hw3 - MS&E 120 Probabilistic Analysis Fall 2007-2008 Prof...

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MS&E 120: Probabilistic Analysis Fall 2007-2008, Prof. Shachter HOMEWORK #3 SOLUTIONS CHAPTER #4 PROBLEMS 32) P(11 tests are made) = P(at least one person has the disease) P(1 test is made) = P(all 10 people do not have the disease) E(# of tests) = 11 x P(11 tests are made) + 1 x P(1 test is made) = ) 9 . 0 ( 1 ) 9 . 0 1 ( 11 10 10 ! " # ! = =7.513 10 9 . 0 10 11 ! # 38) Now Var(X) = E[X ] – (E[X]) =5 2 2 Thus E[X ] = 5 + 1 = 6 2 We use the properties of linearity of expectation to obtain the following results. E[(2+X) ] = E[ 4 + 4X +X ] 2 2 = E[4] + E[4X] + E[X ] 2 = 4 + 4E[X] + E[X ] 2 = 4 + 4 (1) + 6 = 14 Var(4+3X) = E[(4+3X) ] – (E[4+3X]) 2 2 = E [16 + 24X + 9X 2 ] – 7 2 = 16 + 24 + 54 – 49 = 45 or Var(4+3X)=Var(3X)=9Var(X)=45 43) For the message to be received incorrectly, at least 3 of the digits have to be transposed. We assume that the probability of transposing a digit is independent of the other digits in the sequence. P(message received incorrectly) = P(3 digits transposed) + P(4 digits transposed) + P(5 digits transposed) = 5 1 4 2 3 2 . 0 8 . 0 2 . 0 4 5 8 . 0 2 . 0 3 5 " $ $ % & ( ) " $ $ % & ( ) = 0.0579
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48) P(more than one diskette in a package is defective) = * + , - . / " ! ! $ $ % & ( ) # 10 9 99 . 0 99 . 0 01 . 0 1 10 1 = 0.00426 P(exactly one package is returned) = 2 99574 . 0 00426 . 0 1 3 ! ! $ $ % & ( ) = 0.0127 52) We assume that the number of airplane crashes monthly is Poisson distributed with parameter 3.5 Let x be the number of accidents in the next month. a) P(x 2) = 1- P(x 1) 0 1 = 1- ( P(x = 0) + P(x=1) ) = 1- [ ! 0 5 . 3 0 5 . 3 ! # e + ! 1 5 . 3 1 5 . 3 ! # e ] = 0.864 b) P( x 1) = [ 1 ! 0 5 . 3 0 5 . 3 ! # e + ! 1 5 . 3 1 5 . 3 ! # e ] = 0.136 58) a) Difference between Poisson and Bin = 6 2 2 8 . 0 9 . 0 1 . 0 2 8 ! 2 8 . 0 $ $ % & ( ) # ! # e = 0.14378 – 0.1488 = -0.005 b) Difference between Poisson and Bin = 1 9 9 5 . 9 05 . 0 95 . 0
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