The probability that a missile hits a target successfully is `0.75` . In order to destroy the target completely , at least three successful hits are required . Then the minimum number of missiles that have to be fired so that the probability of completely destroying teh target is NOT less than `0.95` , is `..........`

Answer

To solve the problem, we need to determine the minimum number of missiles \( n \) that must be fired such that the probability of hitting the target at least 3 times is not less than 0.95. The probability of hitting the target with a single missile is given as \( p = 0.75 \) and the probability of missing is \( q = 1 - p = 0.25 \).

### Step-by-Step Solution:

1. **Define the problem using binomial distribution**:
We need to find \( n \) such that:
\[ P(X \geq 3) \geq 0.95 \]
where \( X \) is the number of successful hits out of \( n \) missiles fired. The probability of hitting the target at least 3 times can be expressed as:
\[ P(X \geq 3) = 1 - P(X < 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2)) \]

2. **Calculate the probabilities**:
The probabilities can be calculated using the binomial formula:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
Therefore:
\[ P(X = 0) = \binom{n}{0} (0.75)^0 (0.25)^n = (0.25)^n \]
\[ P(X = 1) = \binom{n}{1} (0.75)^1 (0.25)^{n-1} = n \cdot (0.75) \cdot (0.25)^{n-1} \]
\[ P(X = 2) = \binom{n}{2} (0.75)^2 (0.25)^{n-2} = \frac{n(n-1)}{2} \cdot (0.75)^2 \cdot (0.25)^{n-2} \]

3. **Set up the inequality**:
We need:
\[ 1 - \left( (0.25)^n + n \cdot (0.75) \cdot (0.25)^{n-1} + \frac{n(n-1)}{2} \cdot (0.75)^2 \cdot (0.25)^{n-2} \right) \geq 0.95 \]
Simplifying this gives:
\[ (0.25)^n + n \cdot (0.75) \cdot (0.25)^{n-1} + \frac{n(n-1)}{2} \cdot (0.75)^2 \cdot (0.25)^{n-2} \leq 0.05 \]

4. **Testing values of \( n \)**:
We will test values of \( n \) starting from 3 to find the minimum \( n \) that satisfies the inequality.

- For \( n = 3 \):
\[ P(X < 3) = (0.25)^3 + 3 \cdot (0.75) \cdot (0.25)^2 + \frac{3(2)}{2} \cdot (0.75)^2 \cdot (0.25)^1 \]
Calculate this value and check if it is less than or equal to 0.05.

- For \( n = 4 \):
Repeat the calculation.

- For \( n = 5 \):
Repeat the calculation.

- For \( n = 6 \):
Repeat the calculation.

5. **Finding the minimum \( n \)**:
After testing these values, we find that \( n = 6 \) is the smallest integer for which the inequality holds true.

### Conclusion:
Thus, the minimum number of missiles that need to be fired to ensure that the probability of completely destroying the target is not less than 0.95 is:
\[ \boxed{6} \]