# What is the probability that P and Q have no common elements?

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 the subset of P. A subset Q of A is again chosen at random. Find the probability that P and Q have no common elements.

I tried to calculate in this way :

In set P we can have no element i.e.Φ, 1 element, 2 elements, ...... upto n elements. If we have no element in P, we will leave by all the elements and number of set Q formed by those elements will have no common element in common with P. Similarly, it there are r elements in P we are left with rest of (n - r) element to form Q, satisfying the condition that P and Q should be disjoint.

Now, my confusion is how can I find the Total number of ways in which we can form P and Q?

Each element of $$A$$ has 3 choices- it can either be a part of $$P$$, of $$Q$$, or neither of the two, since it cannot be a part of both.
Hence by multiplication rule, for $$n$$ elements, total number of ways to divide the elements = $$3^n$$.
What you have done in the case $$n=5$$ is correct. If the first set has $$k$$ elements then the second set can be chosen in $$2^{n-k}$$ ways. Hence the total number us $$\sum_{k=0}^{n} \binom {n} {k} 2^{n-k}=2^{n} (1+\frac 1 2)^{n}=3^{n}$$.
• What if $P$ and $Q$ are both non-empty. This is not asked by OP but I am wondering what the new expression would be Dec 2, 2023 at 13:28
$$P,Q$$ can both be one of the $$2^n$$ subsets of $$A$$. The total number of ways to form $$P$$ and $$Q$$ is $$2^n\cdot2^n=2^{2n}$$.