Five marbles are distributed at a random among seven urns. What is the expected number of urns with exactly one marble?
To find the expected number of urns with exactly one marble, we can use the linearity of expectation. Let's denote the number of urns as \( n = 7 \) and the number of marbles as \( m = 5 \).

1. **Define Indicator Variables:**
   Let \( X_i \) be an indicator variable that is 1 if the \( i \)-th urn has exactly one marble, and 0 otherwise. The total number of urns with exactly one marble is given by:
   \[
   X = \sum_{i=1}^{7} X_i
   \]

2. **Linearity of Expectation:**
   By the linearity of expectation, we have:
   \[
   \mathbb{E}[X] = \mathbb{E}\left[\sum_{i=1}^{7} X_i\right] = \sum_{i=1}^{7} \mathbb{E}[X_i]
   \]

3. **Calculate \(\mathbb{E}[X_i]\):**
   We need to find the probability that a specific urn (say the \( i \)-th urn) has exactly one marble. This probability can be calculated as follows:
   - The total number of ways to distribute 5 marbles into 7 urns is \( 7^5 \).
   - The number of ways to distribute 4 marbles into the remaining 6 urns (ensuring that the \( i \)-th urn has exactly one marble) is given by \( \binom{5}{1} \cdot 6^4 \). Here, \(\binom{5}{1}\) represents choosing 1 out of 5 marbles to place in the \( i \)-th urn, and \( 6^4 \) represents distributing the remaining 4 marbles into the other 6 urns.

   Therefore, the probability that the \( i \)-th urn has exactly one marble is:
   \[
   \mathbb{P}(X_i = 1) = \frac{\binom{5}{1} \cdot 6^4}{7^5}
   \]

4. **Simplify the Probability:**
   \[
   \mathbb{P}(X_i = 1) = \frac{5 \cdot 6^4}{7^5}
   \]

5. **Calculate the Expected Value:**
   Since \(\mathbb{E}[X_i] = \mathbb{P}(X_i = 1)\), we have:
   \[
   \mathbb{E}[X_i] = \frac{5 \cdot 6^4}{7^5}
   \]
   Therefore, the expected number of urns with exactly one marble is:
   \[
   \mathbb{E}[X] = 7 \cdot \mathbb{E}[X_i] = 7 \cdot \frac{5 \cdot 6^4}{7^5} = \frac{5 \cdot 6^4}{7^4}
   \]

6. **Simplify the Final Expression:**
   \[
   \mathbb{E}[X] = \frac{5 \cdot 1296}{2401} = \frac{6480}{2401}
   \]

The final answer is \(\boxed{\frac{6480}{2401}}\).