Question
Let $$S$$ be the universal set and $$n\left( X \right) = k.$$ The probability of selecting two subsets $$A$$ and $$B$$ of the set $$X$$ such that $$B = \overline A $$ is :
A.
$$\frac{1}{2}$$
B.
$$\frac{1}{{{2^k} - 1}}$$
C.
$$\frac{1}{{{2^k}}}$$
D.
$$\frac{1}{{{3^k}}}$$
Answer :
$$\frac{1}{{{2^k} - 1}}$$
Solution :
The total number of subsets of $$X$$ is $${2^k}.$$ So, $$n\left( S \right) = {}^{{2^k}}{C_2}.$$
$$n\left( E \right) = $$ the number of selections of two nonintersecting
subsets whose union is $$X$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\left( {{}^k{C_0} + {}^k{C_1} + {}^k{C_2} + .....} \right)$$
( $$\because $$ the number of selections in which one subset has $$r$$ elements and the rest are in the other subset $$ = {}^k{C_r}$$ and every selection appears twice in the total number of selections ).
$$\therefore \,P\left( E \right) = \frac{{\left( {\frac{1}{2}} \right){{.2}^k}}}{{{}^{{2^k}}{C_2}}} = \frac{{{2^{k - 1}}}}{{\frac{{{2^k}\left( {{2^k} - 1} \right)}}{2}}} = \frac{1}{{{2^k} - 1}}.$$