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hw4sol - c = 09 091 819 1 09 81 09 091 819 3 P The...

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HOMEWORK #4 SOLUTIONS 1. 2. a) b) 3. = = = 7134 . 0 2866 . 0 2388 . 0 7612 . 0 7864 . 0 2136 . 0 178 . 0 822 . 0 88 . 0 12 . 0 1 . 0 9 . 0 3 2 P P P a) P(x 1 =A) = (0.65)*(0.9)+(0.35)*(0.12) = 0.627 b) P(x 3 =A) = (0.65)*(0.7612)+(0.35)*(0.2866) = 0.595 c) c 545 . 0 1 88 . 0 1 . 0 12 . 0 9 . 0 = + = + = + = A C A C A C C A A π π π π π π π π π d d) e 081 . 0 88 . 0 * 09 . 0 7864 . 0 * 1 . 0 2388 . 0 11 ) 2 ( 01 ) 2 ( 11 ) 1 ( 01 ) 3 ( 01 ) 3 ( 01 = - - = - - = p f p f p f f 4. a) let 0=operational, 1=down, 2=repaired g h then 0 1 . 0 9 . 0 1 0 0 0 1 . 0 9 . 0 = P , we also have + = j k kj ik ij p μ μ 1 i π π π π π π π π π π π π π π π π π π 0 0 1 2 1 0 1 2 2 0 1 2 0 1 2 0 1 2 0 4 0 2 08 03 0 7 0 0 03 01 0 2 1 0 4 0 4 0 2 = + + = + + = + + = + + = = = . . . . . . . . . . , . , . μ μ μ μ 20 20 10 20 1 0 2 0 0 125 = + + = . . . P f ( . . Going to state 0 after at least two steps) = - = - = 1 1 08 0 2 20 1
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j So, = = = + = + = + = 9 / 11 9 / 20 9 / 11 such that 1 . 0 1 1 1 . 0 1 20 10 00 10 20 20 10 10 00 μ μ μ μ μ μ μ μ μ = = = + = + = + = 10 11 10 such that 9 . 0 1 1 9 . 0 1 21 11 01 01 21 21 11 01 01 μ μ μ μ μ μ μ μ μ = = = + + = + = + + = 11 1 11 such that 1 . 0 9 . 0 1 0 1 1 . 0 9 . 0 1 20 10 00 12 02 22 12 12 02 02 μ μ μ μ μ μ μ μ μ μ The expected number of days that elevator will remain operational is
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Unformatted text preview: c) = 09 . 091 . 819 . 1 . 09 . 81 . 09 . 091 . 819 . 3 P The probability that the elevator will be operational at the beginning of the third day given that the elevator is currently operational = ) 3 ( 00 p = 0.819. d) 081 . * 09 . 1 . * 1 . 091 . 11 ) 2 ( 01 ) 2 ( 11 ) 1 ( 01 ) 3 ( 01 ) 3 ( 01 =--=--= p f p f p f...
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