Homework#5_Solutions

# Homework#5_Solutions - ENGINEERING PROBABILITY Spring 2012...

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1 ENGINEERING PROBABILITY Spring 2012 Homework #5 _ Solutions Topic Assignment – Due Next Class Independence and Bayes Rule 2-136, 137, 142, 144, 147, 148 2 136. Let A denote the upper devices function. Let B denote the lower devices function. P(A) = (0.9)(0.8)(0.7) = 0.504 P(B) = (0.95)(0.95)(0.95) = 0.8574 P(A B) = (0.504)(0.8574) = 0.4321 Therefore, the probability that the circuit operates = P(A B) = P(A) +P(B) P(A B) = 0.9293 2 137. P = [1 – (0.1)(0.05)][1 – (0.1)(0.05)][1 – (0.2)(0.1)] = 0.9702 2 142. Because, P( AB ) P(B) = P( ) = P( BA ) P(A), 28 . 0 5 . 0 ) 2 . 0 ( 7 . 0 ) ( ) ( ) ( ) ( A P B P B A P A B P 2 144. Let F denote a fraudulent user and let T denote a user that originates calls from two or more metropolitan areas in a day. Then, 003 . 0 ) 9999 (. 01 . 0 ) 0001 . 0 ( 30 . 0 ) 0001 . 0 ( 30 . 0 ) ' ( ) ' ( ) ( ) ( ) ( ) ( ) ( F P F T

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• Spring '12
• SUSANALBIN
• 3rd millennium, 22nd century, Engineering Probability Spring, devices function, upper devices function, Solutions Topic Independence

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Homework#5_Solutions - ENGINEERING PROBABILITY Spring 2012...

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