A is a set containing n elements. A subset P of A is chosen at random…

The set A has n elements. So, it has $2^n$ subsets. Therefore, set P can be chosen in ${^{2n}C}_1$ ways. Similarly, set Q can also be chosen in ${^{2n}C}_1$ ways.

∴ Sets P and Q can be chosen in ${^{2n}C}_1 × {^{2n}C}_1= 2^n  ×2^n = 4^n $ ways,

Let the subset P of A contains r elements, with 0 ≤ r ≤ n. Then, the number of ways of choosing P is ${^nC}_r$. The subset Q of P can have at most r elements and the number of ways of choosing Q is $2^r$. Therefore, the number of ways of choosing P and Q is ${^nC}_r ×2^r$ when P has r elements. So, P and Q can be chosen in general in

$\sum\limits^{n}_{r=0} {^nC}_r  × 2^r = (1+2)^n = 3^n $ ways

Hence, required probability $=\frac{3^n}{4^n}=\left(\frac{3}{4}\right)^n$