The probability of a bomb hitting a bridge is `(2)/(3)`. Two direct hits are needed to destroy the bridge. The minimum number of bombs required such that the probability of bridge being destroyed is greater than `(3)/(4)`, is

Answer

To solve the problem, we need to determine the minimum number of bombs required such that the probability of the bridge being destroyed (which requires at least 2 direct hits) is greater than \( \frac{3}{4} \).

### Step-by-Step Solution:

1. **Define the Probabilities**:
- Let \( p = \frac{2}{3} \) be the probability of a bomb hitting the bridge.
- The probability of a bomb missing the bridge is \( q = 1 - p = 1 - \frac{2}{3} = \frac{1}{3} \).

2. **Determine the Required Probability**:
- We need the probability of getting at least 2 hits (successes) to be greater than \( \frac{3}{4} \).
- This can be expressed as:
\[ P(X \geq 2) > \frac{3}{4} \]
- Using the complement rule, this can be rewritten as:
\[ 1 - P(X = 0) - P(X = 1) > \frac{3}{4} \]
- Therefore:
\[ P(X = 0) + P(X = 1) < \frac{1}{4} \]

3. **Calculate \( P(X = 0) \) and \( P(X = 1) \)**:
- The probability of getting exactly 0 hits (all bombs miss) when \( n \) bombs are dropped:
\[ P(X = 0) = \binom{n}{0} p^0 q^n = 1 \cdot 1 \cdot \left(\frac{1}{3}\right)^n = \left(\frac{1}{3}\right)^n \]
- The probability of getting exactly 1 hit (1 bomb hits and \( n-1 \) bombs miss):
\[ P(X = 1) = \binom{n}{1} p^1 q^{n-1} = n \cdot \frac{2}{3} \cdot \left(\frac{1}{3}\right)^{n-1} = n \cdot \frac{2}{3^n} \]

4. **Combine the Probabilities**:
- Now, we combine these probabilities:
\[ P(X = 0) + P(X = 1) = \left(\frac{1}{3}\right)^n + n \cdot \frac{2}{3^n} \]
- We want this to be less than \( \frac{1}{4} \):
\[ \left(\frac{1}{3}\right)^n + n \cdot \frac{2}{3^n} < \frac{1}{4} \]

5. **Test Values of \( n \)**:
- Start testing integer values for \( n \) to find the minimum that satisfies the inequality.

- **For \( n = 2 \)**:
\[ \left(\frac{1}{3}\right)^2 + 2 \cdot \frac{2}{3^2} = \frac{1}{9} + \frac{4}{9} = \frac{5}{9} \quad (\text{not less than } \frac{1}{4}) \]

- **For \( n = 3 \)**:
\[ \left(\frac{1}{3}\right)^3 + 3 \cdot \frac{2}{3^3} = \frac{1}{27} + \frac{6}{27} = \frac{7}{27} \quad (\text{not less than } \frac{1}{4}) \]

- **For \( n = 4 \)**:
\[ \left(\frac{1}{3}\right)^4 + 4 \cdot \frac{2}{3^4} = \frac{1}{81} + \frac{8}{81} = \frac{9}{81} = \frac{1}{9} \quad (\text{less than } \frac{1}{4}) \]

- **For \( n = 5 \)**:
\[ \left(\frac{1}{3}\right)^5 + 5 \cdot \frac{2}{3^5} = \frac{1}{243} + \frac{10}{243} = \frac{11}{243} \quad (\text{less than } \frac{1}{4}) \]

6. **Conclusion**:
- The minimum number of bombs required such that the probability of the bridge being destroyed is greater than \( \frac{3}{4} \) is \( n = 4 \).

### Final Answer:
The minimum number of bombs required is **4**.