/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 A family consisting of three peo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 10

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: \(B^{C}\), \(C^{C}, A \cup B, A \cap B, A \cap C\)

Short Answer

Expert verified
Event \(A\) contains the outcomes: (1,1,1), (2,2,2), (3,3,3). Event \(B\) contains: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1). Event \(C\) contains: (1,1,1), (1,1,3), (1,3,1), (1,3,3), (3,1,1), (3,1,3), (3,3,1), (3,3,3). Event \(B^C\) is all outcomes minus \(B\), event \(C^c\) is all outcomes minus \(C\), event \(A \cup B\) are all outcomes in \(A\) and \(B\) without duplicates, event \(A \cap B\) is null (no common outcomes), and \(A \cap C\) is (1,1,1), (3,3,3).

Step by step solution

01

List all outcomes

To list all possible outcomes, start by considering where \(\mathrm{P}_{1}\) can go. For each choice of \(\mathrm{P}_{1}\), there are 3 possibilities for \(\mathrm{P}_{2}\), \(\mathrm{P}_{3}\), creating 3 * 3 = 9 outcomes per choice of \(\mathrm{P}_{1}\). Therefore, there are 27 total outcomes, broken down as follows:\n\nFor \(\mathrm{P}_{1}\) at station 1: (1,1,1), (1,1,2), (1,1,3), (1,2,1), (1,2,2), (1,2,3), (1,3,1), (1,3,2), (1,3,3).\nFor \(\mathrm{P}_{1}\) at station 2: (2,1,1), (2,1,2), (2,1,3), (2,2,1), (2,2,2), (2,2,3), (2,3,1), (2,3,2), (2,3,3).\nFor \(\mathrm{P}_{1}\) at station 3: (3,1,1), (3,1,2), (3,1,3), (3,2,1), (3,2,2), (3,2,3), (3,3,1), (3,3,2), (3,3,3).
02

Identify Event A

Event \(A\) is that all three people go to the same station. This event includes the following outcomes: (1,1,1), (2,2,2), (3,3,3).
03

Identify Event B

Event \(B\) is that all three people go to different stations. This event includes the following outcomes: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1).
04

Identify Event C

Event \(C\) is that no one goes to station 2. This event includes the following outcomes: (1,1,1), (1,1,3), (1,3,1), (1,3,3), (3,1,1), (3,1,3), (3,3,1), (3,3,3).
05

Identify Events Compositions

Here are the outcomes for other events asked:\n\nEvent \(B^c\), i.e., not \(B\): Includes all outcomes that are not in \(B\). This is all outcomes minus the outcomes of \(B\). You simply look at the list of all outcomes and exclude the ones that are listed under \(B\).\n\nEvent \(C^c\), i.e., not \(C\): Includes all outcomes that are not in \(C\). This is all outcomes minus the outcomes of \(C\). You simply look at the list of all outcomes and exclude the ones that are listed under \(C\).\n\nEvent \(A \cup B\) (the union of \(A\) and \(B\)): Includes all outcomes that are in \(A\), in \(B\), or in both. This set can be found by combining the outcome lists for \(A\) and for \(B\), and removing duplicates.\n\nEvent \(A \cap B\) (the intersection of \(A\) and \(B\)): Includes all outcomes that are both in \(A\) and \(B\). It can be found by taking the list of outcomes in \(A\) and removing all that are not in \(B\). As \(A\) and \(B\) do not share any outcomes, this set is empty.\n\nEvent \(A \cap C\) (the intersection of \(A\) and \(C\)): Includes all outcomes that are both in \(A\) and \(C\). It can be found by taking the list of outcomes in \(A\) and removing all that are not in \(C\). The intersection is (1,1,1), (3,3,3).

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 91影视!

Key Concepts

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

Probability
Probability is the way we measure the chance or likelihood of an event occurring. In essence, it tells us how likely something is to happen. It ranges from 0 to 1, where 0 means the event cannot happen, and 1 means it is certain to happen. For example, if we're rolling a fair six-sided die, the probability of rolling a 2 is 1 out of 6. When dealing with probability:
  • You need to identify all possible outcomes (sample space).
  • You calculate the probability of an event by dividing the number of ways the event can happen by the total number of outcomes.
Let's relate this to our exercise. When each family member is assigned randomly to one of the stations, there are 27 possible outcomes. These outcomes represent all possible ways the 3 family members could be assigned. To find the probability of a specific event, like all members going to different stations, you'd count the number of matching outcomes and divide it by the total number of outcomes, which is 27.
Events and Outcomes
In probability, an event is a set of outcomes to which a probability is assigned. Outcomes are the possible results of a probability experiment. A single outcome might be rolling a 4 on a die.

In our exercise, an event might be family members ending up at different stations. Outcomes for this would include sequences such as (1,2,3), (1,3,2), etc.
  • Simple Events: Only one outcome occurs (like a die showing a "2").
  • Compound Events: Consists of more than one outcome. The event that everyone goes to different stations contains multiple outcomes.

Understanding the relationship between events and outcomes helps us compute probabilities correctly. For event B, which is everyone going to different stations, you list all specific outcomes that fit this criteria and analyze them similarly.
Tree Diagram
Tree diagrams are a useful tool in combinatorics to visually map out all possible results of a probability experiment. They make complex problems simpler by allowing us to easily see all potential outcomes and analyze probability events.

A tree diagram begins with a starting point, branching out for each possible outcome at each step. In our example, a good start is to draw branches for the stations each person can choose (1, 2, or 3 for each person, P1, P2, and P3). Each branching path represents a distinct event.
  • Start with P1 and branch into three possibilities (station 1, 2, or 3).
  • For each choice of P1, branch out again for P2's three options.
  • Repeat for P3, resulting in a complete tree with distinct paths for each combination.

Using a tree diagram allows you to visualize that there are 27 potential paths or outcomes. It helps us identify which combination of sequences match specific events like A, B, or C. It's a graphic way of ensuring you haven't missed an outcome and can help compute both simple and compound probabilities effectively.

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 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. 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 \(5 ?\)

Three friends \((\mathrm{A}, \mathrm{B}\), and \(\mathrm{C})\) will participate in a round-robin tournament in which each one plays both of the others. Suppose that \(P(\) A beats \(B)=.7, P(\) A beats \(C)=.8\), \(P(\mathrm{~B}\) beats \(\mathrm{C})=.6\), and that the outcomes of the three matches are independent of one another. a. What is the probability that \(\mathrm{A}\) wins both her matches and that B beats C? b. What is the probability that A wins both her matches? c. What is the probability that A loses both her matches? d. What is the probability that each person wins one match? (Hint: There are two different ways for this to happen.)

The USA Today article referenced in Exercise \(6.37\) also gave information on seat belt usage by age, which is summarized in the following table of counts: $$ \begin{array}{lcc} & \begin{array}{c} \text { Does Not Use } \\ \text { Seat Belt } \\ \text { Regularly } \end{array} & \begin{array}{c} \text { Uses } \\ \text { Seat Belt } \\ \text { Regularly } \end{array} \\ \hline 18-24 & 59 & 41 \\ 25-34 & 73 & 27 \\ 35-44 & 74 & 26 \\ 45-54 & 70 & 30 \\ 55-64 & 70 & 30 \\ 65 \text { and older } & 82 & 18 \\ \hline \end{array} $$ Consider the following events: \(S=\) event that a randomly selected individual uses a seat belt regularly, \(A_{1}=\) event that a randomly selected individual is in age group \(18-24\), and \(A_{6}=\) event that a randomly selected individual is in age group 65 and older. a. Convert the counts to proportions and then use them to compute the following probabilities: i. \(P\left(A_{1}\right)\) ii. \(P\left(A_{1} \cap S\right)\) iii. \(P\left(A_{1} \mid S\right)\) iv. \(P\left(\right.\) not \(\left.A_{1}\right) \quad\) v. \(P\left(S \mid A_{1}\right) \quad\) vi. \(P\left(S \mid A_{6}\right)\) b. Using the probabilities \(P\left(S \mid A_{1}\right)\) and \(P\left(S \mid A_{6}\right)\) computed in Part (a), comment on how \(18-24\) -year-olds and seniors differ with respect to seat belt usage.

In an article that appears on the web site of the American Statistical Association (www.amstat.org), Carlton Gunn, a public defender in Seattle, Washington, wrote about how he uses statistics in his work as an attorney. He states: I personally have used statistics in trying to challenge the reliability of drug testing results. Suppose the chance of a mistake in the taking and processing of a urine sample for a drug test is just 1 in 100 . And your client has a "dirty" (i.e., positive) test result. Only a 1 in 100 chance that it could be wrong? Not necessarily. If the vast majority of all tests given - say 99 in 100 \- are truly clean, then you get one false dirty and one true dirty in every 100 tests, so that half of the dirty tests are false. Define the following events as \(T D=\) event that the test result is dirty, \(T C=\) event that the test result is clean, \(D=\) event that the person tested is actually dirty, and \(C=\) event that the person tested is actually clean. a. Using the information in the quote, what are the values of \mathbf{i} . ~ \(P(T D \mid D)\) iii. \(P(C)\) ii. \(P(T D \mid C) \quad\) iv. \(P(D)\) b. Use the law of total probability to find \(P(T D)\). c. Use Bayes' rule to evaluate \(P(C \mid T D)\). Is this value consistent with the argument given in the quote? Explain.

At a large university, the Statistics Department has tried a different text during each of the last three quarters. During the fall quarter, 500 students used a book by Professor Mean; during the winter quarter, 300 students used a book by Professor Median; and during the spring quarter, 200 students used a book by Professor Mode. A survey at the end of each quarter showed that 200 students were satisfied with the text in the fall quarter, 150 in the winter quarter, and 160 in the spring quarter. a. If a student who took statistics during one of these three quarters is selected at random, what is the probability that the student was satisfied with the textbook? b. If a randomly selected student reports being satisfied with the book, is the student most likely to have used the book by Mean, Median, or Mode? Who is the least likely author? (Hint: Use Bayes' rule to compute three probabilities.)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation
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.