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Unformatted text preview: University of Illinois Fall 2009 ECE 313: Problem Set 3: Solutions Discrete Random Variables: pmf, expectation, LOTUS, and variance 1. [Rolling a die till something different happens] (a) Obviously, X is an integer taking on values 2 , 3 , 4 ,...,n... according as the outcome of the experiment is k,kk,kkk,...,k n 1 ,... where, of course, 6 = k . For any n 2, and for specific values of k and , P { outcome is k n 1 } = ( 1 6 ) n . Since there are 6 choices of k and 5 choices of , we have that p X ( n ) = P { X = n } = 5 6 1 6 n 2 for n 2. Sanity check: X n =2 5 6 1 6 n 2 = 5 6 1 + 1 6 + 1 6 2 + = 5 6 1 1 1 6 = 1 as it should be. P { X is an even number } = X m =1 p X (2 m ) = 5 6 1 + 1 6 2 + 1 6 4 + = 5 6 1 1 1 36 = 6 7 . (b) Y = 0 if and only if X = 2 or X = 3, and so p Y (0) = p X (2) + p X (3) = 5 6 + 5 36 = 35 36 . By symmetry, the remaining 1 36 probability is distributed equally among the 6 other values of Y : p Y ( i ) = 1 216 for 1 i 6. Note that any argument that purports to prove that p Y ( i ) > p Y ( j ) can be transformed (by interchanging i and j everywhere) into an argument that proves p Y ( j ) > p Y ( i ). Those who do not care for such appeals to symmetry can proceed as follows. For any choice of i , 1 i 6, P { Y = i } = P { iiii } + P { jjji : j 6 = i } = 1 6 4 + 5 1 6 4 = 1 6 3 = 1 216 . 2. [The Game of ChuckALuck] (a) Each die has probability 1 6 of matching your chosen number. Thus, the number of dice that match is a binomial random variable with parameters (3 , 1 6 )....
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This note was uploaded on 10/25/2010 for the course ECE 313 taught by Professor Milenkovic,o during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Milenkovic,O

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