Problem 80 - An Introduction to Mathematical Statistics and Its Applications

Question

An underground military installation is fortified to the extent that it can withstand up to three direct hits from air-to-surface missiles and still function. Suppose an enemy aircraft is armed with missiles, each having a \(30 \%\) chance of scoring a direct hit. What is the probability that the installation will be destroyed with the seventh missile fired?

Step-by-Step Solution

Answer

The probability that the installation will be destroyed with the seventh missile fired is 0.0559 or 5.59%.

Step 1: Identifying Variables

In this problem, having a direct hit from a missile is considered a 'success'. The fort can withstand up to three hits and therefore the fourth hit will destroy it. So, the fourth success is the point of interest, which means \(r=4\). The probability of a direct hit, \(p\), is given as $30\%$ or \(0.3\). The total number of trials, \(k\), where the fourth success (destruction of the fort) is reached is seven in this case.

Step 2: Applying the Formula

Substitute the identified variables into the negative binomial distribution formula: \(P(X=7) = C(7-1, 4-1) * (0.3^4) * (1-0.3)^(7-4)\). Here, \(C(7-1, 4-1)\) represents the number of combinations of \(7-1\) items taken \(4-1\) at a time.

Step 3: Computing Combination

Calculate \(C(6, 3)\) which is equal to \(\frac{6!}{3! (6-3)!} = 20\). Substituting this value back into the equation we get: \(P(X=7) = 20 * (0.3^4) * (0.7^3)\).

Step 4: Calculation

Finally, perform the calculations in the equation to find the probability: \(P(X=7) = 20 * (0.0081) * (0.343) = 0.0559\).

Key Concepts

Probability Theory

Probability theory is a branch of mathematics that deals with the likelihood or chance of different outcomes. In any probability problem, we define an event and determine the chance of that event happening.
In this exercise, the event is the military installation being hit by a missile. Each missile has a probability of 0.3, or 30%, to hit the target directly. The installation can endure up to three hits before it's destroyed.
This leads us to evaluate complex probabilities using specific distributions, like the negative binomial distribution, which helps figure out how many trials are needed to achieve a certain number of 'successes'.

• A 'success' is defined by a direct hit.
• The fort withstands up to three hits, so destruction requires a fourth hit.
• Probability calculations consider both successful and failed attempts.

Understanding these points is key for solving this type of probability question.

Combination Formula

The combination formula is crucial when dealing with problems like the one in the exercise. It helps determine how many different ways we can arrange a specified number of successes from a given set of trials.
The formula itself is expressed as \(C(n, k) = \frac{n!}{k!(n-k)!}\), where \(n\) is the total number of items to choose from, and \(k\) is the number of items to choose. In the context of the exercise:
We calculate \(C(6, 3)\), representing the number of ways three successes (hits) can occur in the first six missile firings.
Using the formula:

• \(n = 6\) and \(k = 3\)
• \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1\)
• \(3! = 3 \times 2 \times 1\) and \((6-3)! = 3!\)
• This results in \(\frac{720}{6 \times 6} = 20\)

Thus, there are 20 different ways to carry out these missile strikes with exactly three successes within the first six trials.

Direct Hit Probability

Direct hit probability refers to the likelihood that a missile will successfully hit its intended target. For each missile, this probability is given as 30%, or numerically, 0.3.
Using probability theory, we find the probability of exactly four hits occurring by the time the seventh missile is fired. This uses the direct hit probability raised to the power equal to the number of successes needed (four direct hits), and the probability of a miss \((1-0.3 = 0.7)\) raised to the power of the number of fails (three misses).
Therefore, the calculation includes:

• \((0.3)^4 = 0.0081\) represents the chance of four successive hits.
• \((0.7)^3 = 0.343\) represents the chance of three failures during the first six trials.
• Combine this with the combination count (20), resulting in the probability of the installation being destroyed with the seventh missile fired: \(20 \times 0.0081 \times 0.343\).

This approach helps illustrate the direct hit probability's role in determining specific outcomes in consecutive missile launches.