Step-by-Step Solution:

1. Identify the Probability Distribution Let n be the minimum number of missiles that have to be fired. Let X be the random variable representing the number of successful hits. Each missile firing is an independent Bernoulli trial. The probability of a successful hit (success) is given as p = 0.75 = 3/4. The probability of a miss (failure) is q = 1 - p = 1 - 0.75 = 0.25 = 1/4. Therefore, X follows a binomial distribution with parameters n and p, denoted as X ~ B(n, p). The probability of exactly k successes in n trials is given by the binomial probability formula: $P(X = k) = \binom{n}{k} p^k q^{n-k}$

2. Formulate the Condition The target is completely destroyed if there are at least three successful hits. This means $X \ge 3$. The probability of completely destroying the target must be NOT less than 0.95. This can be written as: $P(X \ge 3) \ge 0.95$

3. Use the Complement Rule for Easier Calculation Calculating $P(X \ge 3)$ directly involves a sum with a variable number of terms (P(X=3) + P(X=4) + ... + P(X=n)). It is simpler to calculate the probability of the complementary event, P(X < 3), which is P(X=0) + P(X=1) + P(X=2). The condition $P(X \ge 3) \ge 0.95$ can be rewritten as: $1 - P(X < 3) \ge 0.95$ $P(X < 3) \le 1 - 0.95$ $P(X=0) + P(X=1) + P(X=2) \le 0.05$

4. Calculate the Probabilities for X=0, 1, 2 Using the binomial formula with p=3/4 and q=1/4:

   • $P(X=0) = \binom{n}{0} (3/4)^0 (1/4)^n = 1 \cdot 1 \cdot (1/4)^n = (1/4)^n$
   • $P(X=1) = \binom{n}{1} (3/4)^1 (1/4)^{n-1} = n \cdot (3/4) \cdot (4/1) \cdot (1/4)^n = 3n (1/4)^n$
   • $P(X=2) = \binom{n}{2} (3/4)^2 (1/4)^{n-2} = \frac{n(n-1)}{2} \cdot (9/16) \cdot (16/1) \cdot (1/4)^n = \frac{9n(n-1)}{2} (1/4)^n$

5. Set up the Inequality and Test Values of n Substitute the expressions back into the inequality: $(1/4)^n + 3n(1/4)^n + \frac{9n(n-1)}{2}(1/4)^n \le 0.05$ $(1/4)^n \left[ 1 + 3n + \frac{9n(n-1)}{2} \right] \le 0.05$ $(1/4)^n \left[ \frac{2 + 6n + 9n^2 - 9n}{2} \right] \le 0.05$ $(1/4)^n \left[ \frac{9n^2 - 3n + 2}{2} \right] \le 0.05$ We need to find the smallest integer n (where $n \ge 3$) that satisfies this inequality. We can test values of n.

   • For n = 3: The condition $P(X \ge 3) \ge 0.95$ becomes $P(X=3) \ge 0.95$. $P(X=3) = \binom{3}{3}(3/4)^3(1/4)^0 = (27/64) = 0.421875$, which is less than 0.95. So, n=3 is not sufficient.

   • For n = 4: $P(X \ge 3) = P(X=3) + P(X=4) = \binom{4}{3}(3/4)^3(1/4)^1 + \binom{4}{4}(3/4)^4(1/4)^0$ $= 4 \cdot (27/64) \cdot (1/4) + 1 \cdot (81/256) = 108/256 + 81/256 = 189/256 \approx 0.738$, which is less than 0.95. So, n=4 is not sufficient.

   • For n = 5: Let's check $P(X < 3) \le 0.05$. P(X < 3) = P(X=0) + P(X=1) + P(X=2) $= (1/4)^5 [1 + 3(5) + \frac{9(5)(4)}{2}] = (1/1024) [1 + 15 + 90] = 106/1024 \approx 0.1035$ Since 0.1035 > 0.05, the condition is not met for n=5.

   • For n = 6: Let's check $P(X < 3) \le 0.05$. P(X < 3) = P(X=0) + P(X=1) + P(X=2) $= (1/4)^6 [1 + 3(6) + \frac{9(6)(5)}{2}] = (1/4096) [1 + 18 + 135] = 154/4096$ $154/4096 \approx 0.03759$ Since $0.03759 \le 0.05$, the condition is satisfied for n=6.

6. Conclusion The smallest integer value of n for which the probability of at least three hits is not less than 0.95 is 6.

Final Answer: The minimum number of missiles that have to be fired is 6.