{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

midreviewsol

# midreviewsol - EE 178 Sunday February 7 2009 Probabilistic...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

midreviewsol - EE 178 Sunday February 7 2009 Probabilistic...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online