Understanding the Missile Hit Probability Problem

The question asks for the minimum number of missiles required to achieve a certain probability of hitting a target at least once. We are given the probability that a single missile hits the target and the desired minimum probability for at least one hit.

Analyzing the Given Information

• Probability of a single missile hitting the target, $p = 0.3$.
• Desired minimum probability of hitting the target at least once = 80% or 0.80.

We need to find the least number of missiles, let's call this number $n$, such that the probability of at least one hit is greater than or equal to 0.80.

Probability of Not Hitting the Target

If the probability of a missile hitting the target is $p = 0.3$, then the probability of a single missile missing the target is $q = 1 - p$.

So, $q = 1 - 0.3 = 0.7$.

Assuming each missile firing is an independent event, the probability that all $n$ missiles miss the target is the product of the individual probabilities of missing:

Probability (all $n$ missiles miss) = $(0.7)^n$.

Probability of At Least One Hit

The event "at least one hit" is the complement of the event "all $n$ missiles miss". Therefore, the probability of at least one hit is:

Probability (at least one hit) = 1 - Probability (all $n$ missiles miss)

Probability (at least one hit) = $1 - (0.7)^n$.

Setting Up the Inequality

We are given that the probability of at least one hit must be at least 80%, which is 0.80. So, we set up the inequality:

$1 - (0.7)^n \ge 0.80$

Solving for the Least Number of Missiles (n)

We need to find the smallest integer value of $n$ that satisfies the inequality. Let's rearrange the inequality:

$- (0.7)^n \ge 0.80 - 1$

$- (0.7)^n \ge -0.20$

Multiply both sides by -1 and reverse the inequality sign:

$(0.7)^n \le 0.20$

Now, we can test integer values for $n$ starting from 1:

• For $n=1$: $(0.7)^1 = 0.7$. Is $0.7 \le 0.20$? No.
• For $n=2$: $(0.7)^2 = 0.7 \times 0.7 = 0.49$. Is $0.49 \le 0.20$? No.
• For $n=3$: $(0.7)^3 = 0.49 \times 0.7 = 0.343$. Is $0.343 \le 0.20$? No.
• For $n=4$: $(0.7)^4 = 0.343 \times 0.7 = 0.2401$. Is $0.2401 \le 0.20$? No.
• For $n=5$: $(0.7)^5 = 0.2401 \times 0.7 = 0.16807$. Is $0.16807 \le 0.20$? Yes.

Since $n=5$ is the smallest integer value for which the inequality $(0.7)^n \le 0.20$ holds true, the least number of missiles that should be fired is 5.

Alternatively, we can use logarithms to solve $(0.7)^n \le 0.20$:

$\log((0.7)^n) \le \log(0.20)$

$n \log(0.7) \le \log(0.20)$

Since $\log(0.7)$ is negative (because $0.7 < 1$), we divide by $\log(0.7)$ and reverse the inequality sign:

$n \ge \frac{\log(0.20)}{\log(0.7)}$

Using a calculator:

$n \ge \frac{-0.69897}{-0.15490} \approx 4.516$

Since $n$ must be an integer (you cannot fire a fraction of a missile), the smallest integer value of $n$ that is greater than or equal to 4.516 is 5.

Conclusion

The least number of missiles that should be fired to have at least an 80% probability that the target is hit is 5.

Probability Calculations Revision

Additional Information on Probability Concepts

This problem involves basic probability concepts and the idea of independent events.

• Independent Events: Two events are independent if the outcome of one event does not affect the outcome of the other. In this problem, it's assumed that whether one missile hits or misses does not influence the outcome of any other missile.
• Complementary Events: Two events are complementary if they are mutually exclusive and together they cover all possible outcomes. The event "at least one hit" and the event "no hits" are complementary. The sum of their probabilities is 1.
• Binomial Probability: While we didn't explicitly use the full binomial probability formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$, the problem is based on a binomial distribution scenario. $X$ represents the number of successes (hits) in $n$ trials, where the probability of success $p$ is constant for each trial. The probability of 0 hits is the specific case $P(X=0) = \binom{n}{0} p^0 q^{n-0} = 1 \times 1 \times q^n = q^n$.