sec_5_1_sol

sec_5_1_sol - APPM 4/5570 Solutions to Problems from...

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Solutions to Problems from Section 5.1 6. a. P ( X = 4 , Y = 2) = P ( Y = 2 | X = 4) · P ( X = 4) = 4 2 ! (0 . 6) 2 (0 . 4) 4 - 2 · 0 . 15 = 0 . 05184 b. P ( X = Y ) = P ( X = 0 , Y = 0) + P ( X = 1 , Y = 1) + ··· + P ( X = 4 , Y = 4) = P ( Y = 0 | X = 0) · P ( X = 0) + P ( Y = 1 | X = 1) · P ( X = 1) + ··· + P ( Y = 4 | X = 4) · P ( X = 4) = 1 · 0 . 1 + 1 1 ! (0 . 6) 1 (0 . 4) 1 - 1 · 0 . 2 + ··· + 4 4 ! (0 . 6) 4 (0 . 4) 4 - 4 · 0 . 15 0 . 4010 c. f ( x, y ) = P ( X = x, Y = y ) = P ( Y = y | X = x ) · P ( X = x ) = x y ! (0 . 6) y (0 . 4) x - y · p X ( x ) for x = 0 , 1 , 2 , 3 , 4 and y = 0 , 1 , 1 , . . . x . The marginal pdf for Y : Since 4 extended warranties can only be sold if 4 cameras are sold, we have p Y (4) = P ( Y = 4) = P ( X = 4 , Y = 4) = f (4 , 4) = 0 . 0194 Since 3 extended warranties can only be sold if either 3 or 4 cameras are sold, we have p Y (3) = P ( Y = 3) = P ( X = 3 , Y = 3)+ P ( X = 4 , Y = 3) = f (3 , 3)+ f (4 , 3) = 0 . 1058 Similarly,
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This note was uploaded on 01/23/2011 for the course APPM 5440 at Colorado.

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sec_5_1_sol - APPM 4/5570 Solutions to Problems from...

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