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Fall 2010 HW01

# Fall 2010 HW01 - University of Illinois Fall 2010 ECE 313...

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University of Illinois Fall 2010 ECE 313: Problem Set 1: Solutions Axioms of probability and calculating the sizes of sets 1. [Defining a set of outcomes] (a) One choice would be Ω = { ( w 1 , w 2 , w 3 ) : w 1 ∈ { 1 , 2 } , w 2 ∈ { 3 , 4 } , w 3 ∈ { w 1 , w 2 }} , where w i denotes which team wins the i th game, for 1 i 3 . Another choice would be Ω = { ( x 1 , x 2 , x 3 ) : x i ∈ { L, H } for 1 i 3 } , where x i indicates which of the two teams playing the i th game wins; x i = H indicates that the higher numbered team wins and x i = L indicates that the lower numbered team wins. For example, ( L, L, H ) indicates that team one wins the first game, team three wins the second game (so team one plays team three in the third game), and team three wins the third game. (b) Eight, because there are two possible outcomes for each of the three games, and 2 3 = 8 . 2. [Possible probability assignments] (This is one of many ways to get the answer.) Since, by De Morgan’s law, the complement of A B is A c B c , the fact P ( A B ) = 0 . 6 is equivalent to P ( A c B c ) = 0 . 4 . Thus, we can fill in the Karnaugh diagram for A and B as shown: 0.3 B c A A c B 0.3 0.4 a ! a We filled in the variable a for P ( AB c ) , and then, since the sum of the probabilities is one, it must be that P ( A c B ) = 0 . 3 - a. The valid values of a are 0 a 0 . 3 , and ( P ( A ) , P ( B )) = (0 . 3 + a, 0 . 6 - a ) . So, in parametric form, the set of possible values of ( P ( A ) , P ( B )) is { (0 . 3 + a, 0 . 6 - a ) : 0 a 0 . 3 } . Equivalent ways to write this set are { ( u, v ) : v = 0 . 9 - u and 0 . 3 u 0 . 6 } or { ( x, 0 . 9 - x ) : 0 . 3 x 0 . 6 } . The set is also represented by the solid line segment in the following sketch: 0.3 P(B) P(A) 0.3 0.6 0.9 0.9 0.6 3. [Grouping students into teams] (a) One solution is the following. Five teams, numbered one through five, can be sequentially selected as follows. To begin, there are ( 10 2 ) ways to choose team one. That leaves eight students, so for any choice of team one, there are ( 8 2 ) ways to choose team two. That leaves

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