Question

Let X be a universal set such that n (X) = 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 1

The correct option is B $\frac{1}{(2^k-1)}$

Number of selection of two subsets =$(2^k{)}_{c_2}$
Let the event E = selection of two subsets A and B such that $A=\overline{B}$ i.e., $A\cap B=\varphi$ and $A\cup B=X$,
n(X) = k
Then $n\left(E\right)=\frac{1}{2}{(}^k{C}_0{+}^k{C}_1{+}^k{C}_2+....{+}^k{C}_k)=\frac{1}{2}{.2}^k$.
$\therefore$ Probability $=\frac{\frac{1}{2}{.2}^K}{\frac{1}{2}{.2}^k.(2^k-1)}=\frac{1}{2^k-1}$