In the game of Avalon, there are $10$ players, $4$ of which are bad. A [i]quest[/i] is a subset of those players. A quest fails if it contains at least one bad player. A randomly chosen quest of $3$ players happens to fail. What is the probability that there is exactly one bad player in the failed quest?

[i]2018 CCA Math Bonanza Team Round #3[/i]
1. **Calculate the total number of ways to choose 3 players out of 10:**
   \[
   \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120
   \]

2. **Calculate the number of ways to choose 3 good players out of 6 (since there are 6 good players):**
   \[
   \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20
   \]

3. **Calculate the number of ways to choose at least one bad player in the quest:**
   \[
   \text{Total ways to choose 3 players} - \text{Ways to choose 3 good players} = \binom{10}{3} - \binom{6}{3} = 120 - 20 = 100
   \]

4. **Calculate the number of ways to choose exactly one bad player in the quest:**
   - Choose 1 bad player out of 4:
     \[
     \binom{4}{1} = 4
     \]
   - Choose 2 good players out of 6:
     \[
     \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15
     \]
   - Total ways to choose exactly one bad player:
     \[
     \binom{4}{1} \times \binom{6}{2} = 4 \times 15 = 60
     \]

5. **Calculate the probability that there is exactly one bad player in the failed quest:**
   \[
   \frac{\text{Number of ways to choose exactly one bad player}}{\text{Number of ways to choose at least one bad player}} = \frac{60}{100} = \frac{3}{5}
   \]

The final answer is \( \boxed{ \frac{3}{5} } \)