Explanation Given Information: The events Y 1 , Y 2 and Y 3 form a partition of S such that P ( X | Y 1 ) = 0.2 , P ( X | Y 2 ) = 0.3 , P ( X | Y 3 ) = 0.6 , P ( Y 1 ) = 0.3 , P ( Y 2 ) = 0.4 . Formula used: Expanded form of Bayes’ theorem: For three events A 1 , A 2 and A 3 which from a partition of the sample space S, Bayes’ theorem is defined as; P ( A 1 | T ) = P ( T | A 1 ) P ( A 1 ) P ( T | A 1 ) P ( A 1 ) + P ( T | A 2 ) P ( A 2 ) + P ( T | A 3 ) P ( A 3 ) Also, P ( A 1 ) + P ( A 2 ) + P ( A 3 ) = 1 Calculation: Consider the events Y 1 , Y 2 and Y 3 form a partition of S such that P ( X | Y 1 ) = 0.2 , P ( X | Y 2 ) = 0.3 , P ( X | Y 3 ) = 0.6 , P ( Y 1 ) = 0.3 , P ( Y 2 ) = 0.4 . Since the events Y 1 , Y 2 and Y 3 form a partition of S , so, P ( Y 3 ) can be obtained as; P ( Y 3 ) = 1 − P ( Y 1 ) − P ( Y 2 ) = 1 − 0.3 − 0.4 = 1 − 0.7 = 0

Finite Mathematics and Applied Calculus (MindTap Course List)

7th Edition

ISBN: 9781337274203

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Chapter 8.6, Problem 8E

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To calculate: The probability P(Y2|X) if Y1,Y2 and Y3 form a partition of S such that P(X|Y1)=0.2, P(X|Y2)=0.3, P(X|Y3)=0.6, P(Y1)=0.3, P(Y2)=0.4.

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Chapter 8.6, Problem 7E

Chapter 8.6, Problem 9E

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Chapter 8 Solutions

Finite Mathematics and Applied Calculus (MindTap Course List)

Ch. 8.1 - In Exercises 118, describe the sample space S of...Ch. 8.1 - Prob. 2E Ch. 8.1 - In Exercises 118, describe the sample space S of...Ch. 8.1 - Prob. 4E Ch. 8.1 - In Exercises 118, describe the sample space S of...Ch. 8.1 - Prob. 6E Ch. 8.1 - In Exercises 118, describe the sample space S of...Ch. 8.1 - Prob. 8E Ch. 8.1 - In Exercises 118, describe the sample space S of...Ch. 8.1 - Prob. 10E

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The probability P ( Y 2 | X ) if Y 1 , Y 2 and Y 3 form a partition of S such that P ( X | Y 1 ) = 0.2 , P ( X | Y 2 ) = 0.3 , P ( X | Y 3 ) = 0.6 , P ( Y 1 ) = 0.3 , P ( Y 2 ) = 0.4 .

Solution Summary: The author calculates the probability P(Y_2|X) for a partition of S.

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To calculate: The probability P(Y2|X) if Y1,Y2 and Y3 form a partition of S such that P(X|Y1)=0.2, P(X|Y2)=0.3, P(X|Y3)=0.6, P(Y1)=0.3, P(Y2)=0.4.

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Chapter 8 Solutions