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=A`, is

Options

A. `(1)/(2^(n))`
B. `((3)/(4))^(n)`
C. `n((3)/(4))^(n)`
D. `(n)/(3)((3)/(4))^(n)`

Explanation

To solve the problem, we need to find the probability that the union of two randomly chosen subsets \( P \) and \( Q \) from a set \( A \) containing \( n \) elements equals the set \( A \) itself.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a set \( A \) with \( n \) elements.
- We randomly choose a subset \( P \) from \( A \).
- After choosing \( P \), we reconstruct \( A \) (i.e., we put back all elements of \( P \)).
- We then randomly choose another subset \( Q \) from \( A \).
- We need to find the probability that \( P \cup Q = A \).

2. **Total Number of Subsets**:
- The total number of subsets of a set with \( n \) elements is \( 2^n \).
- Therefore, the total number of ways to choose subsets \( P \) and \( Q \) is \( 2^n \times 2^n = 4^n \).

3. **Condition for \( P \cup Q = A \)**:
- For \( P \cup Q \) to equal \( A \), every element in \( A \) must be in at least one of the subsets \( P \) or \( Q \).
- If we denote the number of elements in \( P \) as \( r \) (where \( r \) can range from \( 0 \) to \( n \)), then:
- We can choose \( r \) elements from \( A \) to be in \( P \).
- The remaining \( n - r \) elements must be chosen from \( A \) to be in \( Q \).

4. **Counting Favorable Outcomes**:
- For each \( r \) (where \( r \) varies from \( 0 \) to \( n \)):
- The number of ways to choose \( r \) elements for \( P \) is \( \binom{n}{r} \).
- The number of ways to choose the remaining \( n - r \) elements for \( Q \) is \( 2^{n-r} \) (since we can choose any subset of the remaining elements).
- Thus, the total number of favorable outcomes for a specific \( r \) is \( \binom{n}{r} \cdot 2^{n-r} \).

5. **Summing Over All Possible \( r \)**:
- We need to sum this for all \( r \) from \( 0 \) to \( n \):
\[ \sum_{r=0}^{n} \binom{n}{r} \cdot 2^{n-r} = (1 + 2)^n = 3^n \]
- This follows from the binomial theorem.

6. **Calculating the Probability**:
- The probability that \( P \cup Q = A \) is given by the ratio of favorable outcomes to total outcomes:
\[ P(P \cup Q = A) = \frac{3^n}{4^n} = \left(\frac{3}{4}\right)^n \]

7. **Conclusion**:
- The required probability that \( P \cup Q = A \) is \( \left(\frac{3}{4}\right)^n \).

### Final Answer:
The probability that \( P \cup Q = A \) is \( \left(\frac{3}{4}\right)^n \).

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