/*! 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} Q36E An academic department with five... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Q36E (page 74)

An academic department with five faculty members narrowed its choice for department head to either candidate A or candidate B. Each member then voted on a slip of paper for one of the candidates. Suppose there are actually three votes for A and two for B. If the slips are selected for tallying in random order, what is the probability that A remains ahead of B throughout the vote count (e.g., this event occurs if the selected ordering is AABAB, but not for ABBAA)?

Short Answer

Expert verified

The probability is \({\rm{0}}{\rm{.2}}{\rm{. }}\) that A remains ahead of B throughout the vote count.

Step by step solution

01

Definition

The term "probability" simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes鈥攈ow likely they are鈥攚hen we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

02

Finding the probability

We can list all outcomes, see what outcomes are favorable and use

\({\rm{P(A) = }}\frac{{{\rm{ \# favorableoutcomesinA}}}}{{{\rm{ \# outcomesinthesamplespace}}}}{\rm{.}}\)

There are

\(\left( {\begin{aligned}{\rm{5}}\\{\rm{3}}\end{aligned}} \right)\)

ways to select positions for A votes (or equally

\(\left( {\begin{aligned}{\rm{5}}\\{\rm{2}}\end{aligned}} \right)\)

ways to select positions for B votes):

\({\rm{BBAAA,BABAA,BAABA,BAAAB,ABBAA,ABABA,ABAAB,AABBA,AABAB,AAABB}}\)

and only two have \({\rm{A}}\)ahead of \({\rm{B}}\) trough out the vote - \({\rm{AABAB}}\) and\({\rm{AAABB}}\).

All outcomes are equally probable therefore we have that the probability that \({\rm{A}}\) remains ahead of \({\rm{B}}\) trough out the vote is

\(\begin{aligned}{\rm{P(A) = }}\frac{{{\rm{\# favorable outcomes in A }}}}{{{\rm{ \# out comes in the sample space}}}}\\{\rm{ = }}\frac{{\rm{2}}}{{{\rm{10}}}}\\{\rm{ = 0}}{\rm{.2}}{\rm{. }}\end{aligned}\)

Therefore, the probability is \({\rm{0}}{\rm{.2}}{\rm{. }}\)

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!

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 production facility employs 10 workers on the day shift, \({\rm{8}}\) workers on the swing shift, and 6 workers on the graveyard shift. A quality control consultant is to select \({\rm{5}}\) of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of \({\rm{5}}\) workers has the same chance of being selected as does any other group (drawing \({\rm{5}}\) slips without replacement from among \({\rm{24}}\)).

a. How many selections result in all \({\rm{5}}\) workers coming from the day shift? What is the probability that all 5selected workers will be from the day shift?

b. What is the probability that all \({\rm{5}}\) selected workers will be from the same shift?

c. What is the probability that at least two different shifts will be represented among the selected workers?

d. What is the probability that at least one of the shifts will be unrepresented in the sample of workers?

For customers purchasing a refrigerator at a certain appliance store, let A be the event that the refrigerator was manufactured in the U.S., B be the event that the refrigerator had an icemaker, and C be the event that the customer purchased an extended warranty. Relevant probabilities are

\(\begin{aligned}{\rm{P(A) = }}{\rm{.75}}\;\;\;{\rm{P(B}}\mid {\rm{A) = }}{\rm{.9}}\;\;\;{\rm{P}}\left( {{\rm{B}}\mid {{\rm{A}}^{\rm{\cent}}}} \right){\rm{ = }}{\rm{.8}}\\{\rm{P(C}}\mid {\rm{A\c{C}B) = }}{\rm{.8}}\;\;\;{\rm{P}}\left( {{\rm{C}}\mid {\rm{A\c{C}}}{{\rm{B}}^{\rm{\cent}}}} \right){\rm{ = }}{\rm{.6}}\\{\rm{P}}\left( {{\rm{C}}\mid {{\rm{A}}^{\rm{\cent}}}{\rm{\c{C}B}}} \right){\rm{ = }}{\rm{.7}}\;\;\;{\rm{P}}\left( {{\rm{C}}\mid {{\rm{A}}^{\rm{\cent}}}{\rm{\c{C}}}{{\rm{B}}^{\rm{\cent}}}} \right){\rm{ = }}{\rm{.3}}\end{aligned}\)

Construct a tree diagram consisting of first-, second-, and third-generation branches, and place an event label and appropriate probability next to each branch. b. Compute \({\rm{P(A\c{C}B\c{C}C)}}\). c. Compute \({\rm{P(B\c{C}C)}}\). d. Compute \({\rm{P(C)}}\). e. Compute \({\rm{P(A}}\mid {\rm{B\c{C}C)}}\), the probability of a U.S. purchase given that an icemaker and extended warranty are also purchased

The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop machines and the other four have chosen desktop machines. Suppose that only two of the setups can be done on a particular day, and the two computers to be set up are randomly selected from the six (implying 15 equally likely outcomes; if the computers are numbered1, 2,鈥�, 6, then one outcome consists of computers 1 and2, another consists of computers 1 and 3, and so on).

a. What is the probability that both selected setups are for laptop computers?

b. What is the probability that both selected setups are desktop machines?

c. What is the probability that at least one selected setup is for a desktop computer?

d. What is the probability that at least one computer of each type is chosen for setup?

Consider purchasing a system of audio components consisting of a receiver, a pair of speakers, and a \({\rm{CD}}\) player. Let \({{\rm{A}}_{\rm{1}}}\) be the event that the receiver functions properly throughout the warranty period, \({{\rm{A}}_{\rm{2}}}\) be the event that the speakers function properly throughout the warranty period, and \({{\rm{A}}_{\rm{3}}}\) be the event that the \({\rm{CD}}\) player functions properly throughout the warranty period. Suppose that these events are (mutually) independent with \({\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = }}{\rm{.95}}\), \({\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = }}{\rm{.98}}\), and \({\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{ = }}{\rm{.80}}\).

a. What is the probability that all three components function properly throughout the warranty period?

b. What is the probability that at least one component needs service during the warranty period?

c. What is the probability that all three components need service during the warranty period?

d. What is the probability that only the receiver needs service during the warranty period?

e. What is the probability that exactly one of the three components needs service during the warranty period?

f. What is the probability that all three components function properly throughout the warranty period but that at least one fails within a month after the warranty expires?

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 non wetting, 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?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Decision Maths

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.