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Chapter 5: Problem 80

The article "A Crash Course in Probability" from The Economist included the following information: The chance of being involved in an airplane crash when flying on an Airbus 330 from London to New York City on Virgin Atlantic Airlines is 1 in \(5,371,369 .\) This was interpreted as meaning that you "would expect to go down if you took this flight every day for 14,716 years." The article also states that a person could "expect to fly on the route for 14,716 years before plummeting into the Atlantic." Comment on why these statements are misleading.

Short Answer

Expert verified
The statements in the article are misleading because they incorrectly assume that probabilities can be linearly summed over time and imply that a crash becomes inevitable after a certain number of flights. However, the risk of each flight is independent and the overall probability of experiencing a crash at some point during those 14,716 years is not a simple linear calculation. Furthermore, it is not accurate to suggest that crashes become a certainty after a specific number of flights.

Step by step solution

01

Understanding the Probability

The given probability of being involved in an airplane crash when flying on an Airbus 330 from London to New York City on Virgin Atlantic Airlines is 1 in 5,371,369. This means that if there were 5,371,369 flights, we should expect 1 of these flights to be involved in a crash. It is important to understand that this probability is for a single flight and not for multiple flights taken over time.
02

Misleading Statements

The article implies that by taking this flight every day for 14,716 years, a person would "expect to go down" or have a crash. To derive this number, the article assumes that it is safe to fly every day for 14,716 years. The basis of this interpretation is the misconception that individual flight probabilities can be summed linearly over time, which is incorrect. In reality, the risk of each flight is independent, meaning that the overall probability of experiencing a crash at some point during those 14,716 years is not a simple linear calculation. Furthermore, the article suggests that a person could "expect to fly on the route for 14,716 years before plummeting into the Atlantic." This statement is also misleading because it implies that flying for that period guarantees a crash, which is not the case. Once again, each flight is an independent event. While the probability of experiencing a crash over thousands of flights taken would indeed be higher than the probability of a crash on a single flight, it is not accurate to suggest that crashes become a certainty after a certain number of flights. In conclusion, both statements are misleading because they make incorrect assumptions about the nature of probabilities, suggesting the inevitability of experiencing a crash after a certain number of flights. In reality, each flight's risk is independent, and the overall probability of experiencing a crash at some point in time is more complex than simple linear assumptions.

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Key Concepts

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

Risk Assessment
When considering risk, particularly in activities as serious as flying, it's crucial to accurately interpret probabilities. The probability of being involved in a crash on any specific flight is a measure of risk. However, it's a common mistake to misinterpret this number as a guarantee over time.

In the original exercise, the probability was given as 1 in 5,371,369. This doesn't mean that if you fly 5,371,369 times, you're destined to experience an accident. Instead, it highlights the very low likelihood of a crash on each individual flight.

Effective risk assessment involves understanding what these probabilities convey about safety. It means considering each flight's risk independently rather than assuming aggregated probabilities over multiple journeys determine an eventual outcome. This ensures we clearly comprehend how safe or risky an activity truly is without falling into the trap of misleading interpretations.
Independent Events
Each flight operates as an independent event in statistical terms. This concept is essential to grasp why the statements from the article are misleading. In probability theory, independent events mean that the outcome of one flight does not affect the outcomes of following flights.

Given probability as 1 in 5,371,369 on this specific route, every flight has this same risk level, unaffected by previous travel experiences. Even if you traveled this route every day, every single journey carries its distinct risk, which speaks to independence.

This independence is why the notion of flying for 14,716 years leading to a crash is a misconception. Probabilities don't accumulate in a straightforward manner across independent events. Instead, each flight, or independent event, begins with a clean slate in terms of risk.
Probability Calculation
Calculating probability effectively means doing more than simply adding. Each calculation must respect the nature of the events involved, particularly their independence. The chance provided in the exercise, 1 in 5,371,369, must be understood as a probability for a single occurrence, not a cumulative risk over repeated events.

In probability calculations, particularly for independent events, multiplicative rules often apply rather than additive ones. For instance, if two flights are taken, the probability of both having accidents would involve multiplying individual probabilities, not adding them.

Understanding probability calculations ensures an accurate interpretation of risk. It prevents erroneous conclusions that suggest certainty of adverse outcomes after numerous repetitions of a low-probability event. Thus, the probability doesn't guarantee a crash even after numerous flights; it simply remains constant for each independent flight.

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Most popular questions from this chapter

A large cable company reports that \(42 \%\) of its customers subscribe to its Internet service, \(32 \%\) subscribe to its phone service, and \(51 \%\) subscribe to its Internet service or its phone service (or both). a. Use the given probability information to set up a hypothetical 1000 table. b. Use the table to find the following: i. the probability that a randomly selected customer subscribes to both the Internet service and the phone service. ii. the probability that a randomly selected customer subscribes to exactly one of the two services.

An online store offers two methods of shipping-regular ground service and an expedited 2 -day shipping. Customers may also choose whether or not to have the purchase gift wrapped. Suppose that the events \(E=\) event that the customer chooses expedited shipping \(G=\) event that the customer chooses gift wrap are independent with \(P(E)=0.26\) and \(P(G)=0.12\). a. Construct a hypothetical 1000 table with columns corresponding to whether or not expedited shipping was chosen and rows corresponding to whether or not gift wrap was selected. b. Use the information in the table to calculate \(P(E \cup G)\). Give a long- run relative frequency interpretation of this probability.

Three of the most common types of pets are cats, dogs, and fish. Many families have more than one type of pet. Suppose that a family is selected at random and consider the following events and probabilities: \(F=\) event that the selected family has at least one fish \(D=\) event that the selected family has at least one dog \(C=\) event that the selected family has at least one cat $$ \begin{array}{rrr} P(F)=0.20 & P(D)=0.32 & P(C)=0.35 \\ P(F \cap D)=0.18 & P(F \cap C)=0.07 & P(D \mid C)=0.30 \end{array} $$ Calculate the following probabilities. a. \(P(F \mid D)\) b. \(P(F \cup D)\) c. \(P(C \cap D)\)

The article "Obesity, Smoking Damage U.S. Economy," which appeared in the Gallup online Business Journal reported that based on a large representative sample of adult Americans, \(52.7 \%\) claimed that they exercised at least 30 minutes on three or more days per week during \(2015 .\) It also reported that the percentage for millennials (people age \(19-35\) ) was \(57.1 \%,\) and for those over 35 it was \(51.1 \% .\) If an adult American were to be selected at random, are the events selected adult exercises at least 30 minutes three times per week and selected adult is a millennial independent or dependent events? Justify your answer using the given information.

A construction firm bids on two different contracts. Let \(E_{1}\) be the event that the bid on the first contract is successful, and define \(E_{2}\) analogously for the second contract. Suppose that \(P\left(E_{1}\right)=0.4\) and \(P\left(E_{2}\right)=0.3\) and that \(E_{1}\) and \(E_{2}\) are independent events. a. Calculate the probability that both bids are successful (the probability of the event \(E_{1}\) and \(E_{2}\) ). b. Calculate the probability that neither bid is successful (the probability of the event \(\operatorname{not} E_{1}\) and not \(E_{2}\) ). c. What is the probability that the firm is successful in at least one of the two bids?
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