HW03_2011 solution

HW03_2011 solution - Econ 6190, Fall 2011 Problem Set 3...

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Econ 6190, Fall 2011 Problem Set 3 Solutions (1) Yes to both questions. Empty set ; and total set S are mutually exclusive: P ( ;\ S ) = P ( ; ) = 0 : And also independent: P ( ;\ S ) = 0 = P ( ; ) P ( S ) : (2) Example: P ( A ) = P ( B ) = P ( C ) = 1 3 : B = C: P ( A \ B ) = 1 27 :P ( A \ B ) = 1 27 6 = 1 9 = P ( A ) P ( B ) (3) There are 7 7 equally likely sample points. The possible values of X 3 are 0, 1 and 2. Only the pattern 331 (3 balls in one cell, 3 balls in another cell and 1 ball in a third cell) yields X 3 = 2 . The number of sample points with this pattern is 7 2 7 3 4 3 ± 5 = 14,700. So P ( X 3 = 2) = 14 ; 700 = 7 7 : 0178 . There are 4 patterns that yield X 3 = 1 . The number of sample points that give each of these patterns is given below. pattern number of sample points 34 7 7 3 ± 6 = 1470 322 7 7 3 6 2 4 2 2 2 ± = 22050 3211 7 7 3 ± 6 4 2 5 2 ± 2! = 176400 31111 7 7 3 6 4 ± 4! = 88200 sum = 288120 So P ( X 3 = 1) = 288120
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This note was uploaded on 01/07/2012 for the course ECON 6190 taught by Professor Hong during the Fall '07 term at Cornell.

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HW03_2011 solution - Econ 6190, Fall 2011 Problem Set 3...

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