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Chapter 2: Problem 59

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) Probability is approximately 0.067. (b) Probability is approximately 0.509.

Step by step solution

01

Understand the Given Information

We are given the following PROBABILITY DATA: - Probability of an aircraft being discovered is 70%, i.e., \( P(D) = 0.7 \).- Probability that a discovered aircraft has a locator is 60%, i.e., \( P(L|D) = 0.6 \). - Probability that an undiscovered aircraft does not have a locator is 90%, i.e., \( P(L^c | D^c) = 0.9 \).
02

Set Up the Initial Probabilities

From the given information, we extract:1. \( P(D^c) = 0.3 \), since either the aircraft is discovered (70%) or not discovered (30%).2. \( P(L^c|D) = 0.4 \), since 60% have a locator when discovered.3. \( P(L|D^c) = 0.1 \), since 90% do not have a locator when not discovered.
03

Calculate the Probability of Undiscovered with Locator (a)

We are to find \( P(D^c|L) \). Use Bayes' theorem: \[ P(D^c|L) = \frac{P(L|D^c) \times P(D^c)}{P(L)} \] However, we need to first determine \( P(L) \) using the law of total probability:\[ P(L) = P(L|D) \times P(D) + P(L|D^c) \times P(D^c) \]Substitute the values: \[ P(L) = (0.6 \times 0.7) + (0.1 \times 0.3) = 0.42 + 0.03 = 0.45 \]Now use Bayes' theorem with these values:\[ P(D^c|L) = \frac{0.1 \times 0.3}{0.45} = \frac{0.03}{0.45} = \frac{1}{15} \approx 0.067 \]
04

Calculate the Probability of Discovered without Locator (b)

We are to find \( P(D|L^c) \). Use Bayes' theorem:\[ P(D|L^c) = \frac{P(L^c|D) \times P(D)}{P(L^c)} \] Determine \( P(L^c) \) using the law of total probability:\[ P(L^c) = P(L^c|D) \times P(D) + P(L^c|D^c) \times P(D^c) \]Substitute values:\[ P(L^c) = (0.4 \times 0.7) + (0.9 \times 0.3) = 0.28 + 0.27 = 0.55 \]Now use Bayes' theorem:\[ P(D|L^c) = \frac{0.4 \times 0.7}{0.55} = \frac{0.28}{0.55} \approx 0.509 \]

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Key Concepts

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

Conditional Probability
Conditional Probability is a key concept in probability theory that helps us determine the likelihood of an event occurring given that another event has already occurred. Imagine you are trying to guess the chances of it raining, knowing that it is cloudy. Or, in our aircraft situation, discovering chances given if it has a locator or not.

Here is a simple breakdown to understand conditional probability better:
  • We denote conditional probability as \( P(A|B) \). This symbolizes the probability of event A occurring given that event B has occurred.

  • The formula to calculate is \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), where \( P(A \cap B) \) is the joint probability of both A and B occurring.
In our exercise, if we want to assess the probability that a disappeared aircraft is discovered given it does not have a locator, we use conditional probability techniques. Knowing the conditional probabilities helps in weighing possibilities based on given conditions.
Law of Total Probability
The Law of Total Probability is an essential tool in situations where a difficult probability can be broken down based on different, manageable scenarios. It lets us piece together individual probabilities to find a total probability of an event.

To grasp it, think of it as using several different lenses to view a picture to understand it fully:
  • The law can be expressed as \( P(B) = \sum P(B|A_i)P(A_i) \), where \( A_i \) represents all possible outcomes.

  • In the light aircraft problem, to find the overall probability of an aircraft having a locator, \( P(L) \), we consider both scenarios: discovered and not discovered.
This approach allows for calculating probabilities in complex situations efficiently by recognizing every possible contributing scenario. By doing so, we can make more informed estimations about our overall event of interest.
Bayes' Theorem
Bayes' Theorem is a fundamental theorem in probability that provides a way to update our probability estimates after obtaining more evidence. It's named after the famous statistician Thomas Bayes.

Consider Bayes' Theorem as a way of refining our initial guesses with new information.
  • The formula is \( P(A|B) = \frac{P(B|A)P(A)}{P(B)} \).

  • It involves prior probabilities (our initial guess), likelihood (data fitting the model), and posterior probabilities (updated belief).
  • In our exercise, Bayes' Theorem allows determines the probability of an aircraft being undiscovered, given it has a locator. By using the prior data and considering conditional probabilities, we derive a refined probability.
That's the beauty of Bayes' Theorem! It takes new evidence into account to enhance accuracy of predictions or assessments in decision-making.
Probability Theory
Probability Theory is the mathematical framework for quantifying uncertainty. At its core, it's about measuring how likely it is for different outcomes to occur, leaning on both logic and mathematics to guide evidential understanding.

Here's how it connects to everyday reasoning:
  • It incorporates classical events like coin flips to complex scenarios like aircraft discoveries.

  • Key elements include sample spaces, events, probabilities assigned to events, and foundational rules like addition and multiplication rules.
  • Application of probability theory helps address real-world problems, such as predicting the likelihood of an aircraft being found based on conditions.
In short, probability theory forms the bedrock of statistical reasoning, ensuring our decisions are backed by data and logical probabilities. It acts like a compass, guiding us through the uncertain terrains of decision-making based on data and evidence.

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Most popular questions from this chapter

Consider independently rolling two fair dice, one red and the other green. Let \(A\) be the event that the red die shows 3 dots, \(B\) be the event that the green die shows 4 dots, and \(C\) be the event that the total number of dots showing on the two dice is 7 . Are these events pairwise independent (i.e., are \(A\) and \(B\) independent events, are \(A\) and \(C\) independent, and are \(B\) and \(C\) independent)? Are the three events mutually independent?

For customers purchasing a full set of tires at a particular tire store, consider the events \(A=\\{\) tires purchased were made in the United States ] \(B=\\{\) purchaser has tires balanced immediately \(\\}\) \(C=\) \\{purchaser requests front-end alignment \(\\}\) along with \(A^{\prime}, B^{\prime}\), and \(C^{\prime}\). Assume the following unconditional and conditional probabilities: $$ \begin{aligned} &P(A)=.75 \quad P(B \mid A)=.9 \quad P\left(B \mid A^{\prime}\right)=.8 \\ &P(C \mid A \cap B)=.8 \quad P\left(C \mid A \cap B^{\prime}\right)=.6 \\ &P\left(C \mid A^{\prime} \cap B\right)=.7 \quad P\left(C \mid A^{\prime} \cap B^{\prime}\right)=.3 \end{aligned} $$ a. Construct a tree diagram consisting of first-, second-, and third- generation branches and place an event label and appropriate probability next to each branch. b. Compute \(P(A \cap B \cap C)\). c. Compute \(P(B \cap C)\) d. Compute \(P(C)\). e. Compute \(P(A \mid B \cap C)\) the probability of a purchase of U.S. tires given that both balancing and an alignment were requested.

Allan and Beth currently have \(\$ 2\) and \(\$ 3\), respectively. A fair coin is tossed. If the result of the toss is \(\mathrm{H}\), Allan wins \(\$ 1\) from Beth, whereas if the coin toss results in \(\mathrm{T}\), then Beth wins \(\$ 1\) from Allan. This process is then repeated, with a coin toss followed by the exchange of \(\$ 1\), until one of the two players goes broke (one of the two gamblers is ruined). We wish to determine \(a_{2}=P\) (Allan is the winner \(\mid\) he starts with \(\$ 2\) ) To do so, let's also consider \(a_{i}=P\) (Allan wins | he starts with \(\$ i\) ) for \(i=0,1,3,4\), and 5 . a. What are the values of \(a_{0}\) and \(a_{5}\) ? b. Use the law of total probability to obtain an equation relating \(a_{2}\) to \(a_{1}\) and \(a_{3}\). [Hint: Condition on the result of the first coin toss, realizing that if it is a \(\mathrm{H}\), then from that point Allan starts with \$3.] c. Using the logic described in (b), develop a system of equations relating \(a_{i}(i=1,2,3,4)\) to \(a_{i-1}\) and \(a_{i+1}\). Then solve these equations. [Hint: Write each equation so that \(a_{i}-a_{i-1}\) is on the left hand side. Then use the result of the first equation to express each other \(a_{i}-a_{i-1}\) as a function of \(a_{1}\), and add together all four of these expressions \((i=2,3,4,5)\).] d. Generalize the result to the situation in which Allan's initial fortune is \(\$ a\) and Beth's is \(\$ b\). Note: The solution is a bit more complicated if \(p=P(\) Allan wins \(\$ 1) \neq .5 .\)

At a gas station, \(40 \%\) of the customers use regular gas \(\left(A_{1}\right), 35 \%\) use mid-grade gas \(\left(A_{2}\right)\), and \(25 \%\) use premium gas \(\left(A_{3}\right)\). Of those customers using regular gas, only \(30 \%\) fill their tanks (event \(B\) ). Of those customers using mid- grade gas, \(60 \%\) fill their tanks, whereas of those using premium, \(50 \%\) fill their tanks. a. What is the probability that the next customer will request mid-grade gas and fill the tank \(\left(A_{2} \cap B\right)\) ? b. What is the probability that the next customer fills the tank? c. If the next customer fills the tank, what is the probability that regular gas is requested? midgrade gas? Premium gas?

For any events \(A\) and \(B\) with \(P(B)>0\), show that \(P(A \mid B)+P\left(A^{\prime} \mid B\right)=1 .\)
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