HWsolution_2

# HWsolution_2 - Univ Of Maryland at College Park ECE...

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Univ. Of Maryland at College Park, ECE Department ENEE324 Fall 2002 Solution to Assignment #: 2 Posted on: 9/16/02 Problem 1) [GAR] 2.25 a) We had ) ( ) ( ] [ B P A P B A P + U So ) ( ) ( ) ( ) ( ) ( ] ) [( ) ( C P B P A P C P B A P C B A P C B A P + + + = U U U U U Solution 2. using Corollary 7. [ ] ( ) ( ) ( ) ( )( ) ( ) ( ) () Therefor it is enough to prove that ( ) ( ) But so ( ) ( ) Clearly this is true for all three probabiliti PABC P A P B P C PA B PAC P CB P AB C P ABC A B PAB =++ + ---+ ≤++ ⊂≤ U U III II I IIII I I I I es and since these probabilities are non-negative, the statement is true. b) We use induction for 2 = n ) ( ) ( ) ( 2 1 2 1 A P A P A A P + U Let’s assume it is true for k n = i.e. ) ( 1 1 = = k i i k i i A P A P U We will prove it for 1 + = k n or ) ( 1 1 1 1 + = + = k i i k i i A P A P U ( 29 ( 29 ( 29 ( 29 1 1 1 11 1 2 k kk i i k i k i ki b y theassumption ii i o f induction sets PAP A A P A P A P A + + + ++ == =   = + +=     ∑∑ 1442443 U U UU Problem 2: [GAR] 2.31 } 2 / 1 | | ) , {( f y x y x A - = 2 / 1 2 / 1 | | - - - p f y x y x or 2 / 1 f y x - So we draw two lines 2 / 1 & 2 / 1 = - - = - y x y x So 4 / 1 2 / 1 2 / 1 2 / 1 2 / 1 2 / 1 2 / 1 1 ) ( ) ( ) ( 2 1 = × × + × × = + = A Area A Area A P

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Problem 3: [GAR] 2.50 P[ AB C] P[ A ( B C) ] P[A ( B C)] P( B C ) P[A ( B C)] P[B C] P[C] ˙˙ = ˙ ˙= / ˙ ˙ = ˙ Problem 4) Take } 5 , 4 , 3 , 2 , 1 , 0 { = A a) } 5 , , , / ) , , {( = + + = z y x A z A y A x z y x S
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HWsolution_2 - Univ Of Maryland at College Park ECE...

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