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humber of spots showing is five. Follow that method to give a probability model for the total number of spots. The possible outcomes are 2, 3, 4, ..., 12. Then use the probabilities to find the expected value of the total.

humber of spots showing is five. Follow that method to give a probability model for the total number of spots. The possible outcomes are 2, 3, 4, ..., 12. Then use the probabilities to find the expected value of the total.

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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20.16 Rolling two dice. Example 2 of Chapter 18 (page 428) gives a probability model for rolling two casino dice and recording the number of spots on each of the two up-faces. That example also shows how to find the probability that the total number of spots showing is five. Follow that method to give a probability model for the total number of spots. The possible outcomes are 2, 3, 4, ... , 12. Then use the probabilities to find the expected value of the total.
Transcribed Image Text:20.16 Rolling two dice. Example 2 of Chapter 18 (page 428) gives a probability model for rolling two casino dice and recording the number of spots on each of the two up-faces. That example also shows how to find the probability that the total number of spots showing is five. Follow that method to give a probability model for the total number of spots. The possible outcomes are 2, 3, 4, ... , 12. Then use the probabilities to find the expected value of the total.
Arthur-studio10/Shutterstock Rolling two dice is a common way to lose money in casinos. There are 36 possible outcomes when we roll two dice and record the up-faces in order (first die, second die). Figure 18.1 displays these outcomes. What probabilities should we assign? 国国图国 国口 国回 围图 国图 围 Moore/Notz, Statistics: Concepts and Controversies, 10e, © 2020 W. H. Freeman and Company Figure 18.1 The 36 possible outcomes from rolling two dice, Example 2. The outcomes are as follows: (1, 1); (2, 1); (3, 1); (4, 1); (5, 1); (6, 1); (1, 2); (1, 2); (3, 2); (4, 2); (5, 2); (6, 2); (1, 3); (2, 3); (3, 3); (4, 3); (5, 3); (6, 3); (1, 4); (2, 4); (3, 4); (4, 4); (5, 4); (6, 4); (1, 5); (2, 5); (3, 5); (6, 5); (1, 6); (2, 6); (3, 6); (4, 6); (5, 6); and (6, 6). Casino dice are carefully made. Their spots are not hollowed out, which would give the faces different weights, but are filled with white plastic of the same density as the red plastic of the body. For casino dice, it is reasonable to assign the same probability to each of the 36 outcomes in Figure 18.1. Because these 36 probabilities must have sum 1 (Rule B), each outcome must have probability 1/36, or 1-in-36. We are interested in the sum of the spots on the up-faces of the dice. What is the probability that this sum is 5? The event "roll a 5" contains four outcomes, and its probability is the sum of the probabilities of these outcomes: P(roll a 5) +P +P +P 1 1 1 36 36 36 36 4 = 0.111 36 :: :: • 1. 1.:1::::
Transcribed Image Text:Arthur-studio10/Shutterstock Rolling two dice is a common way to lose money in casinos. There are 36 possible outcomes when we roll two dice and record the up-faces in order (first die, second die). Figure 18.1 displays these outcomes. What probabilities should we assign? 国国图国 国口 国回 围图 国图 围 Moore/Notz, Statistics: Concepts and Controversies, 10e, © 2020 W. H. Freeman and Company Figure 18.1 The 36 possible outcomes from rolling two dice, Example 2. The outcomes are as follows: (1, 1); (2, 1); (3, 1); (4, 1); (5, 1); (6, 1); (1, 2); (1, 2); (3, 2); (4, 2); (5, 2); (6, 2); (1, 3); (2, 3); (3, 3); (4, 3); (5, 3); (6, 3); (1, 4); (2, 4); (3, 4); (4, 4); (5, 4); (6, 4); (1, 5); (2, 5); (3, 5); (6, 5); (1, 6); (2, 6); (3, 6); (4, 6); (5, 6); and (6, 6). Casino dice are carefully made. Their spots are not hollowed out, which would give the faces different weights, but are filled with white plastic of the same density as the red plastic of the body. For casino dice, it is reasonable to assign the same probability to each of the 36 outcomes in Figure 18.1. Because these 36 probabilities must have sum 1 (Rule B), each outcome must have probability 1/36, or 1-in-36. We are interested in the sum of the spots on the up-faces of the dice. What is the probability that this sum is 5? The event "roll a 5" contains four outcomes, and its probability is the sum of the probabilities of these outcomes: P(roll a 5) +P +P +P 1 1 1 36 36 36 36 4 = 0.111 36 :: :: • 1. 1.:1::::
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