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 is greater than \( \frac{3}{4} \).

### Step-by-Step Solution:

1. **Identify the probabilities**:
- The probability of a bomb hitting the bridge is given as \( P = \frac{2}{3} \).
- The probability of a bomb missing the bridge (failure) is \( Q = 1 - P = 1 - \frac{2}{3} = \frac{1}{3} \).

2. **Define the event**:
- We need at least 2 direct hits to destroy the bridge. Therefore, we need to find the minimum number of bombs \( n \) such that the probability of getting at least 2 hits is greater than \( \frac{3}{4} \).

3. **Calculate the probability of at least 2 hits**:
- The probability of at least 2 hits can be expressed as:
\[ P(\text{at least 2 hits}) = 1 - P(\text{0 hits}) - P(\text{1 hit}) \]
- Using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} P^k Q^{n-k} \]
- For 0 hits:
\[ P(X = 0) = \binom{n}{0} P^0 Q^n = 1 \cdot 1 \cdot \left(\frac{1}{3}\right)^n = \left(\frac{1}{3}\right)^n \]
- For 1 hit:
\[ P(X = 1) = \binom{n}{1} P^1 Q^{n-1} = n \cdot \frac{2}{3} \cdot \left(\frac{1}{3}\right)^{n-1} = n \cdot \frac{2}{3^n} \]

4. **Set up the inequality**:
- We need:
\[ 1 - \left(\frac{1}{3}\right)^n - n \cdot \frac{2}{3^n} > \frac{3}{4} \]
- Rearranging gives:
\[ \left(\frac{1}{3}\right)^n + n \cdot \frac{2}{3^n} < \frac{1}{4} \]

5. **Combine terms**:
- Factor out \( \frac{1}{3^n} \):
\[ \frac{1}{3^n} \left(1 + 2n\right) < \frac{1}{4} \]
- This leads to:
\[ 1 + 2n < \frac{3^n}{4} \]

6. **Trial and error for minimum \( n \)**:
- Start testing values for \( n \):
- For \( n = 2 \):
\[ 1 + 2 \cdot 2 = 5 \quad \text{and} \quad \frac{3^2}{4} = \frac{9}{4} = 2.25 \quad \text{(not satisfied)} \]
- For \( n = 3 \):
\[ 1 + 2 \cdot 3 = 7 \quad \text{and} \quad \frac{3^3}{4} = \frac{27}{4} = 6.75 \quad \text{(not satisfied)} \]
- For \( n = 4 \):
\[ 1 + 2 \cdot 4 = 9 \quad \text{and} \quad \frac{3^4}{4} = \frac{81}{4} = 20.25 \quad \text{(satisfied)} \]
- For \( n = 5 \):
\[ 1 + 2 \cdot 5 = 11 \quad \text{and} \quad \frac{3^5}{4} = \frac{243}{4} = 60.75 \quad \text{(satisfied)} \]

7. **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**.