mont4e_sm_ch02_sec07

mont4e_sm_ch02_sec07 - 2-112. 10 6 = 10 −10 16 10 1 (b) P...

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Unformatted text preview: 2-112. 10 6 = 10 −10 16 10 1 (b) P = 0.25 × ( ) = 0.020833 12 (a) P = Let A denote the event that a sample is produced in cavity one of the mold. 1 a) By independence, P( A1 ∩ A 2 ∩ A 3 ∩ A 4 ∩ A 5 ) = ( )5 = 0.00003 8 b) Let Bi be the event that all five samples are produced in cavity i. Because the B's are mutually exclusive, P(B1 ∪ B 2 ∪...∪B 8 ) = P(B1) + P(B 2 )+...+P(B 8 ) 2-113. 2-114. 1 1 From part a., P(Bi ) = ( )5 . Therefore, the answer is 8( )5 = 0.00024 8 8 147 ' c) By independence, P( A 1 ∩ A 2 ∩ A 3 ∩ A 4 ∩ A 5 ) = ( ) ( ) . The number of sequences in 8 8 1 7 which four out of five samples are from cavity one is 5. Therefore, the answer is 5( ) 4 ( ) = 0.00107 . 8 8 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 [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 P( A | B) = 2-121. (0.38)(0.53) P( A ∩ B) P( A) P( B | A) = = = 39.3821% P( B) P( A) P( B | A) + P ( A' ) P( B | A' ) (0.38)(0.53) + (0.62)(0.5) Let G denote a product that received a good review. Let H, M, and P denote products that were high, moderate, and poor performers, respectively. a) P(G) = P(G H ) P( H) + P(G M) P( M) + P(G P) P( P) = 0. 95( 0. 40) + 0. 60( 0. 35) + 0. 10( 0. 25) = 0. 615 b) Using the result from part a., P ( G H ) P ( H ) 0. 95( 0. 40) P( H G ) = = = 0. 618 P(G) 0. 615 c) P ( H G ' ) = P ( G ' H ) P ( H ) = 0. 05( 0. 40) = 0. 052 P(G ' ) 1 − 0. 615 2-122. a) P(D)=P(D|G)P(G)+P(D|G’)P(G’)=(.005)(.991)+(.99)(.009)=0.013865 b) P(G|D’)=P(G∩D’)/P(D’)=P(D’|G)P(G)/P(D’)=(.995)(.991)/(1-.013865)=0.9999 2-123.a) P(S) = 0.997(0.60) + 0.9995(0.27) + 0.897(0.13) = 0.9847 b) P(Ch|S) =(0.13)( 0.897)/0.9847 = 0.1184 Section 2-8 2-124. Continuous: a, c, d, f, h, i; Discrete: b, e, and g Supplemental Exercises 2-125. Let B denote the event that a glass breaks. Let L denote the event that large packaging is used. P(B)= P(B|L)P(L) + P(B|L')P(L') = 0.01(0.60) + 0.02(0.40) = 0.014 2-126. Let "d" denote a defective calculator and let "a" denote an acceptable calculator a a) S = {ddd , add , dda, ada, dad , aad , daa, aaa} b) A = {ddd , dda, dad , daa} c) B = {ddd , dda, add , ada} d) A ∩ B = {ddd , dda} e) B ∪ C = {ddd , dda , add , ada, dad , aad } 2-21 ...
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