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.

Step by step solution

01

Define the Events

Let's define the events:- Let A represent the event of her flying on airline \# 1.- Let B represent the event of her flying on airline \# 2.- Let C represent the event of her flying on airline \# 3.- Let D represent the event of arriving late at exactly one destination.Thus, we need to calculate \( P(A|D), P(B|D), \) and \( P(C|D) \).

02

Use Bayes' Theorem

According to Bayes' theorem, the posterior probability can be calculated as: \[ P(X|D) = \frac{P(D|X)P(X)}{P(D)} \]where \(X\) is any of the airlines \# 1, 2, or 3.

03

Calculate the Prior Probabilities

The prior probabilities based on flying on the airlines are:- \( P(A) = 0.5 \)- \( P(B) = 0.3 \)- \( P(C) = 0.2 \).

04

Determine the Probability of Being Late at Exactly One Destination

For each airline:\- For airline \# 1, late at one destination is given by: \[ P(D|A) = P(\text{Late at D.C. and not Late at L.A.}) + P(\text{Not Late at D.C. and Late at L.A.}) \] \[ = 0.3 \times 0.9 + 0.7 \times 0.1 = 0.27 + 0.07 = 0.34 \]- For airline \# 2, late at one destination: \[ P(D|B) = 0.25 \times 0.8 + 0.75 \times 0.2 = 0.2 + 0.15 = 0.35 \]- For airline \# 3, late at one destination: \[ P(D|C) = 0.4 \times 0.75 + 0.6 \times 0.25 = 0.3 + 0.15 = 0.45 \].

05

Find the Total Probability of Being Late at Exactly One Destination

This is calculated as:\[ P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C) \]\[ = 0.34 \times 0.5 + 0.35 \times 0.3 + 0.45 \times 0.2 \]\[ = 0.17 + 0.105 + 0.09 = 0.365 \].

06

Compute the Posterior Probabilities Using Bayes' Theorem

Now calculate \( P(A|D), P(B|D), \) and \( P(C|D) \):\[ P(A|D) = \frac{P(D|A)P(A)}{P(D)} = \frac{0.34 \times 0.5}{0.365} = 0.4658 \]\[ P(B|D) = \frac{P(D|B)P(B)}{P(D)} = \frac{0.35 \times 0.3}{0.365} = 0.2877 \]\[ P(C|D) = \frac{P(D|C)P(C)}{P(D)} = \frac{0.45 \times 0.2}{0.365} = 0.2465 \].

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Key Concepts

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

Posterior Probability

Posterior probability is an essential aspect of Bayes' Theorem used to update our beliefs based on new evidence. In this exercise, posterior probabilities help us determine which airline she likely used, given that she arrived late at exactly one destination.
When new information becomes available, the posterior probability calculates the likelihood of an event given that evidence. Here, we use Bayes' Theorem:

• Posterior Probability: \( P(X|D) \)
• Where \( X \) is the event of flying on a specific airline, and \( D \) is the event of arriving late at one of the destinations.

Using the posterior probabilities in the solution, we get:

• Airline #1: \( P(A|D) = 0.4658 \)
• Airline #2: \( P(B|D) = 0.2877 \)
• Airline #3: \( P(C|D) = 0.2465 \)

This means that given the information she arrived late at one of the destinations, there's a 46.58% chance she flew airline #1, a 28.77% chance for airline #2, and a 24.65% for airline #3.

Prior Probability

Prior probability refers to the initial assessment of the likelihood of an event based on existing knowledge before incorporating new data. In other words, it's the probability we set before accounting for additional evidence.
In our exercise, prior probability represents the general idea of which airline the friend typically flies:

• Airline #1: 50% of the time (\( P(A) = 0.5 \))
• Airline #2: 30% of the time (\( P(B) = 0.3 \))
• Airline #3: 20% of the time (\( P(C) = 0.2 \))

These figures establish the framework for our calculations and provide a foundation upon which Bayes' Theorem builds. By understanding prior probabilities, we recognize how regularly occurring events influence our initial claims, impacting the final conclusions once additional data is included.

Conditional Probability

Conditional probability focuses on the probability of an event happening given that another event has already occurred. This concept is core to Bayesian statistics and problems like our exercise.
In this context, we need to calculate the chances of being late at exactly one destination, given the airline choice:

• Airline #1: \( P(D|A) = 0.34 \)
• Airline #2: \( P(D|B) = 0.35 \)
• Airline #3: \( P(D|C) = 0.45 \)

Conditional probabilities bridge the gap between what we knew before (prior probabilities) and what we've learned with new data (posterior probabilities).
They assist in modeling real-world situations where the occurrence of one event directly influences the likelihood of other events. By calculating precise likelihoods of being late for each airline, we can make informed guesses about her choices based on observable outcomes.