A bag contains 10 balls of which 2 are red and the remaining are either blue or black. If the probability of drawing 3 balls of the same color is `11/120` and if the number of blue balls exceeds the number of black balls, the number of blue balls is_____________ .

Answer

To solve the problem, we need to find the number of blue balls in a bag that contains a total of 10 balls, including 2 red balls. The remaining balls are either blue or black, and we know that the probability of drawing 3 balls of the same color is \( \frac{11}{120} \). Additionally, the number of blue balls exceeds the number of black balls.

### Step-by-Step Solution:

1. **Identify the Total Number of Balls:**
- Total balls = 10
- Red balls = 2
- Therefore, the remaining balls (either blue or black) = 10 - 2 = 8.

2. **Let the Number of Blue Balls be \( b \) and Black Balls be \( k \):**
- We have \( b + k = 8 \) (since the total of blue and black balls is 8).
- We also know that \( b > k \) (the number of blue balls exceeds the number of black balls).

3. **Calculate the Total Ways to Choose 3 Balls from 10:**
- The total number of ways to choose 3 balls from 10 is given by the combination formula \( \binom{n}{r} \):
\[ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120. \]

4. **Calculate the Ways to Choose 3 Balls of the Same Color:**
- The ways to choose 3 blue balls from \( b \) blue balls is \( \binom{b}{3} \).
- The ways to choose 3 black balls from \( k \) black balls is \( \binom{k}{3} \).
- Therefore, the total ways to choose 3 balls of the same color is:
\[ \binom{b}{3} + \binom{k}{3}. \]

5. **Set Up the Probability Equation:**
- The probability of drawing 3 balls of the same color is given as \( \frac{11}{120} \):
\[ \frac{\binom{b}{3} + \binom{k}{3}}{\binom{10}{3}} = \frac{11}{120}. \]
- Since \( \binom{10}{3} = 120 \), we can simplify this to:
\[ \binom{b}{3} + \binom{k}{3} = 11. \]

6. **Substituting Values for \( b \) and \( k \):**
- We can express \( k \) in terms of \( b \): \( k = 8 - b \).
- Substitute \( k \) into the equation:
\[ \binom{b}{3} + \binom{8-b}{3} = 11. \]

7. **Testing Possible Values for \( b \):**
- Since \( b > k \), \( b \) must be greater than 4 (because \( k \) must be less than 4).
- Test \( b = 5 \):
- Then \( k = 8 - 5 = 3 \).
- Calculate \( \binom{5}{3} + \binom{3}{3} \):
\[ \binom{5}{3} = 10, \quad \binom{3}{3} = 1 \quad \Rightarrow \quad 10 + 1 = 11. \]
- This satisfies the equation.

8. **Conclusion:**
- The number of blue balls \( b = 5 \).

### Final Answer:
The number of blue balls is **5**.