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Chapter 6: Problem 6

A college library has four copies of a certain book; the copies are numbered \(1,2,3\), and 4 . Two of these are selected at random. The first selected book is placed on 2 -hr reserve, and the second book can be checked out overnight. a. Construct a tree diagram to display the 12 outcomes in the sample space. b. Let \(A\) denote the event that at least one of the books selected is an even-numbered copy. What outcomes are in \(A ?\) c. Suppose that copies 1 and 2 are first printings, whereas copies 3 and 4 are second printings. Let \(B\) denote the event that exactly one of the copies selected is a first printing. What outcomes are contained in \(B\) ?

Short Answer

Expert verified
There are 12 possible outcomes when selecting the books. For event \(A\), the outcomes are: (1,2), (1,4), (2,1), (2,3), (2,4), (3,2), (3,4), (4,1), (4,2), (4,3). For event \(B\) the outcomes are: (1,3), (1,4), (2,3), (2,4), (3,1), (3,2), (4,1), (4,2).

Step by step solution

01

Create Tree Diagram

To understand the different possible outcomes it's first necessary to construct a tree diagram. In this case, there are four books numbered from 1 to 4, and two books are selected in succession. Thus, for the first selection there are 4 options and for each of these, there are 3 different books that can be selected second, giving a total of \(4 \times 3 = 12\) paths or outcomes.
02

Identifying Outcomes in Event \(A\)

Event \(A\) means that at least one out of the two books selected has an even number; i.e. book number 2 or 4. The outcomes that meet these conditions are: (1,2), (1,4), (2,1), (2,3), (2,4), (3,2), (3,4), (4,1), (4,2), (4,3)
03

Identifying Outcomes in Event \(B\)

Event \(B\) is the event that exactly one of the books selected is a first printing. In this case books numbered 1 and 2 are first printings and books numbered 3 and 4 are second printings. The outcomes for this event where exactly one book is first printing are: (1,3), (1,4), (2,3), (2,4), (3,1), (3,2), (4,1), (4,2)

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Key Concepts

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

Sample Space in Probability
In probability, the sample space is defined as the set of all possible outcomes of a random experiment. When considering the selection of books from the college library example, the sample space consists of all the possible pairs of books that could be selected.

To visualize this, picture each book as a potential choice in a sequence. With four books, labeled 1 through 4, the first choice has four possibilities. After a book is selected for the reserve, the second choice will then have three possibilities, since one of the books is no longer available. This leads to the calculation of the total number of outcomes as the product of the possibilities for each step: 4 choices initially multiplied by 3 choices subsequently gives us 12 possible outcomes in the sample space.

The importance of defining the sample space lies in its role as the foundation of determining probabilities. Any event we consider is a subset of this sample space, and the likelihood of any given event is found by comparing the number of favorable outcomes to the total number of outcomes in the sample space.
Event Outcomes in Statistics
Event outcomes in statistics refer to the results that satisfy the conditions of a specific event within an experiment. In the context of the book selection, an event might be defined as selecting at least one even-numbered copy, which was represented as event A.

To identify the outcomes that belong to event A, we look for all pairs that include either the number 2 or number 4. It's important to note that order matters here since the first book is placed on 2-hour reserve and the second is available for overnight checkout. This results in ten pairs where at least one of the books is an even number, indicating a high probability of event A occurring.

Understanding event outcomes allows us to calculate the probability of an event by comparing the number of event outcomes to the total number of outcomes in the sample space. It also helps to clarify the nature of an event 鈥� whether it's dependent on order, the presence of a particular element, or exclusivity of conditions, for example.
First and Second Printings Probability
In our situation, considering the different printings of the books introduces a new event, event B, which concerns the selection of exactly one first printing copy. This event encapsulates a more specific condition: among the two books, one must be a first printing (copy 1 or 2), and the other must be a second printing (copy 3 or 4).

Here, we've identified that there are eight outcomes that fit this event, as indicated in the step-by-step solution. This event highlights a unique probability situation where we are dealing with a combination of criteria that restrict our event outcomes to specific pairings.

By calculating the probability of selecting exactly one first printing, we also indirectly learn about the likelihood of both or neither of the books being a first printing. This is valuable in statistics as it allows for a deeper understanding of how different events relate to each other within the same sample space and how specific conditions can drastically alter probabilities.

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Most popular questions from this chapter

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: \(P(R)=.06, P(F)=.60\), \(P(R \mid D)=.30, P(F \mid D)=.54\).

In a school machine shop, \(60 \%\) of all machine breakdowns occur on lathes and \(15 \%\) occur on drill presses. Let \(E\) denote the event that the next machine breakdown is on a lathe, and let \(F\) denote the event that a drill press is the next machine to break down. With \(P(E)=.60\) and \(P(F)=.15\), calculate: a. \(P\left(E^{C}\right)\) b. \(P(E \cup F)\) c. \(P\left(E^{C} \cap F^{C}\right)\)

Suppose that, starting at a certain time, batteries coming off an assembly line are examined one by one to see whether they are defective (let \(\mathrm{D}=\) defective and \(\mathrm{N}=\) not defective). The chance experiment terminates as soon as a nondefective battery is obtained. a. Give five possible experimental outcomes. b. What can be said about the number of outcomes in the sample space? c. What outcomes are in the event \(E\), that the number of batteries examined is an even number?

Consider the following information about travelers on vacation: \(40 \%\) check work email, \(30 \%\) use a cell phone to stay connected to work, \(25 \%\) bring a laptop with them on vacation, \(23 \%\) both check work email and use a cell phone to stay connected, and \(51 \%\) neither check work email nor use a cell phone to stay connected nor bring a laptop. In addition \(88 \%\) of those who bring a laptop also check work email and \(70 \%\) of those who use a cell phone to stay connected also bring a laptop. With \(E=\) event that a traveler on vacation checks work email, \(C=\) event that a traveler on vacation uses a cell phone to stay connected, and \(L=\) event that a traveler on vacation brought a laptop, use the given information to determine the following probabilities. A Venn diagram may help. a. \(P(E)\) b. \(P(C)\) c. \(P(L)\) d. \(P(E\) and \(C)\) e. \(P\left(E^{C}\right.\) and \(C^{C}\) and \(L^{C}\) ) f. \(P(\) Eor C or \(L\) ) g. \(P(E \mid L)\) j. \(P(E\) and \(L)\) h. \(P(L \mid C)\) k. \(P(C\) and \(L)\) i. \(P(E\) and \(C\) and \(L)\) 1\. \(P(C \mid E\) and \(L)\)

The following case study was reported in the article "Parking Tickets and Missing Women," which appeared in an early edition of the book Statistics: A Guide to the \(U n\) known. In a Swedish trial on a charge of overtime parking, a police officer testified that he had noted the position of the two air valves on the tires of a parked car: To the closest hour, one was at the one o'clock position and the other was at the six o'clock position. After the allowable time for parking in that zone had passed, the policeman returned, noted that the valves were in the same position, and ticketed the car. The owner of the car claimed that he had left the parking place in time and had returned later. The valves just happened by chance to be in the same positions. An "expert" witness computed the probability of this occurring as \((1 / 12)(1 / 12)=1 / 44\). a. What reasoning did the expert use to arrive at the probability of \(1 / 44\) ? b. Can you spot the error in the reasoning that leads to the stated probability of \(1 / 44\) ? What effect does this error have on the probability of occurrence? Do you think that \(1 / 44\) is larger or smaller than the correct probability of occurrence?
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