Hw1sol - EE 178 Handout #3 Probabilistic Systems Analysis...

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Unformatted text preview: EE 178 Handout #3 Probabilistic Systems Analysis January 22, 2008 Homework Solutions #1 1. (15 points) The Venn diagrams given in Fig. 1,2,3 and 4, for parts (a),(b),(d) and (e), respectively, verify the relations. For part (c), one can express P( F G ) as P( F- G ) + P( G- F ) + P ( F G ). G F ; F G F F G F G F G Figure 1: The Venn diagram for Problem 1-(a). G F 4 G F F F G F G F G Figure 2: The Venn diagram for Problem 1-(b). G F ; G F 4 G F ' F G F G F G Figure 3: The Venn diagram for Problem 1-(d). G F ' H G ' H F ' F G H F G H F G H Figure 4: The Venn diagram for Problem 1-(e). 2. (20 points) a. Consider P( { ( k, m ) : k m } ) = summationdisplaysummationdisplay { ( k,m ): k m } p ( k, m ) = summationdisplay m =1 summationdisplay k = m p 2 (1- p ) k + m- 2 = summationdisplay m =1 summationdisplay k =0 p 2 (1- p ) k +2 m- 2 = p 2 (1- p ) 2 summationdisplay m =1 (1- p ) 2 m summationdisplay k =0 (1- p ) k = p 2 (1- p ) 2 summationdisplay m =1 (1- p ) 2 m 1 1- (1- p ) = p 1 1- (1- p ) 2 = 1 2- p . b. Consider P( { ( k, m ) : k + m = r } ) = summationdisplaysummationdisplay { ( k,m ): k + m = r } p ( k, m ) = r- 1 summationdisplay m =1 p 2 (1- p ) m +( r- m )- 2 = r- 1 summationdisplay m =1 p 2 (1- p ) r- 2 = p 2 (1- p ) r- 2 r- 1 summationdisplay m =1 1 = p 2 (1- p ) r- 2 ( r- 1) . c. Consider P( { ( k, m ) : k odd } ) = summationdisplay k =1 summationdisplay m =1 p 2 (1- p ) (2 k- 1)+ m- 2 = summationdisplay k =0 p (1- p ) 2 k summationdisplay m =0 p (1- p ) m = p 1- (1- p ) 2 = 1 2- p . 3. (15 points) a. Lets check if this P( F ) satisfies the three axioms of probability. It is obviously non- negative just by the definition. Next, P() = 1 because 0 and 1...
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This note was uploaded on 09/29/2008 for the course EE 178 taught by Professor Hewlett during the Spring '08 term at Stanford.

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Hw1sol - EE 178 Handout #3 Probabilistic Systems Analysis...

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