To solve the problem, we need to find the probability that two randomly chosen subsets \( A \) and \( B \) of a set \( X \) with 3 elements have the same number of elements. Let's denote the set \( X = \{ e_1, e_2, e_3 \} \).
### Step-by-Step Solution:
1. **Determine the Total Number of Subsets:**
The total number of subsets of a set with \( n \) elements is given by \( 2^n \). For our set \( X \) with 3 elements:
\[
\text{Total subsets} = 2^3 = 8
\]
The subsets are: \( \emptyset, \{e_1\}, \{e_2\}, \{e_3\}, \{e_1, e_2\}, \{e_1, e_3\}, \{e_2, e_3\}, \{e_1, e_2, e_3\} \).
2. **Count the Favorable Outcomes:**
We need to find the number of ways to choose subsets \( A \) and \( B \) such that they have the same number of elements. The possible sizes of subsets can be 0, 1, 2, or 3.
- **Case 1:** Both subsets have 0 elements.
- There is 1 way: \( A = \emptyset \) and \( B = \emptyset \).
- **Case 2:** Both subsets have 1 element.
- There are 3 subsets of size 1: \( \{e_1\}, \{e_2\}, \{e_3\} \).
- The number of ways to choose 1 element for both \( A \) and \( B \) is \( 3 \times 3 = 9 \).
- **Case 3:** Both subsets have 2 elements.
- There are 3 subsets of size 2: \( \{e_1, e_2\}, \{e_1, e_3\}, \{e_2, e_3\} \).
- The number of ways to choose 2 elements for both \( A \) and \( B \) is \( 3 \times 3 = 9 \).
- **Case 4:** Both subsets have 3 elements.
- There is 1 way: \( A = \{e_1, e_2, e_3\} \) and \( B = \{e_1, e_2, e_3\} \).
Now, we sum the favorable outcomes:
\[
\text{Total favorable outcomes} = 1 + 9 + 9 + 1 = 20
\]
3. **Calculate the Probability:**
The probability \( P \) that subsets \( A \) and \( B \) have the same number of elements is given by the ratio of favorable outcomes to total outcomes:
\[
P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{20}{64} = \frac{5}{16}
\]
4. **Find \( k \):**
From the problem statement, we have \( k = \frac{5}{16} \).
5. **Calculate \( \frac{1}{k} \):**
To find \( \frac{1}{k} \):
\[
\frac{1}{k} = \frac{16}{5}
\]
### Final Answer:
Thus, the value of \( \frac{1}{k} \) is \( \frac{16}{5} \).
