The probability of a bomb hitting a bridge is `1/2` and two direct hits are needed to destroy it. The least number of bombs required so that the probability of the bridge being destroyed is greater than 0.9, is :

Options

A. 8
B. 7
C. 6
D. 9

Answer

To solve the problem, we need to find the least number of bombs required such that the probability of the bridge being destroyed is greater than 0.9.

### Step-by-Step Solution:

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

2. **Understand the Requirement**:
- We need at least 2 hits to destroy the bridge. Therefore, we want to find \( n \) such that:
\[ P(X \geq 2) > 0.9 \]
where \( X \) is the number of hits.

3. **Use the Complement**:
- The complement of the event \( X \geq 2 \) is \( X < 2 \). Thus, we can write:
\[ P(X \geq 2) = 1 - P(X < 2) \]
This means:
\[ 1 - P(X < 2) > 0.9 \implies P(X < 2) < 0.1 \]

4. **Calculate \( P(X < 2) \)**:
- We need to calculate \( P(X = 0) \) and \( P(X = 1) \):
\[ P(X = 0) = \binom{n}{0} p^0 q^n = 1 \cdot \left(\frac{1}{2}\right)^n = \frac{1}{2^n} \]
\[ P(X = 1) = \binom{n}{1} p^1 q^{n-1} = n \cdot \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right)^{n-1} = n \cdot \frac{1}{2^n} \]
Therefore:
\[ P(X < 2) = P(X = 0) + P(X = 1) = \frac{1}{2^n} + n \cdot \frac{1}{2^n} = \frac{1 + n}{2^n} \]

5. **Set Up the Inequality**:
- We need:
\[ \frac{1 + n}{2^n} < 0.1 \]
This can be rearranged to:
\[ 1 + n < 0.1 \cdot 2^n \]

6. **Test Values for \( n \)**:
- We will try different integer values for \( n \) to find the smallest one that satisfies the inequality.

- For \( n = 1 \):
\[ 1 + 1 = 2 < 0.1 \cdot 2^1 = 0.2 \quad \text{(False)} \]
- For \( n = 2 \):
\[ 1 + 2 = 3 < 0.1 \cdot 2^2 = 0.4 \quad \text{(False)} \]
- For \( n = 3 \):
\[ 1 + 3 = 4 < 0.1 \cdot 2^3 = 0.8 \quad \text{(False)} \]
- For \( n = 4 \):
\[ 1 + 4 = 5 < 0.1 \cdot 2^4 = 1.6 \quad \text{(False)} \]
- For \( n = 5 \):
\[ 1 + 5 = 6 < 0.1 \cdot 2^5 = 3.2 \quad \text{(False)} \]
- For \( n = 6 \):
\[ 1 + 6 = 7 < 0.1 \cdot 2^6 = 6.4 \quad \text{(False)} \]
- For \( n = 7 \):
\[ 1 + 7 = 8 < 0.1 \cdot 2^7 = 12.8 \quad \text{(True)} \]
- For \( n = 8 \):
\[ 1 + 8 = 9 < 0.1 \cdot 2^8 = 25.6 \quad \text{(True)} \]

7. **Conclusion**:
- The least number of bombs required such that the probability of the bridge being destroyed is greater than 0.9 is \( n = 7 \).