Question

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

• A. ${\left(3/4\right)}^n$
• B. ${\left(1/4\right)}^n$
• C. ${\left(2/3\right)}^n$
• D. ${\left(1/3\right)}^n$

Answer 1

The correct option is A ${\left(3/4\right)}^n$

Since, set $A$ contains $n$ elements. So, it has $2^n$ subsets.
Therefore set $P$ can be chosen in $2^n$ ways, similarly set $Q$ can be chosen in $2^n$ ways.
Therefore $P$ and $Q$ can be chosen in $\left(2^n\right)\left(2^n\right)=4^n$ ways.
Suppose, $P$ contains $r$ elements, where $r$ varies from $0$ to $n$. Then, $P$ can be chosen in ${}^n{C}_r$ ways, for $0$ to be disjoint from $A$ , it should be chosen from set of all subsets of set consisting of remaining $(n-r)$ elements.This can be done in $2^{n-r}$ ways.
Therefore $P$ and $Q$ can be chosen in ${}^n{C}_r.2^{n-r}$ ways.
But, $r$ can be vary from $0$ to $n$.
Therefore total number of disjoint sets $P$ and $Q$ $={\sum }_{r=0}^n{}^n{C}_r.2^{n-r}={(1+2)}^n=3^n$
Hence required probability $=\frac{3^n}{4^n}={\left(\frac{3}{4}\right)}^n$