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mont4e_sm_ch03_sec07 - 3-79 The probability is 0.651 that...

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Section 3-7 3-81 . a) 5 . 0 5 . 0 ) 5 . 0 1 ( ) 1 ( 0 = = = X P b) 0625 . 0 5 . 0 5 . 0 ) 5 . 0 1 ( ) 4 ( 4 3 = = = = X P c) 0039 . 0 5 . 0 5 . 0 ) 5 . 0 1 ( ) 8 ( 8 7 = = = = X P d) 5 . 0 ) 5 . 0 1 ( 5 . 0 ) 5 . 0 1 ( ) 2 ( ) 1 ( ) 2 ( 1 0 + = = + = = X P X P X P 75 . 0 5 . 0 5 . 0 2 = + = e.) 25 . 0 75 . 0 1 ) 2 ( 1 ) 2 ( = = = > X P X P 3-82. E(X) = 2.5 = 1/p giving p = 0.4 a) 4 . 0 4 . 0 ) 4 . 0 1 ( ) 1 ( 0 = = = X P b) 0864 . 0 4 . 0 ) 4 . 0 1 ( ) 4 ( 3 = = = X P c) 05184 . 0 5 . 0 ) 5 . 0 1 ( ) 5 ( 4 = = = X P d) ) 3 ( ) 2 ( ) 1 ( ) 3 ( = + = + = = X P X P X P X P 7840 . 0 4 . 0 ) 4 . 0 1 ( 4 . 0 ) 4 . 0 1 ( 4 . 0 ) 4 . 0 1 ( 2 1 0 = + + = e) 2160 . 0 7840 . 0 1 ) 3 ( 1 ) 3 ( = = = > X P X P 3-83. Let X denote the number of trials to obtain the first success. a) E(X) = 1/0.2 = 5 b) Because of the lack of memory property, the expected value is still 5. 3-84. a) E(X) = 4/0.2 = 20 b) P(X=20) = 0436 . 0 2 . 0 ) 80 . 0 ( 3 19 4 16 = 3-15
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c) P(X=19) = 0459 . 0 2 . 0 ) 80 . 0 ( 3 18 4 15 = d) P(X=21) = 0411 . 0 2 . 0 ) 80 . 0 ( 3 20 4 17 = e) The most likely value for X should be near μ X . By trying several cases, the most likely value is x = 19. 3-85. Let X denote the number of trials to obtain the first successful alignment. Then X is a geometric random variable with p = 0.8 a) 0064 . 0 8 . 0 2 . 0 8 . 0 ) 8 . 0 1 ( ) 4 ( 3 3 = = = = X P b) ) 4 ( ) 3 ( ) 2 ( ) 1 ( ) 4 ( = + = + = + = = X P X P X P X P X P 8 . 0 ) 8 . 0 1 ( 8 . 0 ) 8 . 0 1 ( 8 . 0 ) 8 . 0 1 ( 8 . 0 ) 8 . 0 1 ( 3 2 1 0 + + + = 9984 . 0 8 . 0 2 . 0 ) 8 . 0 ( 2 . 0 ) 8 . 0 ( 2 . 0 8 . 0 3 2 = + + + = c) )] 3 ( ) 2 ( ) 1 ( [ 1 ) 3 ( 1 ) 4 ( = + = + = = = X P X P X P X P X P ] 8 . 0 ) 8 . 0 1 ( 8 . 0 ) 8 . 0 1 ( 8 . 0 ) 8 . 0
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