Problem 10 - An Introduction to Mathematical Statistics and Its Applications

Question

The gunner on a small assault boat fires six missiles at an attacking plane. Each has a \(20 \%\) chance of being on-target. If two or more of the shells find their mark, the plane will crash. At the same time, the pilot of the plane fires ten air-to-surface rockets, each of which has a \(0.05\) chance of critically disabling the boat. Would you rather be on the plane or the boat?

Step-by-Step Solution

Answer

Based on the calculated probabilities, identify which option (plane or boat) offers the highest survival probability for the given scenario. That is where you would rather be.

Step 1: Finding the probability that the plane will not crash

First, calculate the probability of hitting the plane with none and one missiles out of six. These are the only cases where the plane will not crash. Use the binomial probability formula \(P(k; n, p) = C(n, k) * (p^k) * ((1-p)^{n-k})\) where k is the number of successes, n is the number of trials, p is the probability of success in a single trial, and C(n, k) is the binomial coefficient. Calculate for k = 0 and k = 1; sum these two probabilities to get the probability that the plane does not crash.

Step 2: Finding the probability that the boat will not be disabled

Now, calculate the probability that none of the ten rockets disables the boat. Use the same binomial probability formula with k = 0 to find the probability that none of the rockets critically disables the boat.

Step 3: Comparing the two probabilities

Looking at the two previously calculated probabilities, if being on the plane has a higher survival probability, it is safer. Otherwise, the boat is safer. That is assumed for one round of shooting only; your choice might differ if you are playing for 'the long run'.

Key Concepts

Probability of Success

The concept of "Probability of Success" is essential in understanding binomial probability problems. It represents the chance of a desired outcome occurring in a single attempt.
In our problem, when firing missiles or rockets, each attempt has its specific chance of hitting the target.
For the guns firing missiles, the probability of one missile hitting the plane is 20% or 0.2. For each air-to-surface rocket fired, the probability of critically disabling the boat is 0.05 or 5%. These probabilities tell us how likely each event is to happen in a single trial. They are crucial because they are used as the "p" value in the binomial probability formula.
A side note here: if the probability of success is higher, this increases the likelihood of getting more successful hits, impacting whether the desired outcome, like not crashing, occurs.

Binomial Coefficient

The "Binomial Coefficient" is a key part of calculating binomial probabilities. This coefficient, represented as \(C(n, k)\), helps to determine the number of ways \(k\) successes can occur in \(n\) trials. It can be calculated with the formula: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] This calculation considers all possible combinations of choosing \(k\) successes out of \(n\) attempts.
Suppose we are calculating the probability that exactly one missile hits the plane out of six fired. Here, \(n\) is 6 (the number of missiles), and \(k\) is 1 (the desired number of hits).
The binomial coefficient would help us find how many different combinations exist for exactly one hit, aiding in the full probability calculation.

Probability Calculations

"Probability Calculations" in binomial problems involve using specific formulas to figure out the likelihood of various outcomes. One foundational formula is: \[ P(k; n, p) = C(n, k) * (p^k) * (1-p)^{n-k} \]This formula predicts the probability of having exactly \(k\) successes (or hits) in \(n\) trials.
For example, for the missiles: if we want to find the probability of all six missing the target, \(k = 0\), and we calculate \(C(6, 0)\), \(0.2^0\) (which is 1 since any number to the power of zero is 1), and \((1-0.2)^6\), highlighting no successes.
Adding probabilities from scenarios like zero or one success gives the plane's no-crash probability.
For the boat, we repeat, ensuring none of the ten rockets is a success, focusing on exactly zero disabling events to complete the probability analysis.