Question

Let $A$ be a set containing $n$ elements. A subset $P$ of the set $A$ is chosen at random.
The set $A$ is reconstructed by replacing the element of $P$, and another subsets $Q$ of $A$ is chosen at random. The probability that $(P\cap Q)$ contains exactly $m(m<n)$ elements is

• A. $\frac{3^{n-m}}{4^n}$
• B. $\frac{{}^n{C}_m{3}^m}{4^n}$
• C. $\frac{{}^n{C}_m{3}^{n-m}}{4^n}$
• D. none of these

Answer 1

The correct option is C $\frac{{}^n{C}_m{3}^{n-m}}{4^n}$

We know that the number of subsets of a set containing $n$ elements is $2^n$.
Therefore, the number of ways of choosing $P$ and $Q$ is ${}^{2^n}{C}_1{\times }^{2^n}{C}_1=2^n\times 2^n=4^n$.
Out of $n$ elements, $m$ elements can be chosen in ${}^n{C}_m$ ways.
If $(P\cap Q)$ contains exactly in elements,

then from the remaining $n-m$ elements either an element belongs to $P$ or $Q$ but not both $P$ and $Q$.
Suppose $P$ contains $r$ elements from the remaining $n-m$ elements.
Then $Q$ may contain any number of elements from the remaining $(n-m)-r$ elements.
Therefore, $P$ and $Q$ can be chosen in ${}^{n-m}{C}_r{2}^{(n-m)-r}$
But $r$ can vary from $0$ to $(n-m)$. So, $P$ and $Q$ can be chosen in general in
${\left(\sum _{r=0}^{n-m}{}^{n-m}{C}_r{2}^{(n-m)-r}\right)}^n{C}_m={(1+2)}^{n-m}{\times }^n{C}_m{=}^n{C}_m\times 3^{n-m}$
Hence, required probability $=\frac{{}^n{C}_m\times 3^{n-m}}{4^n}$