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 the target is NOT less than 0.95, is _____

Let $$p = 0.75$$ be the probability that a single missile hits the target successfully.
Fire $$n$$ missiles independently and let $$X$$ be the number of successful hits. Then

$$X \sim \text{Binomial}(n,\;p)$$

To destroy the target completely we need at least three hits, so we require

$$P(X \ge 3) \;\ge\; 0.95$$

Using the complement rule, this is equivalent to

$$P(X \le 2) \;=\;\sum_{k=0}^{2} \binom{n}{k} p^{\,k} (1-p)^{\,n-k} \;\le\; 0.05$$

With $$p = 0.75$$ and $$1-p = 0.25$$, compute the left-hand side for successive integers $$n$$ until the inequality is satisfied.

$$\begin{aligned} P(X \le 2) &= \binom{5}{0}(0.75)^0(0.25)^5 \\[2pt] &\quad+\binom{5}{1}(0.75)^1(0.25)^4 \\[2pt] &\quad+\binom{5}{2}(0.75)^2(0.25)^3 \\[4pt] &= 1 \times 0.25^5 \;+\; 5 \times 0.75 \times 0.25^4 \;+\; 10 \times 0.75^2 \times 0.25^3 \\[4pt] &= 0.0009766 \;+\; 0.0146484 \;+\; 0.0878906 \\[4pt] &= 0.1035156 \end{aligned}$$

$$0.1035 \;>\; 0.05$$, so $$P(X \ge 3)=1-0.1035 = 0.8965 \lt 0.95$$. Five missiles are not enough.

$$\begin{aligned} P(X \le 2) &= \binom{6}{0}(0.75)^0(0.25)^6 \\[2pt] &\quad+\binom{6}{1}(0.75)^1(0.25)^5 \\[2pt] &\quad+\binom{6}{2}(0.75)^2(0.25)^4 \\[4pt] &= 1 \times 0.25^6 \;+\; 6 \times 0.75 \times 0.25^5 \;+\; 15 \times 0.75^2 \times 0.25^4 \\[4pt] &= 0.0002441 \;+\; 0.0043945 \;+\; 0.0329590 \\[4pt] &= 0.0375976 \end{aligned}$$

$$0.0376 \;\le\; 0.05$$, hence

$$P(X \ge 3) = 1 - 0.0375976 = 0.9624024 \;\ge\; 0.95$$

The inequality is satisfied for $$n = 6$$.

Since it fails for $$n = 5$$ and holds for $$n = 6$$, the minimum number of missiles that must be fired is

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