mont4e_sm_ch02_sec06

9293 1 010051 010051 0201 09702 2 115 section

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Unformatted text preview: note 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 [1-(0.1)(0.05)][1-(0.1)(0.05)][1-(0.2)(0.1)] = 0.9702 2-115. Section 2-7 2-116. Because, P( A B ) P(B) = P( A ∩ B ) = P( B A ) P(A), P( B A) = P( A B) P( B) P( A) P( A B) P( B) P( A) = 0.7(0.2) = 0.28 0.5 P( A B) P( B ) P ( A B) P( B) + P( A B ' ) P( B ' ) P( B A) = 2-117. = = 2-118. 0.4 × 0.8 = 0.89 0.4 × 0.8 + 0.2 × 0.2 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, P(T F ) P( F ) 0.30(0.0001) = = 0.003 P( F T ) = P(T F ) P( F ) + P(T F ' ) P( F ' ) 0.30(0.0001) + 0.01(.9999) 2-119. (a) P=(0.31)(0.978)+(0.27)(0.981)+(0.21)(0.965)+(0.13)(0.992)+(0.08)(0.959) = 0.97638 (b) P = (0.21)(0.965) = 0.207552 0.97638 2-120. Let A denote the event that a respondent is a college graduate and let B denote the event that a voter votes for Bush. 2-20...
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