part10

part10 - Univ. Of Maryland at College Park, ECE Department...

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Univ. Of Maryland at College Park, ECE Department ENEE324 Fall 2002 Solution to Assignment #: 4 Posted on: 10/01/02 a). Binomial probability law with parameters n , p : ( 29 ( 29 1 nk K n n P k pp k -  =-   b). n and ap ( 29 ( 29 1 1 1 1 ()1 k n n P k a p ap k - c). n and (1-a)p ( 29 ( 29 2 2 2 2 [( 1 )]1 ( 1) k n n P k a p ap k - -- c). Multinomial probability law with parameters n and p 1 =p, p 2 =ap, and p 3 =(1-a)p: ( 29 1 2 12 1 2 1 2 1 1 , , ( ) [( 1 )] ,, ! ( ! ! ( )! k k nkk n kk n n Pkknk k a p a kknkk n a - - = -= a) [ ] 0 P kp == [ ] ( 29 11 P k = [ ] ( 29 2 21 P k = Problem 1: [GAR]2.76 Problem 2: [GAR]2.78
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[ ] [ ] [ ] [ ] ( 29 3 3 101 21 P k P k P k P kp = = - = - = - ==- b) In general: ( 29 ( 29 1 0 k P k p p km = - ≤< ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 1 11 0 00 1 1 1 111 m m mm kk k p P m P k p p ppp p - -- = ==   = - = = - - =-× ∑∑ ( 29 ( 29 ( 29 1 m pp p = - =-+ - =- a). For a car to pay $k, it needs to be in the lot for (k-1)/2 < t < k/2 22 [ ]= [ ] [ ] [] 2 222 = ( 1) for 0,1,2, k P Payk dollar s Pt P t e e e ek - - < ≤= - = - = -= L P($k) = 1 1 2 2 2 ( 1 ) ( 1 )() ( 1) k e ee - - - - - k=1,2,3,… It is a geometric probability law with p = 1 2 e - . b). Unconditional 1 15 42 2 [ 5 ] ( ) ( P ke e ee - - = = - =- . For all values of k namely ( k ≤≤ ) it is similar to part a. Conditional: Means we have 51 2 2 ( 1 5 ) [ 5 / max 5 ] 2 k P k i s P t - - - -   = = For other values of k 14 k it is the same as part a. You can easily check that sum of all probabilities is equal to 1. P [k tosses required until heads comes up twice] = P [one of the first k-1 tosses is heads AND the k th toss is heads] = () PAB I Problem 3: [GAR]2.79 Problem 4: [GAR]2.80
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A: {k th toss is heads} B: {one of the k-1 tosses is heads} A and B are independent.
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part10 - Univ. Of Maryland at College Park, ECE Department...

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