Question

An airline offers discounted "advancepurchase" fares to customers who buy tickets more than 30 days before travel and charges "regular" fares for tickets purchased during those last 30 days. The company has noticed that \(60 \%\) of its customers take advantage of the advance-purchase fares. The "no- show" rate among people who paid regular fares is \(30 \%\), but only \(5 \%\) of customers with advance-purchase tickets are no-shows. a) What percent of all ticket holders are no-shows? b) What's the probability that a customer who didn't show had an advance- purchase ticket? c) Is being a no-show independent of the type of ticket a passenger holds? Explain.

Short answer

a) 15% are no-shows. b) 20% had advance-purchase tickets. c) No, it's not independent.

Step-by-step solution

Define the Variables

Let's define the variables for clarity: Let \(A\) be the event that a customer buys an advance-purchase ticket, and \(R\) be the event for a regular ticket. Then \(P(A) = 0.6\) and \(P(R) = 1 - P(A) = 0.4\). Let \(N\) represent the event that a customer is a no-show. We are given that \(P(N | R) = 0.3\) and \(P(N | A) = 0.05\).

Calculate the Total Probability of No-Show

Use the law of total probability to find \(P(N)\). The formula is: \[ P(N) = P(N|A) \cdot P(A) + P(N|R) \cdot P(R) \] Substitute the values: \[ P(N) = (0.05 \cdot 0.6) + (0.3 \cdot 0.4) \] Calculate to find \(P(N)\).

Calculate \(P(N)\)

Compute the multiplication:\[ P(N) = (0.05 \times 0.6) + (0.3 \times 0.4) = 0.03 + 0.12 = 0.15 \] Thus, \(15\%\) of all ticket holders are no-shows.

Calculate Probability of Advance-Purchase Given No-Show

Use Bayes' Theorem to find \(P(A|N)\): \[ P(A|N) = \frac{P(N|A) \cdot P(A)}{P(N)} \] Substitute the known values: \[ P(A|N) = \frac{0.05 \cdot 0.6}{0.15} \]

Solve for \(P(A|N)\)

Calculate the above expression: \[ P(A|N) = \frac{0.03}{0.15} = 0.2 \] Thus, the probability that a no-show had an advance-purchase ticket is \(20\%\).

Determine if Being a No-Show is Independent of Ticket Type

Two events \(X\) and \(Y\) are independent if \(P(X|Y) = P(X)\). Check if \(P(N|A) = P(N)\) and \(P(N|R) = P(N)\). Since \(P(N|A) = 0.05 eq 0.15\) and \(P(N|R) = 0.3 eq 0.15\), being a no-show is not independent of the ticket type.

Key concepts

Bayes' Theorem

Bayes' Theorem is a crucial concept in probability, providing us the ability to reverse conditional probabilities. It is especially helpful when we want to determine the probability of an event given that another event has already occurred. In simpler terms, it helps answer questions like, "What's the probability of A happening, given that B has happened?"Bayes' Theorem is expressed through the formula:\[ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} \]This equation allows us to calculate the probability of event \(A\) occurring given event \(B\), using the probability that \(B\) happens if \(A\) is true, and the individual probabilities of \(A\) and \(B\).In the original exercise, we found out the probability that a customer who didn’t show up had an advance-purchase ticket using Bayes’ Theorem. This was calculated as:- \( P(A|N) \), where \(A\) is the event of having an advance-purchase ticket and \(N\) is the no-show event.- By substituting values, we computed \( P(A|N) = 0.2 \), implying that the probability is \( 20\% \).Bayes' Theorem is powerful as it allows transforming prior knowledge into useful predictive statistics, essential in decision-making fields such as finance, medicine, and machine learning.

Law of Total Probability

The Law of Total Probability is an essential theorem in probability theory. It helps in finding the probability of a certain event by breaking down complex systems into simpler parts. Think of it as piecing together a puzzle where each piece represents a part of a complete probability.According to the law, if you have a set of outcomes \( \{ B_1, B_2, \ldots, B_n \} \) that partition the sample space, it's possible to determine the probability of another event \(A\) by summing the joint probabilities of \(A\) occurring with each of those outcomes.The formula is expressed as:\[ P(A) = \sum_{i} P(A|B_i) \cdot P(B_i) \]This means the total probability of event \(A\) is the sum of the probabilities that \(A\) occurs given each of the sub-events \(B_i\), weighed by the probability of each \(B_i\) occurring.In the exercise, the Law of Total Probability was employed to calculate \(P(N)\), the chance of a customer being a no-show:- Calculated as: \( P(N|A) \cdot P(A) + P(N|R) \cdot P(R) \).- This became \( 0.03 + 0.12 = 0.15 \), indicating \( 15\% \) total no-show rate.Understanding this law provides a foundational skill in solving real-world problems involving probabilistic analysis.

Independence of Events

In probability theory, two events are independent if the occurrence of one does not affect the probability of the other. This concept is quite intuitive; if flipping a fair coin and rolling a fair die, the outcome of one doesn’t impact the outcome of the other.The formal rule to determine whether events \(X\) and \(Y\) are independent is:\[ P(X|Y) = P(X) \]This means that the probability of \(X\) happening given \(Y\) has occurred is the same as it would be if \(Y\) had not occurred.In the exercise, we checked the independence of being a no-show from the type of ticket purchased. The steps involved:- Comparing \(P(N|A)\) with \(P(N)\) and \(P(N|R)\) with \(P(N)\).- Since \(P(N|A) eq P(N)\) (0.05 vs. 0.15) and \(P(N|R) eq P(N)\) (0.3 vs. 0.15), the no-show event is dependent on the ticket type.Understanding event independence is essential because it impacts how we calculate probabilities and interpret results. Recognizing dependencies helps refine models to reflect real-world scenarios more accurately.