Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Consider the following events: \(C=\) event that a randomly selected driver is observed to be using a cell phone \(A=\) event that a randomly selected driver is observed driving a car \(V=\) event that a randomly selected driver is observed driving a van or SUV \(T=\) event that a randomly selected driver is observed driving a pickup truck Based on the article "Three Percent of Drivers on Hand-Held Cell Phones at Any Given Time" (San Luis Obispo Tribune, July 24,2001 ), the following probability estimates are reasonable: \(P(C)=0.03, P(C \mid A)=0.026, P(C \mid V)=0.048\) and \(P(C \mid T)=0.019 .\) Explain why \(P(C)\) is not just the average of the three given conditional probabilities.

Short Answer

Expert verified
The reason why \(P(C)\) is not just an average of the three given conditional probabilities is because each type of vehicle does not necessarily have the same number of drivers. The overall probability is influenced by how common each type of vehicle is.

Step by step solution

01

Understanding the Events and Probabilities

To start with, the four events represented are: \(C\): a driver is observed to be using a cell phone, \(A\): a driver is observed driving a car, \(V\): a driver is observed driving a van or SUV, \(T\): a driver is observed driving a pickup truck. The provided probabilities are: \(P(C) = 0.03\), \(P(C|A) = 0.026\), \(P(C|V) = 0.048\), \(P(C|T)=0.019\). These are conditional probabilities, which represent the probability of event C (using a cell phone) given that another event (A, V or T) has occurred.
02

Understanding Conditional Probabilities

Conditional probability changes the total outcomes we consider from the entirety of the probability space to just those outcomes that satisfy some condition. Hence the probability of a driver using a cell phone, given that they are driving a specific type of vehicle, may differ from the overall probability of a driver using a cell phone. The conditional probabilities \(P(C|A)\), \(P(C|V)\), and \(P(C|T)\) are based on the subset of drivers driving a car, a van/SUV, and a pickup truck, respectively.
03

Why \(P(C)\) Is Not the Average

The average of the three given conditional probabilities would give an equal weight to each type of vehicle (car, van/SUV, and pickup truck). However, this doesn't represent the real situation correctly, because these types of vehicles are not equally common on the roads. It would be an incorrect assumption to think that each type of vehicle has the same number of drivers. Thus, without considering the individual probabilities of drivers driving a car, van/SUV, or a pickup truck, \(P(A)\), \(P(V)\), or \(P(T)\), the average of these conditional probabilities won't give us \(P(C)\).

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 Vaia!

Key Concepts

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

Probability Space
The concept of a probability space is fundamental to understanding any probability problem. It includes three main elements: a sample space, a set of events, and a probability function.

A sample space is the set of all possible outcomes of a random experiment. For example, in the case of observing different drivers, the sample space would include all drivers using all kinds of vehicles. Events are subsets of this sample space - like those picking out drivers based on the type of vehicle they're using.

The probability function assigns a value between 0 and 1 to each event from the sample space, representing the likelihood of the event occurring. It adheres to rules, such as the sum of probabilities for all mutually exclusive outcomes equaling one. Understanding this framework helps us to analyze complex problems such as the one with drivers using cell phones while driving different types of vehicles.
Event Independence
Understanding event independence is crucial when studying probability. Two events are independent if the occurrence of one event does not affect the probability of the other event occurring. In statistical terms, events A and B are independent if and only if
\[P(A \cap B)= P(A) \times P(B)\].

In the driving example, if being a driver of a car is independent from using a cell phone, the probability of observing someone driving a car and using a cell phone would be the product of the two separate probabilities. However, this condition is not always met in real-life scenarios, which is a critical concept to keep in mind while calculating probabilities and can affect our interpretations of the given data.
Statistical Reasoning
Statistical reasoning enables us to make sense of data by using the principles of statistics and probability theory. It involves interpreting data represented by probabilities, such as the likelihood of a randomly selected driver being observed using a cell phone.

In exercising statistical reasoning with conditional probabilities, one doesn't simply take the mean of probabilities on a whim. Instead, one considers the context—the likelihood of each type of vehicle being used—as it directly impacts the final probability calculation. Recognizing the importance of contextual factors and how they influence the data is a critical aspect of statistical reasoning, which emphasizes the need for careful interpretation of probabilities.
Probability Theory
Probability theory is the mathematical framework that deals with quantifying uncertainties. It provides the methods and principles for analyzing random events, such as drivers using different types of vehicles.

Applying probability theory to the given problem, it's important to note that, beyond knowing the conditional probabilities, understanding how they contribute to the overall probability entails combining them appropriately. This requires knowledge of the law of total probability, which in this case, would necessitate knowing not just the probability of a driver using a cell phone given the type of vehicle, but also the probability of each type of vehicle being on the road. Probability theory provides the tools to handle such complexities, guiding us through the process of making informed conclusions from uncertain situations.

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 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.

Automobiles that are more than 10 years old must pass a vehicle inspection to be registered in a particular state. The state reports the probability that a car more than 10 years old will fail the vehicle inspection is \(0.09 .\) Give a relative frequency interpretation of this probability.

a. Suppose events \(E\) and \(F\) are mutually exclusive with \(P(E)=0.41\) and \(P(E)=0.23\). i. What is the value of \(P(E \cap F) ?\) ii. What is the value of \(P(E \cup F) ?\) b. Suppose that for events \(A\) and \(B, P(A)=0.26, P(B)=0.34\), and \(P(A \cup B)=0.47 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell?

Phoenix is a hub for a large airline. Suppose that on a particular day, 8,000 passengers arrived in Phoenix on this airline. Phoenix was the final destination for 1,800 of these passengers. The others were all connecting to flights to other cities. On this particular day, several inbound flights were late, and 480 passengers missed their connecting flight. Of these 480 passengers, 75 were delayed overnight and had to spend the night in Phoenix. Consider the chance experiment of choosing a passenger at random from these 8,000 passengers. Calculate the following probabilities: a. the probability that the selected passenger had Phoenix as a final destination. b. the probability that the selected passenger did not have Phoenix as a final destination. c. the probability that the selected passenger was connecting and missed the connecting flight. d. the probability that the selected passenger was a connecting passenger and did not miss the connecting flight. e. the probability that the selected passenger either had Phoenix as a final destination or was delayed overnight in Phoenix. f. An independent customer satisfaction survey is planned. Fifty passengers selected at random from the 8,000 passengers who arrived in Phoenix on the day described above will be contacted for the survey. The airline knows that the survey results will not be favorable if too many people who were delayed overnight are included. Write a few sentences explaining whether or not you think the airline should be worried, using relevant probabilities to support your answer.

A single-elimination tournament with four players is to be held. A total of three games will be played. In Game 1 , the players seeded (rated) first and fourth play. In Game 2 , the players seeded second and third play. In Game \(3,\) the winners of Games 1 and 2 play, with the winner of Game 3 declared the tournament winner. Suppose that the following probabilities are known: \(P(\) Seed 1 defeats Seed 4\()=0.8\) \(P(\) Seed 1 defeats \(\operatorname{Seed} 2)=0.6\) \(P(\) Seed 1 defeats \(\operatorname{Seed} 3)=0.7\) \(P(\) Seed 2 defeats \(\operatorname{Seed} 3)=0.6\) \(P(\) Seed 2 defeats Seed 4\()=0.7\) \(P(\) Seed 3 defeats Seed 4) \(=0.6\) a. How would you use random digits to simulate Game 1 of this tournament? b. How would you use random digits to simulate Game 2 of this tournament? c. How would you use random digits to simulate the third game in the tournament? (This will depend on the outcomes of Games 1 and \(2 .\) ) d. Simulate one complete tournament, giving an explanation for each step in the process. e. Simulate 10 tournaments, and use the resulting information to estimate the probability that the first seed wins the tournament. f. Ask four classmates for their simulation results. Along with your own results, this should give you information on 50 simulated tournaments. Use this information to estimate the probability that the first seed wins the tournament. g. Why do the estimated probabilities from Parts (e) and (f) differ? Which do you think is a better estimate of the actual probability? Explain.

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.

Sign-up for free