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 $\left( P\cap Q \right)$contains exactly \[m\left( m < n \right)\] element is?
(a) $\dfrac{{{3}^{n-m}}}{{{4}^{n}}}$,
(b) $\dfrac{{}^{n}{{C}_{m}}}{{{4}^{n}}}$,
(c) $\dfrac{{}^{n}{{C}_{m}}{{3}^{n-m}}}{{{4}^{n}}}$,
(d) None of these.

Answer 1

Hint: We should know that the number of subsets of a set containing “$n$” elements is “${{2}^{n}}$”.
Since here we are selecting combinations are used to find the required probability. We then
calculate the number ways of choosing P and Q. Therefore, the number of ways of choosing
P and Q is ${}^{{{2}^{n}}}{{C}_{1}}\times {}^{{{2}^{n}}}{{C}_{1}}={{2}^{n}}\times {{2}^{n}}={{4}^{n}}$.

Complete step by step answer:
According to the problem, we are given that A is the set containing “$n$” elements, P is the subset of A and Q is also a subset of A.
Here we have to find the probability of $P\cap Q$ which contains exactly \[m\left( m < n \right)\] elements.
We know that If A has “$n$” elements then P will have ${{2}^{n}}$ elements. Because P is the subset of A.
Similarly, Q also will have ${{2}^{n}}$ elements as it also a subset of A.
Therefore, the number of ways of choosing P and Q is ${}^{{{2}^{n}}}{{C}_{1}}\times {}^{{{2}^{n}}}{{C}_{1}}={{2}^{n}}\times {{2}^{n}}={{4}^{n}}$.
We also need to choose m elements out of the given n elements, which can be done in ${}^{n}{{C}_{m}}$ ways.
If $P\cap Q$ contains exactly “m” elements, then from the remaining $n-m$ elements.an element either lies in set P or set Q but not both.
Let us assume that P has r elements from the remaining $n-m$ elements.
The elements we get from subtracting r elements from the total of $n-m$ elements will present in Q, which means that the set Q has $n-m-r$ elements.
So, we can choose P and Q in ${}^{n-m}{{C}_{r}}{{.2}^{(n-m)-r}}$ ways, as r elements for set P can be chosen in ${}^{n-m}{{C}_{r}}$ ways and the total number of subsets that can be formed with the remaining $n-m-r$ elements is ${{2}^{n-m-r}}$ From which Q is to be chosed.
But the values of r vary from 0 to $\left( n-m \right)$. So, P and Q can be chosen in general in $\sum\limits_{r=0}^{n-m}{{}^{n-m}{{C}_{r}}{{.2}^{(n-m)-r}}}$.
We know that $\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{a}^{n-r}}}={{\left( 1+a \right)}^{n}}$. Using this we get $\sum\limits_{r=0}^{n-m}{{}^{n-m}{{C}_{r}}{{.2}^{(n-m)-r}}}={{\left( 1+2 \right)}^{n-m}}\times {}^{n}{{C}_{m}}$.
$\sum\limits_{r=0}^{n-m}{{}^{n-m}{{C}_{r}}{{.2}^{(n-m)-r}}}={}^{n}{{C}_{m}}{{3}^{n-m}}$.
So, the number of ways to choose the sets P and Q so that $\left( P\cap Q \right)$contains exactly \[m\left( m < n \right)\] element is ${}^{n}{{C}_{m}}{{3}^{n-m}}$.
Now, let us find the required probability and let us assume it be $P\left( x \right)$.
So, we get $P\left( x \right)=\dfrac{\text {Total number of ways to choose subsets P and Q so that P}\cap \text {Q has exactly m elements }\left( m < r \right)}{\text{Total number of ways too choose two subsets P and Q from the subsets of A}}$
$P\left( x \right)=\dfrac{{}^{n}{{C}_{m}}\times {{3}^{n-m}}}{{{4}^{n}}}$.
Hence, the required probability=$\dfrac{{}^{n}{{C}_{m}}\times {{3}^{n-m}}}{{{4}^{n}}}$.
Therefore, the correct option is (C)

Note:
We may do mistakes while choosing the total number of ways to choose both subsets from the set A. We should not make mistakes while making calculations about the total number of elements that may present in the subsets P and Q after taking the intersection of it. We should not confuse with the total number of subsets present in a set of n elements.