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 $Q$ is again chosen at random. The probability that $P \cup Q$ contains just one element is ?
A)$\dfrac{{3n}}{{{4^n}}}$
B)$\dfrac{n}{3}{(\dfrac{3}{4})^n}$
C)${(\dfrac{2}{3})^n}$
D)$\dfrac{n}{{{4^n}}}$

Answer 1

Hint: Probability of an event is found by dividing favourable number of outcomes by total number of outcomes. Total number of ways of choosing two subsets can be calculated using the number of subsets. Number of ways in which this union contains just one element can be calculated by considering all the possible cardinalities of $P$ and $Q$.

Formula used:
Number of subsets for a set with $n$ elements is ${2^n}$.
\[{\text{Probability of an event = }}\dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\]
Number of ways for selecting $r$ items from $n$ items is ${}^nC{}_r = \dfrac{{n!}}{{(n - r)!r!}}$.

Complete step-by-step answer:
Let the set $A$ contains $n$ elements.
So $A$ have ${2^n}$ subsets.
It is said that $P$ and $Q$ were chosen at random.
Number of ways for selecting $r$ items from $n$ items is ${}^nC{}_r = \dfrac{{n!}}{{(n - r)!r!}}$.
This means $P$ can be chosen in ${}^{{2^n}}{C_1}$ ways.
Similarly, $Q$ can be chosen in ${}^{{2^n}}{C_1}$ ways.
Therefore, this choice can be done in ${}^{{2^n}}{C_1} \times {}^{{2^n}}{C_1}$ ways.
So, sets $P$ and $Q$ can be chosen in ${2^n} \times {2^n} = {4^n}$ ways.

We have to consider the case when $P \cup Q$ contains just one element.
Let us check in how many different ways this can happen.
A subset may be empty as well.
So $P$ is empty and $Q$ is a singleton set is a possibility.
If so $P \cup Q$ contains just one element.
Since the set $A$ contains $n$ elements, number of ways $P$ be empty is ${}^n{C_0} = 1$
Also, number of ways $Q$ is a singleton set is ${}^n{C_1} = n$
Therefore both happen together in $1 \times n = n$ ways.

On the other hand if $P$ has one element, $Q$ can be empty or it can be equal to $P$.
So, $Q$ can be chosen in two ways.
Therefore, the number of ways of choosing $P$ and $Q$ in this case is ${}^n{C_1} \times 2 = n \times 2 = 2n$.
So, the total number of ways of choosing $P$ and $Q$ such that $P \cup Q$ contain just one element is $n + 2n = 3n$ ways.
We have, \[{\text{Probability of an event = }}\dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\]
So, the probability of $P \cup Q$ containing just one element is $\dfrac{{3n}}{{{4^n}}}$.
$\therefore $ The answer is option A.

Note: We must be careful while calculating the number of favourable outcomes. Here there are three cases. There is a chance that we avoid the last case; that is, $P$ and $Q$ are identical.