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Two dice are rolled. Let E be the event that the sum of

Fundamentals of Probability, with Stochastic Processes | 3rd Edition | ISBN: 9780131453401 | Authors: Saeed Ghahramani ISBN: 9780131453401 223

Solution for problem 5 Chapter 1.2

Fundamentals of Probability, with Stochastic Processes | 3rd Edition

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Fundamentals of Probability, with Stochastic Processes | 3rd Edition | ISBN: 9780131453401 | Authors: Saeed Ghahramani

Fundamentals of Probability, with Stochastic Processes | 3rd Edition

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Problem 5

Two dice are rolled. Let E be the event that the sum of the outcomes is odd and F be the event of at least one 1. Interpret the events EF, EcF, and EcFc.

Step-by-Step Solution:

Q:Two dice are rolled. Let E be the event that the sum of the outcomes is odd and F be the event of at c c cleast one 1. Interpret the events EF, E F, and E F .AnswerStep 1 of 1(a)The sample space of an experiment consists of all possible outcomes of the experiment.We have given two dice are rolled, then all possible outcomes are as follows, (1,1)(1,2)(1,3)(1,4)(1,5)(1,6) (2,1)(2,2)(2,3)(2,4)(2,5)(2,6) (3,1)(3,2)(3,3)(3,4)(3,5)(3,6) (4,1)(4,2)(4,3)(4,4)(4,5)(4,6) (5,1)(5,2)(5,3)(5,4)(5,5)(5,6) (6,1)(6,2)(6,3)(6,4)(6,5)(6,6)Let E be the event that the sum of the outcomes is odd.Therefore, E = (1,2)(1,4)(1,6) = (2,1)(2,3)(2,5) = (3,2)(3,4)(3,6)\n = (4,1)(4,3)(4,5) = (5,2)(5,4)(5,6) = (6,1)(6,3)(6,5)Let F be the event of at least one.Hence, F = (1,1)(1,2)(1,3)(1,4)(1,5)(1,6) = (2,1)(3,1)(4,1)(5,1)(6,1)We need to findEF or the intersection of both events or common outcomes from both theevents,Hence, EF = E F = common outcomes EF = E F = (1, 2) Hence EF is(1, 2). cWe need to findE F. cE = complimentary event of E = the outcomes which is not there in the sample space of E\n c E = sum of the outcomes is even E = (1,1)(1,3)(1,5) = (2,2)(2,4)(2,6) = (3,1)(3,3)(3,5) = (4,2)(4,4)(4,6) = (5,1)(5,3)(5,5) = (6,2)(6,4)(6,6) F = (1,1)(1,2)(1,3)(1,4)(1,5)(1,6) = (2,1)(3,1)(4,1)(5,1)(6,1) Hence, c c E F = E F = common outcomes c c E F = E F = (1, 1) c Hence E F is (1, 1). c cWe need to find E F . c cF = complimentary event of F = the outcomes which is not there in the sample space of F F c = (2,2)(2,3)(2,4)(2,5)(2,6)\n = (3,2)(3,3)(3,4)(3,5)(3,6) = (4,2)(4,3)(4,4)(4,5)(4,6) = (5,2)(5,3)(5,4)(5,5)(5,6) = (6,2)(6,3)(6,4)(6,5)(6,6) c E = (1,1)(1,3)(1,5) = (2,2)(2,4)(2,6) = (3,1)(3,3)(3,5) = (4,2)(4,4)(4,6) = (5,1)(5,3)(5,5) = (6,2)(6,4)(6,6)Hence, E F = E F = common outcomes c c c c E F = E F = (2,2)(2,4)(2,6)(4,2)(4,4)(4,6)(6,2)(6,4)(6,6) = (3,3)(3,5)(5,3)(5,5)

Step 2 of 1

Chapter 1.2, Problem 5 is Solved
Textbook: Fundamentals of Probability, with Stochastic Processes
Edition: 3
Author: Saeed Ghahramani
ISBN: 9780131453401

Fundamentals of Probability, with Stochastic Processes was written by and is associated to the ISBN: 9780131453401. This textbook survival guide was created for the textbook: Fundamentals of Probability, with Stochastic Processes, edition: 3. The answer to “Two dice are rolled. Let E be the event that the sum of the outcomes is odd and F be the event of at least one 1. Interpret the events EF, EcF, and EcFc.” is broken down into a number of easy to follow steps, and 34 words. This full solution covers the following key subjects: . This expansive textbook survival guide covers 59 chapters, and 1142 solutions. The full step-by-step solution to problem: 5 from chapter: 1.2 was answered by , our top Statistics solution expert on 01/05/18, 06:24PM. Since the solution to 5 from 1.2 chapter was answered, more than 403 students have viewed the full step-by-step answer.

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