Statistics Through Applications
Statistics Through Applications
2nd Edition
ISBN: 9781429219747
Author: Daren S. Starnes, David Moore, Dan Yates
Publisher: Macmillan Higher Education
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Chapter 7, Problem 7.79RE

(a)

To determine

To construct: a diagram the sample space illustrating the possible outcomes on a single turn.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Rolling a fair eight sided dice by teacher and the student rolling a fair six-sided dice

Winners would be the person with highest numbers and if tie is there than no winner

Calculation:

Suppose

T = Teacher wins round

S = Student wins round

N = No winner

8-sided dice has possible outcomes = 1, 2, 3, 4, 5, 6, 7, 8

6-sided dice has possible outcomes = 1, 2, 3, 4, 5, 6

    Teacher
    12345678
    Student1NTTTTTTT
    2SNTTTTTT
    3SSNTTTTT
    4SSSNTTTT
    5SSSSNTTT
    6SSSSSNTT

(b)

To determine

To Calculate: the P(A) .

(b)

Expert Solution
Check Mark

Answer to Problem 7.79RE

  P(A)=0.5625

Explanation of Solution

Formula used:

  Probability=Favourable CasesTotalCases

Calculation:

Assuming that A= teacher wins the first turn of the game.

By seeing the diagram, it is observed that there are 48 possible cases, while 27 of the 48 cases result in the teach T wining

  Probability=Favourable CasesTotalCasesP(A)=2748=0.5625

(c)

To determine

To Calculate: the P(AB)

(c)

Expert Solution
Check Mark

Answer to Problem 7.79RE

  P(AB)=0.625

Explanation of Solution

Formula used:

  Probability=Favourable CasesTotalCases

Calculation:

By seeing the diagram, it is observed that there are 48 possible cases, while 8 of the 48 cases result in the 3 for the student.

  Probability=Favourable CasesTotalCasesP(B)=848=0.1667

  P(AB)=548

General addition rule

  P(AB)=P(B)+P(B)P(AB)=2748+848548=58=0.625

(d)

To determine

To Explain: that the events A and B are independent, justify answer.

(d)

Expert Solution
Check Mark

Answer to Problem 7.79RE

Not Independent

Explanation of Solution

Formula used:

Conditional Probability

  P(A|B)=P(BA)P(B)

Calculation:

  P(B|A)=P(AB)P(A)=5482748=527=0.1852

It is observed that the P(B|A)P(B) it means the event A and B are not independent which is looking forth.

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Chapter 7 Solutions

Statistics Through Applications

Ch. 7.1 - Prob. 7.11ECh. 7.1 - Prob. 7.12ECh. 7.1 - Prob. 7.13ECh. 7.1 - Prob. 7.14ECh. 7.1 - Prob. 7.15ECh. 7.1 - Prob. 7.16ECh. 7.1 - Prob. 7.17ECh. 7.1 - Prob. 7.18ECh. 7.1 - Prob. 7.19ECh. 7.1 - Prob. 7.20ECh. 7.1 - Prob. 7.21ECh. 7.1 - Prob. 7.22ECh. 7.1 - Prob. 7.23ECh. 7.1 - Prob. 7.24ECh. 7.1 - Prob. 7.25ECh. 7.1 - Prob. 7.26ECh. 7.2 - Prob. 7.27ECh. 7.2 - Prob. 7.28ECh. 7.2 - Prob. 7.29ECh. 7.2 - Prob. 7.30ECh. 7.2 - Prob. 7.31ECh. 7.2 - Prob. 7.32ECh. 7.2 - Prob. 7.33ECh. 7.2 - Prob. 7.34ECh. 7.2 - Prob. 7.35ECh. 7.2 - Prob. 7.36ECh. 7.2 - Prob. 7.37ECh. 7.2 - Prob. 7.38ECh. 7.2 - Prob. 7.39ECh. 7.2 - Prob. 7.40ECh. 7.2 - Prob. 7.41ECh. 7.2 - Prob. 7.42ECh. 7.2 - Prob. 7.43ECh. 7.2 - Prob. 7.44ECh. 7.2 - Prob. 7.45ECh. 7.2 - Prob. 7.46ECh. 7.2 - Prob. 7.47ECh. 7.2 - Prob. 7.48ECh. 7.2 - Prob. 7.49ECh. 7.2 - Prob. 7.50ECh. 7.2 - Prob. 7.51ECh. 7.2 - Prob. 7.52ECh. 7.3 - Prob. 7.53ECh. 7.3 - Prob. 7.54ECh. 7.3 - Prob. 7.55ECh. 7.3 - Prob. 7.56ECh. 7.3 - Prob. 7.57ECh. 7.3 - Prob. 7.58ECh. 7.3 - Prob. 7.59ECh. 7.3 - Prob. 7.60ECh. 7.3 - Prob. 7.61ECh. 7.3 - Prob. 7.62ECh. 7.3 - Prob. 7.63ECh. 7.3 - Prob. 7.64ECh. 7.3 - Prob. 7.65ECh. 7.3 - Prob. 7.66ECh. 7.3 - Prob. 7.67ECh. 7.3 - Prob. 7.68ECh. 7.3 - Prob. 7.69ECh. 7.3 - Prob. 7.70ECh. 7.3 - Prob. 7.71ECh. 7.3 - Prob. 7.72ECh. 7.3 - Prob. 7.73ECh. 7.3 - Prob. 7.74ECh. 7.3 - Prob. 7.75ECh. 7.3 - Prob. 7.76ECh. 7.3 - Prob. 7.77ECh. 7.3 - Prob. 7.78ECh. 7 - Prob. 7.79RECh. 7 - Prob. 7.80RECh. 7 - Prob. 7.81RECh. 7 - Prob. 7.82RECh. 7 - Prob. 7.83RECh. 7 - Prob. 7.84RECh. 7 - Prob. 7.85RECh. 7 - Prob. 7.86RECh. 7 - Prob. 7.87RECh. 7 - Prob. 7.88RE
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