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SampleMTsol - EE 178 Probabilistic Systems Analysis Sunday...

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EE 178 Sunday, February 10, 2008 Probabilistic Systems Analysis Handout #10 Sample Midterm Problem Solutions 1. a. (=) since P( A B ) = P( A ) + P( B ) - P( A B ) = P( A ) + P( B ) - P( A )P( B ) = P( A ) + P( B )(1 - P( A )) = P( A ) + P( B )P( A c ) , where the second line follows from the independence of A and B . b. ( ) By DeMorgan’s law, ( A B ) c = A c B c . Thus P (( A B ) c ) = P( A c B c ) P( A c ) + P( B c ) . c. ( ) Since p X ( x ) p X ( x ) p Y | X ( y | x ) = p X,Y ( x, y ). d. ( ) Note that p Z ( g ( y )) = summationdisplay y : g ( y )= g ( y ) p Y ( y ) p Y ( y ) . Thus p Y ( y ) p Z ( g ( y )). e. ( ) This follows from the fact that Var( X ) = E ( X 2 ) - ( E ( X )) 2 E ( X 2 ) . 2. a. The tree: a120 a1 a1 a1 a1 a1 a1 a1 P( E 1 ) = 0 . 2 a120 a0 a0 a0 a0 P( E 2 | E 1 ) = 0 . 4 a120 a64 a64 a64 a64 0 . 6 a120 a65 a65 a65 a65 a65 a65 a65 0 . 8 a120 a0 a0 a0 a0 0 . 1 a120 a64 a64 a64 a64 0 . 9 a120 a8 a8 a8 a8 P( E 3 | E 2 ) = 0 . 4 a72 a72 a72 a72 0 . 6 a8 a8 a8 a8 0 . 1 a72 a72 a72 a72 0 . 9 a8 a8 a8 a8 0 . 4 a72 a72 a72 a72 0 . 6 a8 a8 a8 a8 0 . 1 a72 a72 a72 a72 0 . 9 a120 E 1 ,E 2 ,E 3 a120 E 1 ,E 2 ,E c 3 a120 E 1 ,E c 2 ,E 3 a120 E 1 ,E c 2 ,E c 3 a120 E c 1 ,E 2 ,E 3 a120 E c 1 ,E 2 ,E c 3 a120 E c 1 ,E c 2 ,E 3 a120 E c 1 ,E c 2 ,E c 3
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Page 2 of 6 EE 178, Winter 2008 b. The probability of the second bit being in error is: P( E 2 ) = P( E 1 )P( E 2 | E 1 ) + P( E c 1 )P( E 2 | E c 1 ) = 0 . 2 × 0 . 4 + 0 . 8 × 0 . 1 = 0 . 16 . c. We use Bayes rule P( E 2 | E 3 ) = P( E 3 | E 2 )P( E 2 ) P( E 2 )P( E 3 | E 2 ) + P( E c 2 )P( E 3 | E c 2 ) = 16 / 37 . 3. The a priori probabilities are P( d i ) = 1 / 3 , i = 1 , 2 , 3 .
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