Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015
Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015
1st Edition
ISBN: 9781680331165
Author: HOUGHTON MIFFLIN HARCOURT
Publisher: Houghton Mifflin Harcourt
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 10.2, Problem 20E

(a)

To determine

To calculate:

The probability value of the selection of three red marbles when every marble is gets replaced before the selection

(a)

Expert Solution
Check Mark

Answer to Problem 20E

The probability value of the selection of three red marbles when every marble is gets replaced before the selection is 0.091125 .

Explanation of Solution

Given information:

Number of red marble: nR=9

Number of yellow marbles: ny=7

Number of blue marbles: nb=4

Calculation:

Here, total numbers of marbles in bag is calculated as:

  n=nR+ny+nb=9+4+7=20

In the first draw, need to select the three marbles of red colored when every marble is getting replaced before the selection.

Consider this event as event P(A)=P(3R) .

Probability values of events A of three red colored marbles selection is calculated as:

  P(A)=P(3R)P(A)=920×920×920P(A)=7298000P(A)=0.091125

(b)

To determine

To calculate:

The probability value of the selection of three red marbles when every marble is not replaced before the selection of the next marble

(b)

Expert Solution
Check Mark

Answer to Problem 20E

The probability value of the selection of three red marbles when every marble is not replaced before the selection of the next marble is 0.0737 .

Explanation of Solution

Given information:

Number of red marble: nR=9

Number of yellow marbles: ny=7

Number of blue marbles: nb=4

Calculation:

Here, total numbers of marbles in bag is calculated as:

  n=nR+ny+nb=9+4+7=20

In the draw, need to select the three marbles of red colored when every marble is not replaced before the selection of the next marble.

Consider this event as event P(A)=P(3R) .

Here, in the first draw nine marbles of red colored selected, then next one marble is not replaced and there is remaining of eight red colored marbles. Now, need to select one red marble again without replacement that means there is remaining of seven red colored marbles.

Probability values of events A of three red colored marbles selection without replacement are calculated as:

  P(A)=P(3R)P(A)=920×820×720P(A)=5048000P(A)=0.0737

Chapter 10 Solutions

Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015

Ch. 10.1 - Prob. 11ECh. 10.1 - Prob. 12ECh. 10.1 - Prob. 13ECh. 10.1 - Prob. 14ECh. 10.1 - Prob. 15ECh. 10.1 - Prob. 16ECh. 10.1 - Prob. 17ECh. 10.1 - Prob. 18ECh. 10.1 - Prob. 19ECh. 10.1 - Prob. 20ECh. 10.1 - Prob. 21ECh. 10.1 - Prob. 22ECh. 10.1 - Prob. 23ECh. 10.1 - Prob. 24ECh. 10.1 - Prob. 25ECh. 10.1 - Prob. 26ECh. 10.1 - Prob. 27ECh. 10.1 - Prob. 28ECh. 10.1 - Prob. 29ECh. 10.1 - Prob. 30ECh. 10.1 - Prob. 31ECh. 10.1 - Prob. 32ECh. 10.1 - Prob. 33ECh. 10.1 - Prob. 34ECh. 10.2 - Prob. 1ECh. 10.2 - Prob. 2ECh. 10.2 - Prob. 3ECh. 10.2 - Prob. 4ECh. 10.2 - Prob. 5ECh. 10.2 - Prob. 6ECh. 10.2 - Prob. 7ECh. 10.2 - Prob. 8ECh. 10.2 - Prob. 9ECh. 10.2 - Prob. 10ECh. 10.2 - Prob. 11ECh. 10.2 - Prob. 12ECh. 10.2 - Prob. 13ECh. 10.2 - Prob. 14ECh. 10.2 - Prob. 15ECh. 10.2 - Prob. 16ECh. 10.2 - Prob. 17ECh. 10.2 - Prob. 18ECh. 10.2 - Prob. 19ECh. 10.2 - Prob. 20ECh. 10.2 - Prob. 21ECh. 10.2 - Prob. 22ECh. 10.2 - Prob. 23ECh. 10.2 - Prob. 24ECh. 10.2 - Prob. 25ECh. 10.2 - Prob. 26ECh. 10.2 - Prob. 27ECh. 10.2 - Prob. 28ECh. 10.2 - Prob. 29ECh. 10.2 - Prob. 30ECh. 10.2 - Prob. 31ECh. 10.2 - Prob. 32ECh. 10.2 - Prob. 33ECh. 10.3 - Prob. 1ECh. 10.3 - Prob. 2ECh. 10.3 - Prob. 3ECh. 10.3 - Prob. 4ECh. 10.3 - Prob. 5ECh. 10.3 - Prob. 6ECh. 10.3 - Prob. 7ECh. 10.3 - Prob. 8ECh. 10.3 - Prob. 9ECh. 10.3 - Prob. 10ECh. 10.3 - Prob. 11ECh. 10.3 - Prob. 12ECh. 10.3 - Prob. 13ECh. 10.3 - Prob. 14ECh. 10.3 - Prob. 15ECh. 10.3 - Prob. 16ECh. 10.3 - Prob. 17ECh. 10.3 - Prob. 18ECh. 10.3 - Prob. 19ECh. 10.3 - Prob. 20ECh. 10.3 - Prob. 21ECh. 10.3 - Prob. 22ECh. 10.3 - Prob. 23ECh. 10.3 - Prob. 24ECh. 10.3 - Prob. 25ECh. 10.3 - Prob. 26ECh. 10.3 - Prob. 27ECh. 10.3 - Prob. 28ECh. 10.3 - Prob. 29ECh. 10.3 - Prob. 1QCh. 10.3 - Prob. 2QCh. 10.3 - Prob. 3QCh. 10.3 - Prob. 4QCh. 10.3 - Prob. 5QCh. 10.3 - Prob. 6QCh. 10.3 - Prob. 7QCh. 10.3 - Prob. 8QCh. 10.3 - Prob. 9QCh. 10.3 - Prob. 10QCh. 10.3 - Prob. 11QCh. 10.4 - Prob. 1ECh. 10.4 - Prob. 2ECh. 10.4 - Prob. 3ECh. 10.4 - Prob. 4ECh. 10.4 - Prob. 5ECh. 10.4 - Prob. 6ECh. 10.4 - Prob. 7ECh. 10.4 - Prob. 8ECh. 10.4 - Prob. 9ECh. 10.4 - Prob. 10ECh. 10.4 - Prob. 11ECh. 10.4 - Prob. 12ECh. 10.4 - Prob. 13ECh. 10.4 - Prob. 14ECh. 10.4 - Prob. 15ECh. 10.4 - Prob. 16ECh. 10.4 - Prob. 17ECh. 10.4 - Prob. 18ECh. 10.4 - Prob. 19ECh. 10.4 - Prob. 20ECh. 10.4 - Prob. 21ECh. 10.4 - Prob. 22ECh. 10.4 - Prob. 23ECh. 10.4 - Prob. 24ECh. 10.4 - Prob. 25ECh. 10.4 - Prob. 26ECh. 10.5 - Prob. 1ECh. 10.5 - Prob. 2ECh. 10.5 - Prob. 3ECh. 10.5 - Prob. 4ECh. 10.5 - Prob. 5ECh. 10.5 - Prob. 6ECh. 10.5 - Prob. 7ECh. 10.5 - Prob. 8ECh. 10.5 - Prob. 9ECh. 10.5 - Prob. 10ECh. 10.5 - Prob. 11ECh. 10.5 - Prob. 12ECh. 10.5 - Prob. 13ECh. 10.5 - Prob. 14ECh. 10.5 - Prob. 15ECh. 10.5 - Prob. 16ECh. 10.5 - Prob. 17ECh. 10.5 - Prob. 18ECh. 10.5 - Prob. 19ECh. 10.5 - Prob. 20ECh. 10.5 - Prob. 21ECh. 10.5 - Prob. 22ECh. 10.5 - Prob. 23ECh. 10.5 - Prob. 24ECh. 10.5 - Prob. 25ECh. 10.5 - Prob. 26ECh. 10.5 - Prob. 27ECh. 10.5 - Prob. 28ECh. 10.5 - Prob. 29ECh. 10.5 - Prob. 30ECh. 10.5 - Prob. 31ECh. 10.5 - Prob. 32ECh. 10.5 - Prob. 33ECh. 10.5 - Prob. 34ECh. 10.5 - Prob. 35ECh. 10.5 - Prob. 36ECh. 10.5 - Prob. 37ECh. 10.5 - Prob. 38ECh. 10.5 - Prob. 39ECh. 10.5 - Prob. 40ECh. 10.5 - Prob. 41ECh. 10.5 - Prob. 42ECh. 10.5 - Prob. 43ECh. 10.5 - Prob. 44ECh. 10.5 - Prob. 45ECh. 10.5 - Prob. 46ECh. 10.5 - Prob. 47ECh. 10.5 - Prob. 48ECh. 10.5 - Prob. 49ECh. 10.5 - Prob. 50ECh. 10.5 - Prob. 51ECh. 10.5 - Prob. 52ECh. 10.5 - Prob. 53ECh. 10.5 - Prob. 54ECh. 10.5 - Prob. 55ECh. 10.5 - Prob. 56ECh. 10.5 - Prob. 57ECh. 10.5 - Prob. 58ECh. 10.5 - Prob. 59ECh. 10.5 - Prob. 60ECh. 10.5 - Prob. 61ECh. 10.5 - Prob. 62ECh. 10.5 - Prob. 63ECh. 10.5 - Prob. 64ECh. 10.5 - Prob. 65ECh. 10.5 - Prob. 66ECh. 10.5 - Prob. 67ECh. 10.5 - Prob. 68ECh. 10.5 - Prob. 69ECh. 10.5 - Prob. 70ECh. 10.5 - Prob. 71ECh. 10.5 - Prob. 72ECh. 10.5 - Prob. 73ECh. 10.5 - Prob. 74ECh. 10.5 - Prob. 75ECh. 10.5 - Prob. 76ECh. 10.5 - Prob. 77ECh. 10.5 - Prob. 78ECh. 10.5 - Prob. 79ECh. 10.5 - Prob. 80ECh. 10.5 - Prob. 81ECh. 10.6 - Prob. 1ECh. 10.6 - Prob. 2ECh. 10.6 - Prob. 3ECh. 10.6 - Prob. 4ECh. 10.6 - Prob. 5ECh. 10.6 - Prob. 6ECh. 10.6 - Prob. 7ECh. 10.6 - Prob. 8ECh. 10.6 - Prob. 9ECh. 10.6 - Prob. 10ECh. 10.6 - Prob. 11ECh. 10.6 - Prob. 12ECh. 10.6 - Prob. 13ECh. 10.6 - Prob. 14ECh. 10.6 - Prob. 15ECh. 10.6 - Prob. 16ECh. 10.6 - Prob. 17ECh. 10.6 - Prob. 18ECh. 10.6 - Prob. 19ECh. 10.6 - Prob. 20ECh. 10.6 - Prob. 21ECh. 10.6 - Prob. 22ECh. 10.6 - Prob. 23ECh. 10.6 - Prob. 24ECh. 10 - Prob. 1CRCh. 10 - Prob. 2CRCh. 10 - Prob. 3CRCh. 10 - Prob. 4CRCh. 10 - Prob. 5CRCh. 10 - Prob. 6CRCh. 10 - Prob. 7CRCh. 10 - Prob. 8CRCh. 10 - Prob. 9CRCh. 10 - Prob. 10CRCh. 10 - Prob. 11CRCh. 10 - Prob. 12CRCh. 10 - Prob. 13CRCh. 10 - Prob. 14CRCh. 10 - Prob. 15CRCh. 10 - Prob. 16CRCh. 10 - Prob. 17CRCh. 10 - Prob. 1CTCh. 10 - Prob. 2CTCh. 10 - Prob. 3CTCh. 10 - Prob. 4CTCh. 10 - Prob. 5CTCh. 10 - Prob. 6CTCh. 10 - Prob. 7CTCh. 10 - Prob. 8CTCh. 10 - Prob. 9CTCh. 10 - Prob. 10CTCh. 10 - Prob. 11CTCh. 10 - Prob. 12CTCh. 10 - Prob. 13CTCh. 10 - Prob. 14CTCh. 10 - Prob. 1CACh. 10 - Prob. 2CACh. 10 - Prob. 3CACh. 10 - Prob. 4CACh. 10 - Prob. 5CACh. 10 - Prob. 6CACh. 10 - Prob. 7CACh. 10 - Prob. 8CACh. 10 - Prob. 9CA
Knowledge Booster
Background pattern image
Algebra
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, algebra and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Algebra and Trigonometry (6th Edition)
Algebra
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:PEARSON
Text book image
Contemporary Abstract Algebra
Algebra
ISBN:9781305657960
Author:Joseph Gallian
Publisher:Cengage Learning
Text book image
Linear Algebra: A Modern Introduction
Algebra
ISBN:9781285463247
Author:David Poole
Publisher:Cengage Learning
Text book image
Algebra And Trigonometry (11th Edition)
Algebra
ISBN:9780135163078
Author:Michael Sullivan
Publisher:PEARSON
Text book image
Introduction to Linear Algebra, Fifth Edition
Algebra
ISBN:9780980232776
Author:Gilbert Strang
Publisher:Wellesley-Cambridge Press
Text book image
College Algebra (Collegiate Math)
Algebra
ISBN:9780077836344
Author:Julie Miller, Donna Gerken
Publisher:McGraw-Hill Education
Statistics 4.1 Point Estimators; Author: Dr. Jack L. Jackson II;https://www.youtube.com/watch?v=2MrI0J8XCEE;License: Standard YouTube License, CC-BY
Statistics 101: Point Estimators; Author: Brandon Foltz;https://www.youtube.com/watch?v=4v41z3HwLaM;License: Standard YouTube License, CC-BY
Central limit theorem; Author: 365 Data Science;https://www.youtube.com/watch?v=b5xQmk9veZ4;License: Standard YouTube License, CC-BY
Point Estimate Definition & Example; Author: Prof. Essa;https://www.youtube.com/watch?v=OTVwtvQmSn0;License: Standard Youtube License
Point Estimation; Author: Vamsidhar Ambatipudi;https://www.youtube.com/watch?v=flqhlM2bZWc;License: Standard Youtube License