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Unformatted text preview: EE 178 Sunday, February 7, 2009 Probabilistic Systems Analysis Handout #10 Sample Midterm Problem Solutions 1. a. ( ≥ ). Consider P( A ) = P( A  C )P( C ) + P( A  C c )P( C c ) ≥ P( B  C )P( C ) + P( B  C c )P( C c ) = P( B ) . b. (=). Consider P( ∪ n i =1 A i ) = P(( ∩ n i =1 A c i ) c ) , By DeMorgan = 1 P( ∩ n i =1 A c i ) = 1 n productdisplay i =1 P( A c i ) = 1 n productdisplay i =1 (1 P( A i )) . c. ( ≤ ). This is because integraldisplay b a f X ( x ) dx = F X ( b ) F X ( a ) ≤ F X ( b ) . d. ( ≥ ). Consider P { g ( X ) ∈ { y 1 ,y 2 }} = summationdisplay { x : g ( x )= y 1 } p X ( x ) + summationdisplay { x : g ( x )= y 2 } p X ( x ) ≥ summationdisplay { x : g ( x )= y 1 } p X ( x ) . e. (NONE). To see why, let X ∼ U[0 , 1] and Y ∼ U[ 1 , 1]. Then, E( X ) = 0 . 5 > E( Y ), but E( X 2 ) = 1 / 3 < E( Y 2 ) = 4 / 3. Page 2 of 7 EE 178, Winter 2009 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 . 6 a120 a65 a65 a65 a65 a65 a65 a65 . 8 a120 a0 a0 a0 a0 . 1 a120 a64 a64 a64 a64 . 9 a120 a8 a8 a8 a8 P( E 3  E 2 ) = 0 . 4 a72 a72 a72 a72 . 6 a8 a8 a8 a8 . 1 a72 a72 a72 a72 . 9 a8 a8 a8 a8 . 4 a72 a72 a72 a72 . 6 a8 a8 a8 a8 . 1 a72 a72 a72 a72 . 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 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 × . 4 + 0 . 8 × . 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 . Sample Midterm Solutions Page 3 of 7 3. a. The tree: a120 a1 a1 a1 a1 a1 a1 a1 P( A ) = 0 . 8 a120 a0 a0 a0 a0 P( F 1  A ) = 0 . 1 a120 a64 a64...
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This note was uploaded on 10/26/2011 for the course MATH 180C taught by Professor Eggers during the Spring '09 term at Aarhus Universitet.
 Spring '09
 Eggers
 Systems Analysis

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