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Unformatted text preview: Solutions to PS 1 * Hongyan Zhao September 14, 2009 1 Question 1 (a) Let X be a RV that represents the value of your hand after your last card is dealt. The third card can be chosen from 2,3,4,...,Q,K,A. We know a 6 and two 5 cards are missing. Thus there are only 49 cards left for you to choose. X=12 means the last card is 2,the probability is 4 49 , X=13 means the last card is 3,the probability is 4 49 , X=14 means the last card is 4,the probability is 4 49 , X=15 means the last card is 5,the probability is 2 49 , X=16 means the last card is 6,the probability is 3 49 , X=17 means the last card is 7,the probability is 4 49 , X=18 means the last card is 8,the probability is 4 49 , X=19 means the last card is 9,the probability is 4 49 , X=20 means the last card can be 10,J,Q,K,the probability is 16 49 , X=21 means the last card is A,the probability is 4 49 , Therefore, we can draw the CDF of X. (b) Based on the probability distribution, we can calculate the expectation and variance. E ( X ) = 4 49 · 12 + 4 49 · 13 + ··· = 17 . 43 (c) V ar ( X ) = (12 E ( X )) 2 · 4 49 + (13 E ( X )) 2 · 4 49 + ··· = 8 . 78 * If you find any mistake or typo, please email me [email protected] 1 (d) Now there are two 5, a 6, a 7, a J cards missing, the number of the left cards for the dealer to choose is 47....
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This note was uploaded on 02/17/2010 for the course ECON 141 taught by Professor Staff during the Fall '08 term at Berkeley.
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