f2006Solns8 - ECE 190 - Fall, 2006 Assignment 8 Solutions...

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ECE 190 — Fall, 2006 Assignment 8 Solutions Section 6.9 1 See text. 2 P ( X | Y )= P ( X Y ) P ( Y ) P ( X Y P ( X | Y ) P ( Y ) = 1 3 × 1 4 = 1 12 4 See text. 5 See text. 7 (a) See text. Since both of the people chosen are treated equally (they both become Fnalists), the order of choosing is immaterial. ±urthermore, since names are drawn at random, we are dealing with equally likely outcomes. Thus, to determine probabilities, we can use ratios of cardinalities. | S | = ± 10 2 ² =45 (b) P (2 men) = number of ways to choose 2 men | S | = ( 7 2 ) 45 = 21 45 = 7 15 (c) P (1 woman, 1 man) = number of ways to choose 1 woman, 1 man | S | = ( 3 1 )( 7 1 ) 45 = 21 45 = 7 15 10 See text. 13 See text. 19 S = { HHH,HHT,HTH,THH,HTT,HTH,THT,TTT } A = { HHH,HHT,HTH,HTT } B = { HHT,HTH,THH,TTT } A B = { HHT,HTH } 1
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(a) P ( A | B )= P ( A B ) P ( B ) = | A B | | B | because the coin is fair = 2 4 = 1 2 But P ( A | A | | S | = 4 8 = 1 2 Thus P ( A | B P ( A ) (b) P ( B | A P ( B A ) P ( A ) = | A B | | A | because the coin is fair = 2 4 = 1 2 But P ( B | B | | S | = 4 8 = 1 2 Thus P ( B | A P ( B ) 20 Since A B = ,P ( A B )=0 Since A and B are independent, P ( A B P ( A ) P ( B )
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f2006Solns8 - ECE 190 - Fall, 2006 Assignment 8 Solutions...

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