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Chapter 6: Problem 59

A certain university has 10 vehicles available for use by faculty and staff. Six of these are vans and four are cars. On a particular day, only two requests for vehicles have been made. Suppose that the two vehicles to be assigned are chosen in a completely random fashion from among the 10 . a. Let \(E\) denote the event that the first vehicle assigned is a van. What is \(P(E) ?\) b. Let \(F\) denote the event that the second vehicle assigned is a van. What is \(P(F \mid E)\) ? c. Use the results of Parts (a) and (b) to calculate \(P(E\) and \(F)\) (Hint: Use the definition of \(P(F \mid E)\).)

Short Answer

Expert verified
The probability that the first vehicle assigned is a van is 0.6. The conditional probability that the second vehicle assigned is a van, given that a van was assigned first, is \(\frac{5}{9}\). The probability of both events occurring is \(0.6 \times \frac{5}{9} = 0.33\) or approximately 33%.

Step by step solution

01

Calculating the probability of event E

Event E is the event that the first vehicle assigned is a van. There are 10 vehicles in total of which 6 are vans. So, the probability of this event is the number of vans over the total number of vehicles. This translates into \(P(E)=\frac{6}{10}\) or reduced to \(P(E)=0.6\)
02

Calculating the conditional probability of event F given E

Event F is the event that the second vehicle assigned is a van, given that a van was assigned first. Since a van was assigned first, there are now 9 vehicles left of which 5 are vans. So the conditional probability is the number of vans left over the total number of vehicles left. This translates into \(P(F|E)=\frac{5}{9}\)
03

Calculating the probability of both E and F occurring

The final step is to calculate the probability of both E and F occurring. According to the definition of conditional probability, the probability of both events occurring is the probability of E times the conditional probability of F given E. This translates into \(P(E \text{ and } F)=P(E) \times P(F|E) = 0.6 \times \frac{5}{9}\)

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Key Concepts

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

Conditional Probability
Understanding conditional probability is critical in the world of statistics. It deals with determining the likelihood of an event given that another event has occurred. In our exercise, we looked at the probability of assigning a van as the second vehicle, given that the first vehicle assigned was also a van.

Conditional probability is notated as \(P(F\mid E)\), where \(E\) represents the first event and \(F\) represents the second, dependent event. The formula for conditional probability is \(\frac{P(E \text{ and } F)}{P(E)}\), but since we're looking for \(P(F\mid E)\), we rearrange this equation to solve for \(P(E \text{ and } F)\), as shown in the original solution.

In everyday scenarios, conditional probability can be applied to various situations, such as predicting weather patterns or making business decisions based on current market trends. It's an invaluable concept that teaches us that the occurrence of events can be interconnected and dependent on one another.
Probability Calculation
Probability calculation is a fundamental process in statistics, involving quantifying the likelihood of different outcomes. To calculate the probability of an event, we divide the number of favorable outcomes by the total number of possible outcomes. In our example, \(P(E)\) represents the probability of selecting a van first, and we calculated it by dividing the total vans available by the total number of vehicles.

It's essential to remember that probabilities range from 0 to 1, where 0 indicates impossibility and 1 indicates certainty. To further simplify probability calculations, fractions should be reduced to their simplest form, and whenever possible, percentages or decimals are used for a clearer understanding.

In statistics education, mastering the art of probability calculation is crucial. It allows students to effectively analyze risk and uncertainty, predict outcomes, and make informed decisions in both academic pursuits and real-life situations.
Statistics Education
Statistics education is an empowering field that equips learners with the skills to collect, analyze, and interpret data. Engaging with exercises similar to the vehicle assignment problem enhances students' understanding of statistical concepts, such as conditional probability and probability calculation.

Statistics education encompasses a wide range of topics from basic probability to more advanced concepts like hypothesis testing and regression analysis. Emphasizing clear step-by-step solutions helps in building a strong foundational knowledge for students.

To ensure students fully grasp these concepts, educators often employ real-world examples, interactive activities, and comprehensive explanations. Empowering students with a strong command of statistics enables them to become critical thinkers and data-savvy professionals, capable of navigating through the complex landscape of today's data-driven world.

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Most popular questions from this chapter

The report "Twitter in Higher Education: Usage Habits and Trends of Today's College Faculty" (Magna Publications, September 2009) describes results of a survey of nearly 2000 college faculty. The report indicates the following: \- \(30.7 \%\) reported that they use Twitter and \(69.3 \%\) said that they did not use Twitter. \- Of those who use Twitter, \(39.9 \%\) said they sometimes use Twitter to communicate with students. \- Of those who use Twitter, \(27.5 \%\) said that they sometimes use Twitter as a learning tool in the classroom. Consider the chance experiment that selects one of the study participants at random and define the following events: \(T=\) event that selected faculty member uses Twitter \(C=\) event that selected faculty member sometimes uses Twitter to communicate with students \(L=\) event that selected faculty member sometimes uses Twitter as a learning tool in the classroom a. Use the given information to determine the following probabilities: i. \(\quad P(T)\) ii. \(P\left(T^{C}\right)\) iii. \(P(C \mid T)\) iv. \(P(L \mid T)\) v. \(P(C \cap T)\) b. Interpret each of the probabilities computed in Part (a). c. What proportion of the faculty surveyed sometimes use Twitter to communicate with students? [Hint: Use the law of total probability to find \(P(C) .]\) d. What proportion of faculty surveyed sometimes use Twitter as a learning tool in the classroom?

The student council for a school of science and math has one representative from each of the 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 (by placing five slips of paper in a bowl, mixing, and drawing out two of them). a. What are the 10 possible outcomes (simple events)? b. From the description of the selection process, all outcomes are equally likely. What is the probability of each simple event? 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 article "Birth Beats Long Odds for Leap Year Mom, Baby" (San Luis Obispo Tribune, March 2 , 1996) reported that a leap year baby (someone born on February 29) became a leap year mom when she gave birth to a baby on February 29,1996 . The article stated that a hospital spokesperson said that only about 1 in \(2.1\) million births is a leap year baby born to a leap year mom (a probability of approximately .00000047). a. In computing the given probability, the hospital spokesperson used the fact that a leap day occurs only once in 1461 days. Write a few sentences explaining how the hospital spokesperson computed the stated probability. b. To compute the stated probability, the hospital spokesperson had to assume that the birth was equally likely to occur on any of the 1461 days in a four- year period. Do you think that this is a reasonable assumption? Explain. c. Based on your answer to Part (b), do you think that the probability given by the hospital spokesperson is too small, about right, or too large? Explain.

A mutual fund company offers its customers several different funds: a money market fund, three different bond funds, two stock funds, and a balanced fund. Among customers who own shares in just one fund, the percentages of customers in the different funds are as follows: Money market \(20 \%\) Short-term bond \(15 \%\) \(\begin{aligned}&\text { Intermediate-term bond } & 10 \%\end{aligned}\) Long- term bond \(5 \%\) High-risk stock \(18 \%\) \(\begin{array}{ll}\text { Moderate-risk stock } & 25 \%\end{array}\) \(\begin{array}{ll}\text { Balanced fund } & 7 \%\end{array}\) A customer who owns shares in just one fund is to be selected at random. a. What is the probability that the selected individual owns shares in the balanced fund? b. What is the probability that the individual owns shares in a bond fund? c. What is the probability that the selected individual does not own shares in a stock fund?

Suppose that a six-sided die is "loaded" so that any particular even-numbered face is twice as likely to land face up as any particular odd-numbered face. Consider the chance experiment that consists of rolling this die. 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 condition on the sum of these probabilities to determine \(p\).) b. What is the probability that the number showing is an odd number? at most three? 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.
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