Minimum Fired Missiles for Given Target Destruction Probability

Let $n$ be the total number of missiles fired at the target.

Let the random variable $X$ denote the number of successful hits on the target. Since each trial is independent, $X$ follows a Binomial distribution: $X \sim \text{Binomial}(n, p)$ where $p = 0.75 = \dfrac{3}{4}$ is the probability of a successful hit, and $q = 1 - p = 0.25 = \dfrac{1}{4}$ is the probability of a miss.

To completely destroy the target, at least $3$ successful hits are required. We are given that the probability of completely destroying the target must be at least $0.95$: $P(X \ge 3) \ge 0.95$

Using the complement rule, this can be written as: $1 - P(X < 3) \ge 0.95 \implies P(X < 3) \le 0.05$

The probability $P(X < 3)$ is the sum of the probabilities of getting $0$, $1$, or $2$ successful hits: $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

Using the binomial probability formula $P(X = k) = \binom{n}{k} p^k q^{n-k}$: $P(X = 0) = \binom{n}{0} \left(\frac{3}{4}\right)^0 \left(\frac{1}{4}\right)^n = \frac{1}{4^n}$ $P(X = 1) = \binom{n}{1} \left(\frac{3}{4}\right)^1 \left(\frac{1}{4}\right)^{n-1} = \frac{3n}{4^n}$ $P(X = 2) = \binom{n}{2} \left(\frac{3}{4}\right)^2 \left(\frac{1}{4}\right)^{n-2} = \frac{9n(n-1)}{2 \cdot 4^n}$

Summing these terms gives: $P(X < 3) = \frac{1}{4^n} \left[ 1 + 3n + \frac{9n(n-1)}{2} \right]$

We need to find the smallest integer $n \ge 3$ such that: $\frac{1}{4^n} \left[ 1 + 3n + \frac{9n(n-1)}{2} \right] \le 0.05 = \frac{1}{20}$

Let's test integer values of $n$:

1. For $n = 4$: $P(X < 3) = \frac{1}{4^4} \left[ 1 + 3(4) + \frac{9 \times 4 \times 3}{2} \right] = \frac{1 + 12 + 54}{256} = \frac{67}{256} \approx 0.2617 > 0.05$

2. For $n = 5$: $P(X < 3) = \frac{1}{4^5} \left[ 1 + 3(5) + \frac{9 \times 5 \times 4}{2} \right] = \frac{1 + 15 + 90}{1024} = \frac{106}{1024} \approx 0.1035 > 0.05$

3. For $n = 6$: $P(X < 3) = \frac{1}{4^6} \left[ 1 + 3(6) + \frac{9 \times 6 \times 5}{2} \right] = \frac{1 + 18 + 135}{4096} = \frac{154}{4096} = \frac{77}{2048} \approx 0.0376 \le 0.05$

Since $0.0376 \le 0.05$, the condition $P(X \ge 3) \ge 0.95$ is satisfied for $n = 6$.

Thus, the minimum number of missiles that must be fired is $6$.