Suppose we have n different elements say (x1,x2,x3,....................xn) and two subsets P and Q.
now either the chosen element will belong to p or not or it may belong to q or not.
means 1) x1 belongs to both P and Q.
2) to only P not Q.
3) to only Q not P.
4) none of P and Q.
so an element can be placed in 4 different ways in two sets.
so for n elements it can be done in 4n ways(total ways).
favourable outcomes = P ∩ Q = ϕ , which can be selected in 3 ways (i.e. option 2 ,3 or 4).
for n elements it can be done in 3n ways.
probability = $\large \frac{3^n}{4^n}$ = $\large (\frac{3}{4})^n$