/*! 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 5 A family consisting of three per... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A family consisting of three persons- \(A, B\), and \(C-\) goes to a medical clinic that always has a doctor at each of stations 1 , 2 , and 3. During a certain week, each member of the family visits the clinic once and is assigned at random to a station. The experiment consists of recording the station number for each member. One outcome is \((1,2,1)\) for \(A\) to station \(1, B\) to station 2 , and \(C\) to station 1 . a. List the 27 outcomes in the sample space. b. List all outcomes in the event that all three members go to the same station. c. List all outcomes in the event that all members go to different stations. d. List all outcomes in the event that no one goes to station 2 .

Short Answer

Expert verified
There are 27 total outcomes. 3 have the same station; 6 have different stations; 8 avoid station 2.

Step by step solution

01

Identify the Total Number of Outcomes

Each family member can be assigned to one of three stations. Therefore, the total number of possible outcomes is obtained by multiplying the individual choices: \(3\) choices for A, \(3\) choices for B, and \(3\) choices for C: \(3 \times 3 \times 3 = 27\) outcomes.
02

List All Possible Outcomes

The sample space consists of all possible combinations of (station for A, station for B, station for C). The outcomes are: (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), (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), (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).
03

List Outcomes Where All Members Visit the Same Station

For all members to visit the same station, they must all be assigned to either station 1, 2, or 3. The outcomes are: (1,1,1), (2,2,2), (3,3,3).
04

List Outcomes Where All Members Visit Different Stations

Each member can be assigned to a different station, resulting in permutations of three people in three spaces. The outcomes are: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1).
05

List Outcomes Where No Member Goes to Station 2

Station 2 must not be assigned to any member. Therefore, each member can only choose between stations 1 and 3. The outcomes are: (1,1,1), (1,1,3), (1,3,1), (1,3,3), (3,1,1), (3,1,3), (3,3,1), (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 91Ó°ÊÓ!

Key Concepts

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

Sample Space
When dealing with probability, a fundamental concept that provides the basis for any analysis is the sample space. A sample space is the complete set of all possible outcomes of a probability experiment. In our exercise, the sample space involves three family members visiting a clinic with three stations available for each one.

Each family member, labeled as A, B, and C, can independently be assigned to one of the three stations numbered 1, 2, or 3. As there are 3 choices for A, 3 choices for B, and 3 choices for C, the total number of possible outcomes is a result of multiplying these choices together: \[3 \times 3 \times 3 = 27\] outcomes.

This forms the sample space, which includes outcomes such as:
  • (1,1,1), which means A, B, and C all go to station 1
  • (1,2,3), where A goes to station 1, B goes to station 2, and C goes to station 3
  • (3,2,1), where A goes to station 3, B to station 2, and C to station 1
The full list of all 27 outcomes is systematically constructed by considering every possible combination, embracing the complete range of possibilities under the given conditions.
Permutations
Permutations come into play when considering the different possible arrangements of a set of items. In our case involving the family, permutations are used when we want every member to go to a different station.

For three members going to three stations, each member must visit a unique station. This constitutes arranging three distinct items in three spaces. So, we calculate the number of permutations of three items, which happens to be \[3! = 3 \times 2 \times 1 = 6\] permutations.

These permutations are:
  • (1,2,3): A assigned to 1, B to 2, C to 3
  • (1,3,2): A to 1, B to 3, C to 2
  • (2,1,3): A to 2, B to 1, C to 3
  • (2,3,1): A to 2, B to 3, C to 1
  • (3,1,2): A to 3, B to 1, C to 2
  • (3,2,1): A to 3, B to 2, C to 1
The permutations give us all the possible ways to arrange A, B, and C across the stations, where no two family members will share the same station.
Event Outcomes
An event is a specific set of outcomes from the sample space that we are interested in examining. In context, an event might be something like all members visiting different stations, or none of them going to a specific station.

Let's focus on one interesting event: no member visiting station 2. For this to occur, each family member must be assigned to the remaining stations 1 or 3. This narrows down the potential outcomes as:
  • (1,1,1): Everyone at station 1
  • (1,1,3): A and B at station 1, C at station 3
  • (1,3,1): A and C at station 1, B at station 3
  • (1,3,3): A at station 1, B and C at station 3
  • (3,1,1): B and C at station 1, A at station 3
  • (3,1,3): B at station 1, A and C at station 3
  • (3,3,1): C at station 1, A and B at station 3
  • (3,3,3): Everyone at station 3
Each outcome meets the criteria of the event, displaying the flexibility and utility of understanding events within sample spaces. Events help in targeting specific conditions or probabilities we wish to calculate.

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

Each of a sample of four home mortgages is classified as fixed rate \((F)\) or variable rate \((V)\). a. What are the 16 outcomes in \(\mathcal{S}\) ? b. Which outcomes are in the event that exactly three of the selected mortgages are fixed rate? c. Which outcomes are in the event that all four mortgages are of the same type? d. Which outcomes are in the event that at most one of the four is a variable- rate mortgage? e. What is the union of the events in parts (c) and (d), and what is the intersection of these two events? f. What are the union and intersection of the two events in parts (b) and (c)?

Components of a certain type are shipped to a supplier in batches of ten. Suppose that \(50 \%\) of all such batches contain no defective components, \(30 \%\) contain one defective component, and \(20 \%\) contain two defective components. Two components from a batch are randomly selected and tested. What are the probabilities associated with 0,1 , and 2 defective components being in the batch under each of the following conditions? a. Neither tested component is defective. b. One of the two tested components is defective.

Again consider a Little League team that has 15 players on its roster. a. How many ways are there to select 9 players for the starting lineup? b. How many ways are there to select 9 players for the starting lineup and a batting order for the 9 starters? c. Suppose 5 of the 15 players are left-handed. How many ways are there to select 3 left-handed outfielders and have all 6 other positions occupied by right-handed players?

Human visual inspection of solder joints on printed circuit boards can be very subjective. Part of the problem stems from the numerous types of solder defects (e.g., pad nonwetting, knee visibility, voids) and even the degree to which a joint possesses one or more of these defects. Consequently, even highly trained inspectors can disagree on the disposition of a particular joint. In one batch of 10,000 joints, inspector A found 724 that were judged defective, inspector B found 751 such joints, and 1159 of the joints were judged defective by at least one of the inspectors. Suppose that one of the 10,000 joints is randomly selected. a. What is the probability that the selected joint was judged to be defective by neither of the two inspectors? b. What is the probability that the selected joint was judged to be defective by inspector \(B\) but not by inspector A?

Consider the following information about travelers on vacation (based partly on a recent Travelocity poll): \(40 \%\) check work email, \(30 \%\) use a cell phone to stay connected to work, \(25 \%\) bring a laptop with them, \(23 \%\) 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 out of every 100 who bring a laptop also check work email, and 70 out of every 100 who use a cell phone to stay connected also bring a laptop. a. What is the probability that a randomly selected traveler who checks work email also uses a cell phone to stay connected? b. What is the probability that someone who brings a laptop on vacation also uses a cell phone to stay connected? c. If the randomly selected traveler checked work email and brought a laptop, what is the probability that he/she uses a cell phone to stay connected?

See all solutions

Recommended explanations on Math Textbooks

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.