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Chapter 6: Problem 19

The article "Anxiety Increases for Airline Passengers After Plane Crash" (San Luis Obispo Tribune. November 13,2001 ) reported that air passengers have a 1 in 11 million chance of dying in an airplane crash. This probability was then interpreted as "You could fly every day for 26,000 years before your number was up." Comment on why this probability interpretation is misleading.

Short Answer

Expert verified
The probability interpretation is misleading because it suggests that the risk of dying in an airplane crash accumulates over time, whereas in reality, the risk for each flight is constant and independent, remaining at 1 in 11 million each time no matter how often you fly.

Step by step solution

01

Understanding The Problem

First, understand that probability, in this case, refers to the likelihood of a passenger dying in an airplane crash. The probability given by the article is 1 in 11 million. This does not change no matter how many times somebody has travelled by airplane. Probability pertains to the relative frequency of an event's occurrence in a long series of repetitive experiences.
02

Identifying The Misinterpretation

The misinterpretation 'You could fly every day for 26,000 years before your number was up' suggests a cumulative property of probability. This suggestion implies that the risk accumulates over time, creating an overall increasing risk. In reality, the risk remains the same (1 in 11 million) each time somebody boards a plane.
03

Evaluating The Statement

The statement provided by the article in error can be corrected as 'Even if you flew every day for 26,000 years, your chance of being in a fatal crash on any given flight would still be 1 in 11 million.' The probability stays constant regardless of how often an individual flies.

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Key Concepts

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

Probability Theory
Probability theory examines the likelihood of events happening. It uses numbers to represent the chances of something occurring. In the case of airplane travel, the probability of dying in a crash is given as 1 in 11 million. This means that for each flight, the odds are extremely low that it will result in a fatal crash.

One vital idea in probability theory is that each event in a sequence of events (such as flights) is often independent. This means that the outcome of one flight does not affect the outcomes of future flights.
  • The probability of an event, such as a plane crash for a given flight, remains constant.
  • Independence of events is crucial to understanding why the probability does not change over repeated trials.
  • Probability is not cumulative; each event has its own chance of occurring.
For air travel, each flight should be considered a unique event with the same probability, regardless of past occurrences.
Risk Assessment
Risk assessment uses probabilities to determine the potential risks of certain actions or events, such as flying. It helps individuals and organizations identify and manage potential threats. By understanding the risks involved, decisions can be made more rationally and confidently.

Misinterpretations of probabilities, like thinking that flying every day increases the likelihood of a fatal crash, can lead to unnecessary anxiety and fear. Risk assessment needs to be based on accurate probabilistic understanding:
  • The risk of a fatal crash per flight is always the same, whether it is your first or thousandth flight.
  • Understanding the fixed risk helps reduce overestimating dangers in regular air travel.
  • Proper risk assessment involves recognizing that repeated actions don't automatically increase the chances of rare events.
Clarity in how risks are communicated and understood is crucial to addressing any misinterpretations.
Statistics Education
Statistics education plays a central role in helping people correctly understand probabilities and avoid common pitfalls. One of the major challenges in teaching statistics is helping learners understand the difference between independent and cumulative events.

With proper statistics education, individuals can better comprehend real-world applications such as air travel safety.
  • Understanding independent events is essential to avoid misinterpretations like "You could fly every day for 26,000 years..."
  • Education should focus on the stable nature of probabilities in repeated trials of independent events.
  • Teaching methods need to clarify misconceptions about probability and help individuals apply statistical knowledge accurately in everyday life.
When people are well-versed in statistics, they can interpret information about risks more accurately and make informed decisions based on that understanding.

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Most popular questions from this chapter

Jeanie is a bit forgetful, and if she doesn't make a "to do" list, the probability that she forgets something she is supposed to do is .1. Tomorrow she intends to run three errands, and she fails to write them on her list. a. What is the probability that Jeanie forgets all three errands? What assumptions did you make to calculate this probability? b. What is the probability that Jeanie remembers at least one of the three errands? c. What is the probability that Jeanie remembers the first errand but not the second or third?

A certain company sends \(40 \%\) of its overnight mail parcels by means of express mail service \(A_{1}\). Of these parcels, \(2 \%\) arrive after the guaranteed delivery time (use \(L\) to denote the event late delivery). If a record of an overnight mailing is randomly selected from the company's files, what is the probability that the parcel went by means of \(A_{1}\) and was late?

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 \(P\left(E_{1}\right)=.4\) and \(P\left(E_{2}\right)=.3\) 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 \(\left(\right.\) not \(\left.E_{1}\right)\) and \(\left(\right.\) not \(\left.E_{2}\right)\). c. What is the probability that the firm is successful in at least one of the two bids?

The National Public Radio show Car Talk has a feature called "The Puzzler." Listeners are asked to send in answers to some puzzling questions-usually about cars but sometimes about probability (which, of course, must account for the incredible popularity of the program!). Suppose that for a car question, 800 answers are submitted, of which 50 are correct. a. Suppose that the hosts randomly select two answers from those submitted with replacement. Calculate the probability that both selected answers are correct. (For purposes of this problem, keep at least five digits to the right of the decimal.) b. Suppose now that the hosts select the answers at random but without replacement. Use conditional probability to evaluate the probability that both answers selected are correct. How does this probability compare to the one computed in Part (a)?

An article in the New york Times (March 2. 1994) 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 \()=.01\) for people who suffer a cardiac arrest in New York City. (The article attributed this poor survival rate to factors common in large cities: traffic congestion and the difficulty of finding victims in large buildings.) a. Give a relative frequency interpretation of the given probability. b. The research that was the basis for the New York Times article was a study of 2329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2329 cardiac arrest sufferers do you think survived? Explain.
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