{m}}-1\right)}^{\mathrm{n}}{\mathrm{n}}_{\mathrm{C}}}{2^{\mathrm{mn}}}$