A is a set containg n elements. A subset P of A is chosen at random and the set A is reconstructed by replacing the random. Fond the probability that `Pcup Q` contains exactoly r elements with ` 1 le r le n`.

Answer

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

### Step-by-Step Solution:

1. **Understanding the Sets**:
Let \( A = \{ a_1, a_2, a_3, \ldots, a_n \} \) be the set containing \( n \) elements.

2. **Choosing Subsets**:
We are choosing a subset \( P \) from \( A \) at random. The total number of subsets of \( A \) is \( 2^n \) since each element can either be included or excluded.

3. **Union of Subsets**:
We need to analyze the union \( P \cup Q \). The union will contain elements from both subsets \( P \) and \( Q \).

4. **Cases for Elements**:
Each element \( a_i \) in \( A \) can fall into one of four cases regarding its membership in \( P \) and \( Q \):
- Case 1: \( a_i \) is in both \( P \) and \( Q \).
- Case 2: \( a_i \) is in \( P \) but not in \( Q \).
- Case 3: \( a_i \) is in \( Q \) but not in \( P \).
- Case 4: \( a_i \) is in neither \( P \) nor \( Q \).

5. **Counting the Cases**:
For each of the \( n \) elements, there are 4 choices (the four cases above). Therefore, the total number of ways to choose subsets \( P \) and \( Q \) is \( 4^n \).

6. **Choosing Exactly \( r \) Elements**:
We want the union \( P \cup Q \) to contain exactly \( r \) elements. The number of ways to choose \( r \) elements from \( n \) is given by \( \binom{n}{r} \).

7. **Distributing the Remaining Elements**:
After choosing \( r \) elements to be in \( P \cup Q \), the remaining \( n - r \) elements can be in any of the four cases (either in \( P \), in \( Q \), in both, or in neither). Thus, for each of the \( n - r \) elements, we have 4 choices.

8. **Total Ways for Exactly \( r \) Elements**:
The total number of ways to achieve exactly \( r \) elements in \( P \cup Q \) is:
\[ \binom{n}{r} \cdot 4^{n - r} \]

9. **Calculating the Probability**:
The probability that \( P \cup Q \) contains exactly \( r \) elements is given by the ratio of the favorable outcomes to the total outcomes:
\[ P(P \cup Q \text{ has exactly } r \text{ elements}) = \frac{\binom{n}{r} \cdot 4^{n - r}}{4^n} \]

10. **Simplifying the Probability**:
This simplifies to:
\[ P(P \cup Q \text{ has exactly } r \text{ elements}) = \frac{\binom{n}{r}}{4^r} \]

### Final Answer:
The probability that \( P \cup Q \) contains exactly \( r \) elements is:
\[ \frac{\binom{n}{r}}{4^r} \]