A is a set containing n elements. A subset P of A is chosen at random. The set A is reconstructed by replacing the elements of P. A subset Q is again chosen at random. The probability that  Q contains just one element more than P, is

$\frac{^{2n}C_{n-1}}{4^n}$

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.

If P contains r elements, then Q must contain (r+1) elements. In this case the number of ways of choosing P and Q is ${^nC}_r × {^nC}_{r+1}$, where 0 ≤ r ≤n-1.

Thus, the number of ways of choosing P and Q in general, is

$\sum\limits^{n-1}_{r=0} {^nC}_r  × {^nC}_{r+1}  = {^{2n}C}_{n-1}$

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