/*! 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 4 For the 10-year period between 2... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For the 10-year period between 2000 and \(2010,\) the average number of deaths due to accidents involving U.S. commercial airline carriers has been about 46 per year. Over that same period, the average number of passengers has been more than 600 million per year. a. Can you consider this a long run or short run of trials? Explain. b. Estimate the probability of dying on a particular flight. (By contrast, for a trip by auto in a Western country, the probability of death in a 1000 -mile trip is about 1 in \(42,000,\) or more than 50 times the flight's estimated probability.) c. As of April 2011, the last fatal accident involving a U.S. carrier occurred in Buffalo, New York, in February \(2009,\) a span of more than two years. Mary comments that she is currently afraid of flying because the airlines are "due for an accident." Comment on Mary's reasoning.

Short Answer

Expert verified
a. It's a long run. b. Probability is about \(7.67 \times 10^{-8}\). c. Mary's reasoning is the gambler's fallacy.

Step by step solution

01

Determine the Type of Trial

A long run of trials refers to a scenario where enough repetitions have been completed to give probabilities their accurate empirical values. Over a 10-year period, consistent data points such as 46 deaths per year make this a long run of trials. Thus, it is a long run.
02

Calculate Probability of Death per Flight

To estimate this probability, first find the average number of deaths per year (46) and compare it to the number of passengers (600 million per year). Assume each passenger takes one flight per year. The probability of dying on a flight is given by \( \frac{46}{600,000,000} \approx 7.67 \times 10^{-8} \).
03

Comparing Probabilities with Auto Trips

For context, compare the flight's probability of death to the probability of death on a 1000-mile auto trip, which is \( \frac{1}{42,000} \approx 2.38 \times 10^{-5} \). The flight's probability is approximately 50 times smaller, confirming its relative safety.
04

Analyze Mary’s Reasoning

Mary's statement reflects the "gambler's fallacy," the belief that a long run of no accidents means one is 'due.' Since each flight is independent of previous incidents, preceding two years without an accident do not influence the probability of a future crash. Thus, her reasoning is flawed.

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.

Long Run of Trials
When analyzing statistical data, the term 'long run of trials' refers to a significant number of repetitions that allow empirical probabilities to stabilize around their true values.
In this case, the analysis spans a 10-year period, which offers a substantial dataset for observing patterns in airline accident data.

Here's why this is considered a long run:
  • The data comes from a consistent period where conditions, like safety protocols and aircraft usage, are relatively stable.
  • A span of 10 years allows us to average out anomalies and assess long-term trends rather than short-term fluctuations.
  • The number of incidents, like 46 airline deaths annually, provides a solid foundation to evaluate probabilities with greater accuracy.
Therefore, the period in question qualifies as a long run of trials, ensuring computations of probabilities are reliable and meaningful.
Flight Safety Statistics
Understanding the probability of certain events, like accidents, is crucial to assessing risk.
With over 600 million passengers annually and an average of 46 fatalities per year, the probability of dying in a flight accident is extremely low.

Let's break it down:
  • To calculate the probability, divide the average number of deaths by the total number of passengers.
  • This results in a probability of approximately \( 7.67 \times 10^{-8} \) for each flight, indicating the rarity of such events.
  • When compared to auto travel with a probability of \( 2.38 \times 10^{-5} \) per 1000-mile trip, flying appears much safer, being roughly 50 times safer than driving for the equivalent distance.
These statistics reinforce the perception of air travel as one of the safest modes of transport.
Gambler's Fallacy
The gambler's fallacy is a common error in reasoning, where one mistakenly believes that the outcome of a random event is affected by previous outcomes.
In other words, if a particular event hasn't occurred in a while, we might think it's 'due' to happen soon.

Here's how it applies to Mary's case:
  • She thinks that because there's been a hiatus since the last accident, a new one is imminent.
  • Each flight, however, is independent of the last; the probability doesn't increase just because a certain event (like an accident) has not happened recently.
  • This illustrates the gambler's fallacy, where assumptions about 'due' events are misconceived.
To avoid such an error, it's important to understand that past events don't dictate future probabilities, especially in independent scenarios like flights.

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 2007 study by the National Center on Addiction and Substance Abuse at Columbia University reported that for college students, the estimated probability of being a binge drinker was 0.50 for males and 0.34 for females. Using notation, express each of these as a conditional probability.

A standard deck of poker playing cards contains four suits (clubs, diamonds, hearts, and spades) and 13 different cards of each suit. During a hand of poker, 5 of the 52 cards have been exposed. Of the exposed cards, 3 were diamonds. Tony will have the opportunity to draw two more cards, and he has surmised that in order to win the hand, each of those two cards will need to be diamonds. What is Tony's probability of winning the hand? (Assume the two unexposed cards are not diamonds.)

Wisconsin has 5.4 million residents. On any given day, the probability is \(1 / 5000\) that a randomly selected Wisconsin resident decides to visit DisneyWorld in Florida. a. Find the probability that they all will decide to go tomorrow, in which case DisneyWorld has more than 5.4 million people in line when it opens in the morning. b. What assumptions did your solution in part a make? Are they realistic? Explain.

A standard card deck has 52 cards consisting of 26 black and 26 red cards. Three cards are dealt from a shuffled deck, without replacement. a. True or false: The probability of being dealt three black cards is \((1 / 2) \times(1 / 2) \times(1 / 2)=1 / 8 .\) If true, explain why. If false, show how to get the correct probability. b. Let \(A=\) first card red and \(B=\) second card red. Are A and B independent? Explain why or why not. c. Answer parts a and b if each card is replaced in the deck after being dealt.

State an event that has happened to you or to someone you know that seems highly coincidental (such as seeing a friend while on vacation). Explain why that event may not be especially surprising, once you think of all the similar types of events that could have happened to you or someone that you know, over the course of several years.

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.