91Ó°ÊÓ

1,30 \%\( of the time on airline \)\… # A friend who lives in Los Angeles makes frequent consulting trips to Washington, D.C.; \(50 \%\) of the time she travels on airline \(\\# 1,30 \%\) of the time on airline \(\\# 2\), and the remaining \(20 \%\) of the time on airline #3. For airline #1, flights are late into D.C. \(30 \%\) of the time and late into L.A. \(10 \%\) of the time. For airline \(\\# 2\), these percentages are \(25 \%\) and \(20 \%\), whereas for airline \(\\# 3\) the percentages are \(40 \%\) and \(25 \%\). If we learn that on a particular trip she arrived late at exactly one of the two destinations, what are the posterior probabilities of having flown on airlines #1, #2, and #3? Assume that the chance of a late arrival in L.A. is unaffected by what happens on the flight to D.C. [Hint: From the tip of each first-generation branch on a tree diagram, draw three second-generation branches labeled, respectively, 0 late, 1 late, and 2 late.]

Step by step solution

01

Understand the Problem and Concepts Involved

We are asked to find the posterior probabilities of flying on a specific airline given that the traveler arrives late at exactly one of two destinations. This involves using Bayes' Theorem and requires understanding the probabilities of late arrivals and prior probabilities of choosing each airline.

02

Define Events and Probabilities

* Let F1, F2, and F3 represent flying on airlines #1, #2, and #3 respectively. * Let A mean arriving late at one destination only (either D.C. or L.A.). * Use the given probabilities: P(F1) = 0.5, P(F2) = 0.3, P(F3) = 0.2; conditional probabilities for each airline's late arrivals to D.C. and L.A.

03

Calculate Probability of 'Arrive Late to Exactly One Destination'

For each airline, calculate P(A | Fi). - For Airline #1: P(A | F1) = P(Late at D.C., Not Late at L.A.) + P(Not Late at D.C., Late at L.A.) = (0.3 * 0.9) + (0.7 * 0.1) = 0.27 + 0.07 = 0.34. - For Airline #2: P(A | F2) = (0.25 * 0.8) + (0.75 * 0.2) = 0.20 + 0.15 = 0.35. - For Airline #3: P(A | F3) = (0.4 * 0.75) + (0.6 * 0.25) = 0.3 + 0.15 = 0.45.

04

Apply Total Probability Theorem

Calculate the total probability P(A) by summing the weighted conditional probabilities:\[ P(A) = P(A|F1)P(F1) + P(A|F2)P(F2) + P(A|F3)P(F3) = (0.34)(0.5) + (0.35)(0.3) + (0.45)(0.2) = 0.17 + 0.105 + 0.09 = 0.365. \]

05

Calculate Posterior Probabilities Using Bayes' Theorem

Use Bayes' Theorem to find the posterior probability for each airline:\[ P(F1|A) = \frac{P(A|F1)P(F1)}{P(A)} = \frac{0.34 \times 0.5}{0.365} \approx 0.4658, \]\[ P(F2|A) = \frac{P(A|F2)P(F2)}{P(A)} = \frac{0.35 \times 0.3}{0.365} \approx 0.2877, \]\[ P(F3|A) = \frac{P(A|F3)P(F3)}{P(A)} = \frac{0.45 \times 0.2}{0.365} \approx 0.2466. \]

06

Verify and Interpret the Results

Check calculations for each probability, ensuring they sum to 1 and correctly interpret the results: P(F1|A) = 0.4658, P(F2|A) = 0.2877, P(F3|A) = 0.2466. These values represent the conditional probabilities of flying on each airline given the specific condition of arriving late at only one destination.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Posterior Probability

Posterior probability is an important concept in probability theory, especially in the context of Bayes' Theorem. It refers to the probability of an event occurring after taking into account new evidence or information. In the context of airline flights, we are interested in determining the probability of traveling on a specific airline given that the traveler has arrived late at exactly one destination.

To calculate this, we use the formula from Bayes' Theorem:

• Let \( P(F_i|A) \) represent the probability of flying on airline \( i \) given that the passenger arrived late at one destination.
• \( P(F_i) \) is the prior probability of choosing airline \( i \).
• \( P(A|F_i) \) is the probability of arriving late at one destination given the flight was on airline \( i \).
• \( P(A) \) is the total probability of arriving late at one destination, considering all airlines.

Bayes' Theorem tells us that:\[ P(F_i|A) = \frac{P(A|F_i)P(F_i)}{P(A)} \]This helps us update our beliefs about which flight she probably took, based on the fact that she arrived late at one destination.

Total Probability Theorem

The Total Probability Theorem is crucial when evaluating the overall probability of an outcome that can occur through several different pathways or scenarios. In this problem, the theorem helps us determine the probability that the traveler arrived late at exactly one destination using any of the three airlines.

When using the Total Probability Theorem, it's important to first identify all possible scenarios:

• The probability she arrived late at one destination while flying on each airline, \( P(A|F_i) \), was already determined in the exercise steps.
• The probability of her choosing each airline, \( P(F_i) \), is based on her past travel choices.

The Total Probability Theorem combines these to find \( P(A) \), the probability of arriving late at one destination, no matter the airline:\[ P(A) = P(A|F_1)P(F_1) + P(A|F_2)P(F_2) + P(A|F_3)P(F_3) \]This gives us a comprehensive view of the event 'late arrival at one destination' by considering all the probabilities associated with different flight choices.

Conditional Probability

Conditional probability is a fundamental part of many calculations involving probability, including those in Bayes' Theorem and the Total Probability Theorem. It captures the idea of finding the probability of one event occurring given that another has already occurred.

In the airline problem, we define the following:

• \( P(A|F_i) \), the conditional probability of arriving late at exactly one destination given that airline \( i \) was flown.

This involves combining the probabilities of being late on different legs of the trip. For instance:

• For airline #1, \( P(A|F_1) \) is the sum of being late into D.C. but not L.A., and being late into L.A. but not D.C.
• This is calculated by:\[ P(A|F_1) = P(Late \text{ D.C.}, \text{ Not Late L.A.}) + P(\text{Not Late D.C.}, \text{ Late L.A.}) \]
• Such analysis is repeated for each airline using the respective probabilities.

Conditional probabilities allow us to isolate and evaluate each airline scenario in the context of the given event, "arriving late at exactly one destination."

Tree Diagram

Tree diagrams are visual tools used for organizing and solving probability problems. They help in breaking down complex probability scenarios into simpler, more manageable parts by illustrating all possible outcomes and their probabilities.

For our exercise with flights and late arrivals, a tree diagram can clearly represent:

• Each initial branch representing an airline choice: Airline #1, #2, and #3.
• From each of these branches, further split into scenarios like '0 late', '1 late', and '2 late'.
• The probabilities associated with being late or not at D.C. or L.A. for each airline.

By using a tree diagram, we can visually track the flow of different events (possible combinations of late arrivals) and calculate conditional probabilities at each stage. It's especially useful for determining the right conditional probabilities required for applying formulas like those in Bayes' Theorem.

A well-detailed tree diagram aligns with the probabilities computed through steps, reinforcing understanding and ensuring no scenario is missed.