Question

Airline tickets can be purchased online, by telephone, or by using a travel agent. Passengers who have a ticket sometimes don't show up for their flights. Suppose a person who purchased a ticket is selected at random. Consider the following events: \(O=\) event selected person purchased ticket online \(N=\) event selected person did not show up for flight $$\text { Suppose } P(O)=0.70, P(N)=0.07, \text { and } P(O \cap N)=0.04$$ a. Are the events \(N\) and \(O\) independent? How can you tell? b. Construct a "hypothetical 1000 " table with columns corresponding to \(N\) and not \(N\) and rows corresponding to \(O\) and not \(O\). c. Use the table to find \(P(O \cup N)\). Give a relative frequency interpretation of this probability.

Short answer

a. The events \(N\) and \(O\) are not independent because \(P(O \cap N)\) is not equal to \(P(O)P(N)\). b. In our hypothetical 1000 person sample, 700 buy their tickets online, 70 do not show up for the flight, and 40 do both. c. \(P(O \cup N) = 0.73\), meaning 73% of ticket buyers will either buy their ticket online or not show up for their flight (or both).

Step-by-step solution

Check if the events \(N\) and \(O\) are independent

In probability, two events are independent if the occurrence of one does not influence the occurrence of another. This can be mathematically stated as \( P(A \cap B) = P(A)P(B) \). Now, comparing \( P(N \cap O) = P(N)P(O) \). If both sides are equal then events are independent. It is known that \( P(N \cap O) = 0.04, P(N) = 0.07 \) and \( P(O) = 0.70 \), so \( P(N)P(O) = 0.07*0.70 = 0.049 \). Since \( P(N \cap O) \neq P(N)P(O) \), the events \(N\) and \(O\) are not independent.

Construct a hypothetical 1000 table

A hypothetical 1000 table means assuming a sample space of 1000 individuals. Then we distribute this according to the probabilities of \(N\) and \(O\). From 1,000 people, 70%, or 700 people buy tickets online (event \(O\)), the rest, 300 people do not. Of those 1,000, 7%, or 70 people, do not show up. Meanwhile, 4% or 40 people who bought tickets online do not show up (event \(N \cap O\)). Consequently, we have 30 people who didn't buy online and didn't show up. The rest, 930 people, are those who showed up.

Finding \(P(O \cup N)\) and its interpretation

The probability of the union of two events, \(O\) and \(N\), can be found by using the principle of inclusion-exclusion, \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). Applying this, we get \( P(O \cup N) = P(N) + P(O) - P(N \cap O) = 0.07 + 0.70 - 0.04 = 0.73 \). Interpreting this, it means that, out of 1,000 individuals, 730 will either buy tickets online or not show up (or both).

Key concepts

Independent Events

When studying probability, the concept of independent events is paramount in determining how one event affects the occurrence of another. For instance, if we designate Event A and Event B as two outcomes, they are said to be independent if the likelihood of Event A happening is unaffected by the occurrence of Event B, and vice versa. This is mathematically expressed as \( P(A \cap B) = P(A)P(B) \).

In the context of our airline ticket problem, if someone purchasing a ticket online (Event O) and that same individual not showing up for the flight (Event N) were independent, you would expect \( P(O \cap N) \) to equal \( P(O)P(N) \). However, since \( P(O \cap N) \eq P(O)P(N) \), we deduce that the two events are not independent – the method of purchasing a ticket does seem to affect the likelihood of a no-show.

Hypothetical 1000 Table

To better visualize probabilities, sometimes a hypothetical 1000 table is a useful tool. Such a table presupposes a sample size of 1000 occurrences to mirror the given probabilities in more concrete terms. It crystallizes abstract probabilities into an easily digestible number of cases.

In the airline example, imagine lining up 1000 passengers: 700 of them bought their tickets online (since 70% of tickets are purchased online), and 70 of them failed to show up for their flights (reflecting the 7% no-show rate). Of those, we are told 40 individuals both bought their tickets online and did not show up. It fills out the table with concrete numbers and helps to visualize the overlap and distinct portions of each event.

Inclusion-Exclusion Principle

A key principle in probability is the inclusion-exclusion principle, which helps us calculate the probability that at least one of two events will happen. The formula is elegantly simple: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).

This principle ensures we do not double-count the joint occurrence of A and B in our probability calculation. Applying this to the flight scenario, we calculate the probability of a randomly selected person either buying a ticket online or not showing up (or both). After applying the principle, we gather that 73% of our hypothetical 1000 passengers fit into this combined category, effectively demonstrating the power of inclusion-exclusion in handling compound probabilities.

Relative Frequency Interpretation

The relative frequency interpretation of probability is a way to understand the likelihood of an event in terms of long-run relative frequencies. By defining probability as the ratio of the number of times an event occurs to the total number of trials, we can translate it to something akin to 'real-world' terms.

From our airplane seats example, \( P(O \cup N) = 0.73 \) indicates that if we repeat the random selection of ticket holders 1000 times, about 730 of them are expected to have bought tickets online or not shown up for their flights (or done both). This frame of reference often gives a more tangible grasp of abstract probability numbers.