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.

If P ∩ Q contains exactly one element, both P and Q must be non-empty. Thus, if P has r elements, Q must have exactly one of these r elements and any number of elements from among the remaining (n-r) elements in A, so that the number of ways of choosing Q is ${^rC}_1 ×2^{n-r}$.

But, P can be chosen in ${^nC}_r$ ways. Therefore, P and Q can be chosen in ${^rC}_r × {^rC}_1×2^{n-r}$ when P contains r elements.

Therefore, P and Q in general can be chosen in

$\sum\limits^{n}_{r=1}×{^nC}_r × {^rC}_1×2^{n-r}$  ways $= \sum\limits^{n}_{r=1}{^nC}_r \,\, r \,\, 2^{n-r}$ ways

$= \sum\limits^{n}_{r=1}\frac{n}{r}{^{n-1}C}_{r-1}.r. 2^{n-r}$             $[∵ {^nC}_r =\frac{n}{r} {^{n-1}C}_{r-1}]$

$= n\sum\limits^{n}_{r=1}{^{n-1}C}_{r-1} 2^{n-r}$

$= n\sum\limits^{n}_{r=1}{^{n-1}C}_{r-1} 2^{(n-1)-(r-1)}=n(1+2)^{n+1}= n3^{n-1}$

Hence, required probability $=\frac{n×3^{n-1}}{4^n}=\frac{n}{3}\left(\frac{3}{4}\right)^n$