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Seventy percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of the aircraft that are discovered, \(60 \%\) have an emergency locator, whereas \(90 \%\) of the aircraft not discovered do not have such a locator. Suppose a light aircraft has disappeared. a. If it has an emergency locator, what is the probability that it will not be discovered? b. If it does not have an emergency locator, what is the probability that it will be discovered?

Short Answer

Expert verified
a. 0.067; b. 0.509.

Step by step solution

01

Define Probabilities and Notations

Let's define the events to make problem solving easier. Let \( D \) be the event that an aircraft is discovered, \( L \) the event the aircraft has a locator, and \( \overline{D} \) the event the aircraft is not discovered. We are given that \( P(D) = 0.7 \), \( P(L | D) = 0.6 \), and \( P(\overline{L} | \overline{D}) = 0.9 \). We also know that \( P(\overline{D} | L) \) and \( P(D | \overline{L}) \) need to be calculated.
02

Calculate the Probability of Not Having a Locator if Not Discovered

Using the law of total probability, \( P(\overline{D}) = 0.3 \). We know \( P(\overline{L} | \overline{D}) = 0.9 \). We can also infer \( P(L | \overline{D}) = 1 - P(\overline{L} | \overline{D}) = 0.1 \). These are related events of having a locator or not when discovered or not-discovered.
03

Apply Bayes' Theorem for Non-discovery with a Locator

By Bayes' Theorem: \(P(L) = P(L|D) \cdot P(D) + P(L|\overline{D}) \cdot P(\overline{D}) = 0.6 \cdot 0.7 + 0.1 \cdot 0.3 = 0.42 + 0.03 = 0.45\). Now, \( P(\overline{D} | L) = \frac{P(L | \overline{D}) \cdot P(\overline{D})}{P(L)} = \frac{0.1 \cdot 0.3}{0.45} = \frac{0.03}{0.45} \approx 0.067 \).
04

Calculate Probability of Discovery without a Locator

Since \( P(D) = 0.7 \), by complementarity \( P(\overline{D}) = 0.3 \). Using Bayes' Theorem, \( P(\overline{L}) = P(\overline{L} | D) \cdot P(D) + P(\overline{L} | \overline{D}) \cdot P(\overline{D}) = 0.4 \cdot 0.7 + 0.9 \cdot 0.3 = 0.28 + 0.27 = 0.55 \). Thus \( P(D | \overline{L}) = \frac{P(\overline{L} | D) \cdot P(D)}{P(\overline{L})} = \frac{0.4 \cdot 0.7}{0.55} = \frac{0.28}{0.55} \approx 0.509 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conditional Probability
Conditional probability helps us determine the likelihood of an event given that another event has already occurred. It is expressed as \( P(A|B) \), which is the probability of event A happening given that event B has occurred. In the context of the exercise, we use conditional probability to understand how the likelihood of an aircraft being discovered changes depending on whether it has an emergency locator.
For example, when trying to find the probability of an aircraft not being discovered even though it has an emergency locator, we are essentially looking for \( P(\overline{D} | L) \). It's important to note that conditional probability is a core component of Bayes' Theorem, which further helps in swapping probabilistic dependencies to provide insights into reverse conditions.
Probability Theory
Probability theory is the mathematical framework that quantifies uncertainty. It allows us to model random events and reason about their likelihood.
In this exercise, probability theory plays a crucial role as it outlines the structure to deal with both discovered and non-discovered aircraft, breaking down the scenarios involving emergency locators through probabilities like \( P(D) \) and \( P(\overline{D}) \).
Probability theory builds upon foundational principles such as the sum of probabilities of all possible outcomes equaling 1 and concepts like complements, intersections, and unions of events. This fundamental approach allows us to break down complex problems into more manageable steps, ultimately leading to solutions through logical inference and calculations.
Emergency Locator
An emergency locator is a device that helps search and rescue teams find an aircraft more easily when it has disappeared.
In probability calculations, having an emergency locator influences the likelihood of the aircraft being discovered. For example, the exercise indicates that 60% of the aircraft that are discovered have emergency locators, reflecting the effectiveness of these devices in aiding discovery efforts. If an aircraft does not have a locator, there's a 90% chance that it will not be discovered, highlighting their importance.
Understanding the impact of an emergency locator on probability allows one to better appreciate how critical these devices are in search and rescue operations.
Discovered Aircraft Calculations
Calculations related to discovered aircraft help assess the probability of different outcomes associated with the presence or absence of emergency locators.
Bayes' Theorem plays a pivotal role in these calculations, allowing us to reverse the conditional probabilities and derive insights such as the probability of non-discovery for an aircraft with a locator \( P(\overline{D} | L) \) or discovery probability without a locator \( P(D | \overline{L}) \).
The exercise ultimately leads us to understand that, although having an emergency locator increases the chances of discovery, there are scenarios where other factors may play a role, which is elegantly quantified through these calculated probabilities.

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Most popular questions from this chapter

Professor Stan der Deviation can take one of two routes on his way home from work. On the first route, there are four railroad crossings. The probability that he will be stopped by a train at any particular one of the crossings is .1, and trains operate independently at the four crossings. The other route is longer but there are only two crossings, independent of one another, with the same stoppage probability for each as on the first route. On a particular day, Professor Deviation has a meeting scheduled at home for a certain time. Whichever route he takes, he calculates that he will be late if he is stopped by trains at at least half the crossings encountered. a. Which route should he take to minimize the probability of being late to the meeting? b. If he tosses a fair coin to decide on a route and he is late, what is the probability that he took the four-crossing route?

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One box contains six red balls and four green balls, and a second box contains seven red balls and three green balls. A ball is randomly chosen from the first box and placed in the second box. Then a ball is randomly selected from the second box and placed in the first box. a. What is the probability that a red ball is selected from the first box and a red ball is selected from the second box? b. At the conclusion of the selection process, what is the probability that the numbers of red and green balls in the first box are identical to the numbers at the beginning?

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