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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 professor assigns five problems to be completed as homework. At the next class meeting, two of the five problems will be selected at random and collected for grading. You have only completed the first three problems. a. What is the sample space for the chance experiment of selecting two problems at random? (Hint: You can think of the problems as being labeled \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and \(\mathrm{E}\). One possible selection of two problems is \(\mathrm{A}\) and \(\mathrm{B}\). If these two problems are selected and you did problems \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\), you will be able to turn in both problems. There are nine other possible selections to consider.) b. Are the outcomes in the sample space equally likely? c. What is the probability that you will be able to turn in both of the problems selected? d. Does the probability that you will be able to turn in both problems change if you had completed the last three problems instead of the first three problems? Explain. e. What happens to the probability that you will be able to turn in both problems selected if you had completed four of the problems rather than just three?

5.57 There are two traffic lights on Shelly's route from home to work. Let \(E\) denote the event that Shelly must stop at the first light, and define the event \(F\) in a similar manner for the second light. Suppose that \(P(E)=0.4, P(F)=0.3\) and \(P(E \cap F)=0.15 .\) In Exercise \(5.25,\) you constructed a hypothetical 1000 table to calculate the following probabilities. Now use the probability formulas of this section to find these probabilities. a. The probability that Shelly must stop for at least one light (the probability of the event \(E \cup F\) ). b. The probability that Shelly does not have to stop at either light. c. The probability that Shelly must stop at exactly one of the two lights. d. The probability that Shelly must stop only at the first light.

The student council for a school of science and math has one representative from each of five academic departments: Biology (B), Chemistry (C), Mathematics (M), Physics (P), and Statistics (S). Two of these students are to be randomly selected for inclusion on a university-wide student committee. a. What are the 10 possible outcomes? b. From the description of the selection process, all outcomes are equally likely. What is the probability of each outcome? c. What is the probability that one of the committee members is the statistics department representative? d. What is the probability that both committee members come from laboratory science departments?

The probability of getting a king when a card is selected at random from a standard deck of 52 playing cards is \(\frac{1}{13}\). a. Give a relative frequency interpretation of this probability. b. Express the probability as a decimal rounded to three decimal places. Then complete the following statement: If a card is selected at random, I would expect to see a king about times in 1000 .

A large cable company reports the following: \(80 \%\) of its customers subscribe to cable TV service \(42 \%\) of its customers subscribe to Internet service \(32 \%\) of its customers subscribe to telephone service \(25 \%\) of its customers subscribe to both cable TV and Internet service \(21 \%\) of its customers subscribe to both cable TV and phone service \(23 \%\) of its customers subscribe to both Internet and phone service \(15 \%\) of its customers subscribe to all three services Consider the chance experiment that consists of selecting one of the cable company customers at random. In Exercise \(5.53,\) you constructed a hypothetical 1000 table to calculate the following probabilities. Now use the probability formulas of this section to find these probabilities. a. \(P(\) cable TV only) b. \(P\) (Internet \(\mid\) cable TV) c. \(P(\) exactly two services \()\) d. \(P\) (Internet and cable TV only)
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