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Chapter 6: Problem 42

The article "SUVs Score Low in New Federal Rollover Ratings" (San Luis Obispo Tribune, January 6,2001 ) gave information on death rates for various kinds of accidents by vehicle type for accidents reported to the police. Suppose that we randomly select an accident reported to the police and consider the following events: \(R=\) event that the selected accident is a single-vehicle rollover, \(F=\) event that the selected accident is a frontal collision, and \(D=\) event that the selected accident results in a death. Information in the article indicates that the following probability estimates are reasonable: \(P(R)=.06, P(F)=.60\), \(P(R \mid D)=.30, P(F \mid D)=.54\).

Short Answer

Expert verified
The probability of an accident reported to the police resulting in a death is approximately 0.1190.

Step by step solution

01

Understanding and interpreting the given information

The first task is to understand the symbols and the given probabilities. Event R is a single-vehicle rollover, F is a frontal collision, and D is an accident results in a death. The given probabilities are: \(P(R)=.06\), \(P(F)=.60\), \(P(R \mid D)=.30\), and \(P(F \mid D)=.54\).
02

Solve for the probability of an accident resulting in death

We know that the total probability of an accident resulting in death is the sum of the probabilities of a death occurring in a rollover and a frontal collision. Using the formula for conditional probability \(P(A \mid B) = P(A \cap B) / P(B)\), we can express \(P(D) = P(R \cap D) + P(F \cap D)\). We can also express \(P(R \cap D) = P(R \mid D) * P(D)\) and \(P(F \cap D) = P(F \mid D) * P(D)\), and substituting these into our formula for \(P(D)\) leaves us with \(P(D) = P(R \mid D) * P(D) + P(F \mid D) * P(D)\). Therefore, to isolate \(P(D)\) on one side, we can factor it out and obtain \(P(D) = 1 / (P(R \mid D) + P(F \mid D))\).
03

Calculate the probability

Substitute the given values for \(P(R \mid D)\) and \(P(F \mid D)\), into the equation obtained in step 2 to calculate \(P(D)\). Hence the probability of an accident resulting in a death \(P(D) = 1 / (.30 + .54) = 1 / .84 = 0.1190\).

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Key Concepts

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

Probability Theory
Probability theory is a branch of mathematics concerned with analyzing random phenomena and modeling the random events through mathematical expressions. To understand concepts in probability theory, one must be familiar with terms such as events, outcomes, and probabilities. Events are occurrences that can have different outcomes, and each outcome has a probability associated with it, representing the likelihood of that outcome occurring.

For example, when we talk about the event of a traffic accident, outcomes can be a rollover, frontal collision, or other types of accidents. The probabilities given in our exercise, such as \(P(R) = .06\) and \(P(F) = .60\), represent the likelihood of each event occurring. Moreover, the exercise incorporates the concept of conditional probability, denoted as \(P(A \mid B)\), which defines the probability of event \(A\) occurring given that event \(B\) has already occurred. This idea is crucial in various fields including risk assessment and statistical analysis of events.
Statistical Analysis
Statistical analysis involves collecting, presenting, and interpreting data to make informed conclusions. In the context of the given problem, we're using statistical analysis to interpret the risk of death in different types of accidents reported to the police. The calculation of conditional probabilities is an example of the application of statistical analysis.

By analyzing these probabilities, statisticians can help public authorities make data-driven decisions regarding traffic safety, regulations, and vehicle design improvements. The computation done in our exercise also reveals how to isolate certain variables, such as the probability of death \(P(D)\), when only knowing the probabilities of specific types of accidents occurring given that a death has already happened. This kind of isolation is a fundamental procedure in statistical analysis, enabling professionals to understand and predict outcomes based on the various influencing factors.
Probability Estimation
Probability estimation involves determining the likelihood of a specific event based on available data or statistics. In our example, we are given estimates such as \(P(R \mid D) = .30\) and \(P(F \mid D) = .54\). These estimates likely came from the analysis of previous accident data. To make these estimates useful, one must understand how to correctly apply them, as shown in the exercise's step by step solution.

The final step requires plugging in our conditional probability estimates into a formula to solve for \(P(D)\), the probability of death. This is a prime instance of using probability estimation in practice; combining prior knowledge of probabilities with a mathematical formula to deduce an unknown probability. Such estimates help in creating models that predict the chance of future occurrences and are of immense value in fields such as insurance, healthcare, or any area that relies on risk assessment.

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Most popular questions from this chapter

The newspaper article "Folic Acid Might Reduce Risk of Down Syndrome" (USA Today, September 29 , 1999) makes the following statement: "Older women are at a greater risk of giving birth to a baby with Down Syndrome than are younger women. But younger women are more fertile, so most children with Down Syndrome are born to mothers under \(30 .\) " Let \(D=\) event that a randomly selected baby is born with Down Syndrome and \(Y=\) event that a randomly selected baby is born to a young mother (under age 30 ). For each of the following probability statements, indicate whether the statement is consistent with the quote from the article, and if not, explain why not. a. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.7\) b. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.001, \quad P(Y)=.7\) c. \(P(D \mid Y)=.004, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.7\) d. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.4\) e. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.001, \quad P(Y)=.4\) f. \(P(D \mid Y)=.004, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.4\)

Suppose that a six-sided die is "loaded" so that any particular even-numbered face is twice as likely to be observed as any particular odd-numbered face. a. What are the probabilities of the six simple events? (Hint: Denote these events by \(O_{1}, \ldots, O_{6}\). Then \(P\left(O_{1}\right)=p\), \(P\left(O_{2}\right)=2 p, P\left(O_{3}\right)=p, \ldots, P\left(O_{6}\right)=2 p .\) Now use a condi- tion on the sum of these probabilities to determine \(p .\) ) b. What is the probability that the number showing is an odd number? at most \(3 ?\) c. Now suppose that the die is loaded so that the probability of any particular simple event is proportional to the number showing on the corresponding upturned face; that is, \(P\left(O_{1}\right)=c, P\left(O_{2}\right)=2 c, \ldots, P\left(O_{6}\right)=6 c .\) What are the probabilities of the six simple events? Calculate the probabilities of Part (b) for this die.

Suppose that an individual is randomly selected from the population of all adult males living in the United States. Let \(A\) be the event that the selected individual is oyer 6 ft in height, and let \(B\) be the event that the selected individual is a professional basketball player. Which do you think is larger, \(P(A \mid B)\) or \(P(B \mid A) ?\) Why?

Suppose that we define 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 passenger automobile, \(V=\) event that a randomly selected driver is observed driving a van or SUV, and \(T=\) 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: \(P(C)=.03\), \(P(C \mid A)=.026, P(C \mid V)=.048\), and \(P(C \mid T)=.019 .\) Ex- plain why \(P(C)\) is not just the average of the three given conditional probabilities.

A shipment of 5000 printed circuit boards contains 40 that are defective. Two boards will be chosen at random, without replacement. Consider the two events \(E_{1}=\) event that the first board selected is defective and \(E_{2}=\) event that the second board selected is defective. a. Are \(E_{1}\) and \(E_{2}\) dependent events? Explain in words. b. Let not \(E_{1}\) be the event that the first board selected is not defective (the event \(E_{1}^{C}\) ). What is \(P\left(\right.\) not \(\left.E_{1}\right)\) ? c. How do the two probabilities \(P\left(E_{2} \mid E_{1}\right)\) and \(P\left(E_{2} \mid\right.\) not \(\left.E_{1}\right)\) compare? d. Based on your answer to Part (c), would it be reasonable to view \(E_{1}\) and \(E_{2}\) as approximately independent?
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