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Two fair six-sided dice are tossed independently. Let M =

Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye ISBN: 9780321629111 32

Solution for problem 18E Chapter 3

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Problem 18E

Two fair six-sided dice are tossed independently. Let M = the maximum of the two tosses (so M (1,5) =5, M(3,3,) = 3, etc.). a.? ?What is the pmf of M? [Hint: First determine p(1), then p(2), and so on.] b.? ?Determine the cdf of M and graph it.

Step-by-Step Solution:

Solution : Step 1 : It is given that two fair- dice are tossed independently. And let M be the maximum of two tosses. (M (1,5) =5, M(3,3,) = 3) , We have to find the pmf and cdf of M based on the probability pattern. Step 2 : a) We have to find the probability pmf of M , for that we need to find the probability pattern for different values of M. Since the probability of getting a face in the die is = 1/6 , and the dice are tossed independently, that means outcome of one die is not depend on the other so proposition P(A B) = P(A) * P(B) So P(M=1) = P(1,1), which means probability that the outcome of two fair dice are one. = ()*( ) = 1/36 P(M=2)= P(1,2)+P(2,1)+P(2,2), it is the probability that the maximum outcome is 2 ,which can happen in three ways ( (1st die shows 1 ,second shows 2), (1st shows 2, second shows 1),(both shows two)). = 3*(1/36) = 3/36 Similarly we can find others, P(M=3)=P(1,3)+P(2,3)+P(3,1)+P(3,2)+P(3,3) probability that maximum outcome is 3, which can happen in 5 ways. P(M=3) = 5* 1/36 = 5/36 Similarly, P(M=4)= P( 1,4)+P(4,1)+P(2,4)+P(4,2)+P(3,4)+P(4,3)+P(4,4) P(M=4)= 7*(1/36) = 7/36 P(M=5) =P( 1,5)+P(5,1)+P(2,5)+P(5,2)+P(3,5)+P(5,3)+P(4,5)+P(5,4)+P(5,5) = 9/36 P(M=6) =P( 1,6)+P(6,1)+P(2,6)+P(6,2)+P(3,6)+P(6,3)+P(4,6)+P(6,4)+P(5,6)+P(6,5)+P(6,6) = 11/36 So from the above observations we can find that , there is probability pattern is following, and which is M values P(M) 1 1/36 2 3/36 3 5/36 4 7/36 5 9/36 6 11/36

Step 3 of 3

Chapter 3, Problem 18E is Solved
Textbook: Probability and Statistics for Engineers and the Scientists
Edition: 9
Author: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye
ISBN: 9780321629111

Probability and Statistics for Engineers and the Scientists was written by and is associated to the ISBN: 9780321629111. This textbook survival guide was created for the textbook: Probability and Statistics for Engineers and the Scientists, edition: 9. The answer to “Two fair six-sided dice are tossed independently. Let M = the maximum of the two tosses (so M (1,5) =5, M(3,3,) = 3, etc.). a.? ?What is the pmf of M? [Hint: First determine p(1), then p(2), and so on.] b.? ?Determine the cdf of M and graph it.” is broken down into a number of easy to follow steps, and 49 words. This full solution covers the following key subjects: determine, independently, dice, ETC, fair. This expansive textbook survival guide covers 18 chapters, and 1582 solutions. The full step-by-step solution to problem: 18E from chapter: 3 was answered by , our top Statistics solution expert on 05/06/17, 06:21PM. Since the solution to 18E from 3 chapter was answered, more than 887 students have viewed the full step-by-step answer.

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