Two subsets A and B of a set S consisting of ‘n’ elements are constructed randomly. The probability that $A \cap B=\phi$ and $A \cup B=S$ is equal to

$\frac{1}{2^n}$

Let ‘A’ has ‘r’ elements, then ‘B’ have all the remaining (n – r) elements and none of the elements that are already present in A.

Thus, total number of favourable ways

$=\sum\limits_{r=0}^n{ }^n C_r=2^n$

Hence, required probability = $\frac{2^n}{4^n}=\frac{1}{2^n}$