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Chapter 6: Problem 7

A library has five copies of a certain textbook on reserve of which two copies ( 1 and 2 ) are first printings and the other three \((3,4\), and 5\()\) are second printings. \(\mathrm{A}\) student examines these books in random order, stopping only when a second printing has been selected. a. Display the possible outcomes in a tree diagram. b. What outcomes are contained in the event \(A\), that exactly one book is examined before the chance experiment terminates? c. What outcomes are contained in the event \(C\), that the chance experiment terminates with the examination of book 5 ?

Short Answer

Expert verified
For event \(A\), the possible outcomes are \(1-3\), \(1-4\), \(1-5\), \(2-3\), \(2-4\), and \(2-5\). For event \(C\), the possible outcomes would involve any combination of books \(1, 2, 3,\) and \(4\) (in any order), ending with book \(5\).

Step by step solution

01

Understanding Outcomes and Creating Tree Diagram

In creating the tree diagram, start with a branch for each book that the student could examine first. From each of these, draw additional branches representing the possible second books the student could examine, and so on, until all five books have been represented. The possible outcomes of the experiment are sequences in which the books might be examined. For instance, an outcome could be \(3\), or \(1-3\), or \(1-2-3\), etc., ending when the student picks up a second printing.
02

Identifying Outcomes in Event \(A\)

Event \(A\) is the event that exactly one book is examined before the experiment terminates. This means the second book picked is from the second printing. The possible outcomes for event \(A\) are therefore \(1-3\), \(1-4\), \(1-5\), \(2-3\), \(2-4\), and \(2-5\).
03

Identifying Outcomes in Event \(C\)

Event \(C\) is the event that the experiment terminates with the examination of book 5. This would mean that the student examines any combination of books \(1, 2, 3, and 4\), in any order, and finally ends with book 5. The possible outcomes include \(1-2-3-4-5\), \(1-2-4-3-5\), \(2-1-3-4-5\), up to \(5\) (where the student picks book 5 right away).

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Key Concepts

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

Tree Diagram
A tree diagram is a useful tool in probability to visualize all possible outcomes of an experiment. In this scenario, we begin by imagining each book as a branch. Each branch represents a possible choice the student could make initially. From there, we extend additional branches to represent subsequent choices.
For this exercise:
  • Begin with five initial branches, each standing for one of the books: 1, 2, 3, 4, and 5.
  • Next, from each branch associated with books 1 and 2 (the first printings), draw branches for books 3, 4, and 5, which are the second printings.
  • Continue this pattern until every book has been considered as a possible choice.
The tree diagram demonstrates sequences like 1-3 (first examine book 1, then book 3) or 3 (immediately selecting a second printing book). This visualization helps us see that the student stops examining as soon as they select a book from the second printing.
Random Selection
The concept of random selection is central to this exercise. When the student chooses a book without any preference and continues until a condition is met, that is random selection in action. Here, we assume all books have an equal chance of being picked, which adds randomness to the process.

Randomness implies the unpredictability of the order in which the student may examine the books. This is important because it influences the probability of each sequence of selections occurring. For example:
  • The chance of picking book 3 before others depends on the random turn-taking of the subsequent books.
  • If the student freely chooses a book, any book might be the first choice, which drives the need for careful event identification.
Through random selection, the learning exercise emphasizes understanding diverse combinations and their impact, showcasing the varied paths one can follow in real-life choices.
Event Identification
Event identification helps in pinpointing specific outcomes or sequences of interest within a broader set of possibilities. In this task, we explore two particular events, "Event A" and "Event C".

**Event A: One Book Examined** Event A occurs when the student stops examining books immediately after selecting the first second printing book. Possible sequences for this event include:
  • 1-3 (pick book 1, then book 3)
  • 1-4, 1-5 (similar logic with book 1 first)
  • 2-3, 2-4, 2-5 (similar logic with book 2 first)
Each outcome shows the termination just after the first second printing book is chosen. **Event C: Terminating with Book 5** Event C describes sequences where the student eventually picks book 5. Book 5 could be the first selection, or it might come after any number of other books. Possibilities include selecting all books in some order ending with book 5 like:
  • 1-2-3-4-5
  • and any combination ending with book 5
Identifying these events is vital for understanding which sequences satisfy given conditions, aiding in probability calculations.

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Most popular questions from this chapter

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.

A family consisting of three people- \(\mathrm{P}_{1}, \mathrm{P}_{2}\), and \(\mathrm{P}_{3}\) -belongs to a medical clinic that always has a physician at each of stations 1,2, and \(3 .\) During a certain week, each member of the family visits the clinic exactly once and is randomly assigned to a station. One experimental outcome is \((1,2,1)\), which means that \(\mathrm{P}_{1}\) is assigned to station \(1, \mathrm{P}_{2}\) to station 2, and \(\mathrm{P}_{3}\) to station \(1 .\) a. List the 27 possible outcomes. (Hint: First list the nine outcomes in which \(\mathrm{P}_{1}\) goes to station 1, then the nine in which \(\mathrm{P}_{1}\) goes to station 2, and finally the nine in which \(\mathrm{P}_{1}\) goes to station 3 ; a tree diagram might help.) b. List all outcomes in the event \(A\), that all three people go to the same station. c. List all outcomes in the event \(B\), that all three people go to different stations. d. List all outcomes in the event \(C\), that no one goes to station 2 . e. Identify outcomes in each of the following events: \(B^{C}, C^{C}, A \cup B, A \cap B, A \cap C\).

The paper "Good for Women, Good for Men, Bad for People: Simpson's Paradox and the Importance of Sex-Spedfic Analysis in Observational Studies" (Journal of Women's Health and Gender-Based Medicine [2001]: \(867-872\) ) described the results of a medical study in which one treatment was shown to be better for men and better for women than a competing treatment. However, if the data for men and women are combined, it appears as though the competing treatment is better. To see how this can happen, consider the accompanying data tables constructed from information in the paper. Subjects in the study were given either Treatment \(\mathrm{A}\) or Treatment \(\mathrm{B}\), and survival was noted. Let \(S\) be the event that a patient selected at random survives, \(A\) be the event that a patient selected at random received Treatment \(\mathrm{A}\), and \(B\) be the event that a patient selected at random received Treatment \(\mathrm{B}\). a. The following table summarizes data for men and women combined: $$ \begin{array}{l|ccc} & \text { Survived } & \text { Died } & \text { Total } \\ \hline \text { Treatment A } & 215 & 85 & \mathbf{3 0 0} \\ \text { Treatment B } & 241 & 59 & \mathbf{3 0 0} \\ \text { Total } & \mathbf{4 5 6} & \mathbf{1 4 4} & \\ \hline \end{array} $$ i. Find \(P(S)\). ii. Find \(P(S \mid A)\). iii. Find \(P(S \mid B)\). iv. Which treatment appears to be better? b. Now consider the summary data for the men who participated in the study: $$ \begin{array}{l|rrr} & \text { Survived } & \text { Died } & \text { Total } \\ \hline \text { Treatment A } & 120 & 80 & \mathbf{2 0 0} \\ \text { Treatment B } & 20 & 20 & 40 \\ \text { Total } & \mathbf{1 4 0} & \mathbf{1 0 0} & \\ \hline \end{array} $$ i. Find \(P(S)\). ii. Find \(P(S \mid A)\). iii. Find \(P(S \mid B)\). iv. Which treatment appears to be better? c. Now consider the summary data for the women who participated in the study: $$ \begin{array}{l|rrc} & \text { Survived } & \text { Died } & \text { Total } \\ \hline \text { Treatment A } & 95 & 5 & \mathbf{1 0 0} \\ \text { Treatment B } & 221 & 39 & \mathbf{2 6 0} \\ \text { Total } & \mathbf{3 1 6} & \mathbf{1 4 4} & \\ \hline \end{array} $$ i. Find \(P(S)\). ii. Find \(P(S \mid A)\). iii. Find \(P(S \mid B)\). iv. Which treatment appears to be better? d. You should have noticed from Parts (b) and (c) that for both men and women, Treatment \(A\) appears to be better. But in Part (a), when the data for men and women are combined, it looks like Treatment \(\mathrm{B}\) is better. This is an example of what is called Simpson's paradox. Write a brief explanation of why this apparent inconsistency occurs for this data set. (Hint: Do men and women respond similarly to the two treatments?)

A company uses three different assembly lines \(A_{1}, A_{2}\), and \(A_{3}\) -to manufacture a particular component. Of those manufactured by \(A_{1}, 5 \%\) need rework to remedy a defect, whereas \(8 \%\) of \(A_{2}\) 's components and \(10 \%\) of \(A_{3}\) 's components need rework. Suppose that \(50 \%\) of all components are produced by \(A_{1}\), whereas \(30 \%\) are produced by \(A_{2}\) and \(20 \%\) come from \(A_{3}\). a. Construct a tree diagram with first-generation branches corresponding to the three lines. Leading from each branch, draw one branch for rework (R) and another for no rework (N). Then enter appropriate probabilities on the branches. b. What is the probability that a randomly selected component came from \(A_{1}\) and needed rework? c. What is the probability that a randomly selected component needed rework?

Define the term chance experiment, and give an example of a chance experiment with four possible outcomes.
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