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 cap Q` contain just one element, is

Options

A. `((3)/(4))^(n)`
B. `n((3)/(4))^(n)`
C. `(n)/(3)((3)/(4))^(n)`
D. `(n)/(4)((3)/(4))^(n)`

Explanation

To solve the problem, we need to find the probability that the intersection of two randomly chosen subsets \( P \) and \( Q \) from a set \( A \) containing \( n \) elements contains exactly one element.

### Step-by-step Solution:

1. **Understanding the Set and Subsets**:
- Let \( A \) be a set with \( n \) elements. The total number of subsets of \( A \) is \( 2^n \).

2. **Choosing Subset \( P \)**:
- A subset \( P \) of \( A \) is chosen at random. The number of ways to choose a non-empty subset \( P \) is \( 2^n - 1 \) (since we exclude the empty set).

3. **Choosing Subset \( Q \)**:
- After reconstructing \( A \) (which means replacing the elements of \( P \)), we choose another subset \( Q \) from \( A \). Similarly, the number of ways to choose a non-empty subset \( Q \) is also \( 2^n - 1 \).

4. **Calculating Total Ways to Choose \( P \) and \( Q \)**:
- The total number of ways to choose both subsets \( P \) and \( Q \) is \( (2^n - 1)(2^n - 1) = (2^n - 1)^2 \).

5. **Condition for Intersection**:
- We want \( P \cap Q \) to contain exactly one element. Let’s denote the number of elements in \( P \) as \( r \). For \( P \cap Q \) to have exactly one element, we must choose one element from \( P \) and any combination of the remaining elements in \( A \).

6. **Choosing the Intersection Element**:
- We can choose 1 element from \( P \) in \( r \) ways (where \( r \) is the number of elements in \( P \)).

7. **Choosing Remaining Elements for \( Q \)**:
- After choosing 1 element from \( P \), we can choose any combination of the remaining \( n - r \) elements (those not in \( P \)). The number of ways to choose any subset from these \( n - r \) elements is \( 2^{n - r} \).

8. **Summing Over All Possible Sizes of \( P \)**:
- We need to sum over all possible sizes of \( P \) (from 1 to \( n \)):
\[ \sum_{r=1}^{n} \binom{n}{r} \cdot r \cdot 2^{n - r} \]

9. **Using Combinatorial Identities**:
- The expression \( \binom{n}{r} \cdot r \) can be rewritten as \( n \cdot \binom{n-1}{r-1} \). Thus, we can rewrite the sum:
\[ n \sum_{r=1}^{n} \binom{n-1}{r-1} \cdot 2^{n - r} \]
- This sum simplifies to \( n \cdot 3^{n-1} \) (using the binomial theorem).

10. **Final Probability Calculation**:
- The total number of favorable outcomes (where \( P \cap Q \) has exactly one element) is \( n \cdot 3^{n-1} \).
- The total number of outcomes is \( (2^n - 1)^2 \).
- Therefore, the probability \( P(P \cap Q \text{ has exactly one element}) \) is:
\[ \frac{n \cdot 3^{n-1}}{(2^n - 1)^2} \]

### Final Answer:
The probability that \( P \cap Q \) contains exactly one element is:
\[ \frac{n \cdot 3^{n-1}}{(2^n - 1)^2} \]