/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 Airline tickets can be purchased... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Expert verified
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

01

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.
02

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.
03

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).

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.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A Gallup survey found that \(46 \%\) of women and \(37 \%\) of men experience pain on a daily basis (San Luis Obispo Tribune, April 6,2000 ). Suppose that this information is representative of U.S. adults. If a U.S. adult is selected at random, are the events selected adult is male and selected adult experiences pain on a daily basis independent or dependent? Explain.

Suppose you want to estimate the probability that a randomly selected customer at a particular grocery store will pay by credit card. Over the past 3 months, 80,500 payments were made, and 37,100 of them were by credit card. What is the estimated probability that a randomly selected customer will pay by credit card?

Roulette is a game of chance that involves spinning a wheel that is divided into 38 equal segments, as shown in the accompanying picture. A metal ball is tossed into the wheel as it is spinning, and the ball eventually lands in one of the 38 segments. Each segment has an associated color. Two segments are green. Half of the other 36 segments are red, and the others are black. When a balanced roulette wheel is spun, the ball is equally likely to land in any one of the 38 segments. a. When a balanced roulette wheel is spun, what is the probability that the ball lands in a red segment? b. In the roulette wheel shown, black and red segments alternate. Suppose instead that all red segments were grouped together and that all black segments were together. Does this increase the probability that the ball will land in a red segment? Explain. c. Suppose that you watch 1000 spins of a roulette wheel and note the color that results from each spin. What would be an indication that the wheel was not balanced?

An article in the New York Times reported that people who suffer cardiac arrest in New York City have only a 1 in 100 chance of survival. Using probability notation, an equivalent statement would be \(P(\) survival \()=0.01\) for people who suffer cardiac arrest in New York City. (The article attributed this poor survival rate to factors common in large cities: traffic congestion and difficulty finding victims in large buildings. Similar studies in smaller cities showed higher survival rates.) a. Give a relative frequency interpretation of the given probability. b. The basis for the New York Times article was a research study of 2,329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2,329 cardiac arrest sufferers do you think survived? Explain.

A large retail store sells MP3 players. A customer who purchases an MP3 player can pay either by cash or credit card. An extended warranty is also available for purchase. Suppose that the events \(M=\) event that the customer paid by cash \(E=\) event that the customer purchased an extended warranty are independent with \(P(M)=0.47\) and \(P(E)=0.16\). a. Construct a "hypothetical 1000 " table with columns corresponding to cash or credit card and rows corresponding to whether or not an extended warranty is purchased. (Hint: See Example 5.9) b. Use the table to find \(P(M \cup E)\). Give a long-run relative frequency interpretation of this probability.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.