$A$ is a set containing $n$ elements. $A$ subset $P$ of $A$ is chosen at random. The set $A$ reconst

Question

$A$ is a set containing $n$ elements. $A$ subset $P$ of $A$ is chosen at random. The set $A$ 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.

Professor PC Thomas Classes · Answer

The set $A$ contains $n$ elements. So, it has $2^{n}$ subsets.\ $	herefore \quad$ Set $P$ can be chosen in $2^{n}$ ways, similarly set $Q$ can be chosen in $2^{n}$ ways.\ $	herefore \quad P$ and $Q$ can be chosen in $\left(2^{n}ight)\left(2^{n}ight)=4^{n}$ 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 the set of all subsets of set consisting of remaining $(n-r)$ elements. This can be done in $2^{n-r}$ ways.\ $	herefore P$ and $Q$ can be chosen in ${ }^{n} C_{r} \cdot 2^{n-r}$ ways.\ But $r$ can vary from 0 to $n$.\ $	herefore$ Total number of disjoint sets $P$ and $Q$ are\ $=\sum_{r=0}^{n}{ }^{n} C_{r} 2^{n-r}=(1+2)^{n}=3^{n}$\ Hence, required probability $=\frac{3^{n}}{4^{n}}=\left(\frac{3}{4}ight)^{n}$\