hw6vqn - Vinh Nguyen Stat 120A(Fall 2007): HW5 Solutions...

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Vinh Nguyen Stat 120A(Fall 2007): HW5 Solutions November 28, 2007 Text: Probabilty and Statistics , by Degroot and Schervish (3 rd ed) Problems: 5.2.3, 5.2.4, 5.3.6, 5.3.7, 5.5.2, 5.4.1, 5.4.2, and Additional Problems 5.2.3 Let X be the number of heads in 10 tosses. We have X Bin (10 , 1 / 2). In order to get more heads than tails, we must have at least six heads. Thus, P ( X 6) = 1 - P ( X 5) = 1 - 5 X x =0 ± 10 x ² (1 / 2) x (1 / 2) 10 - x = 0 . 3770 . 5.2.4 We have X Bin (15 , 0 . 4). Thus, P (6 X 9) = 9 X x =6 ± 15 x ² (0 . 4) x (0 . 6) 15 - x = 0 . 5630 . 5.3.6 Suppose X 1 Bin ( n 1 ,p ) and X 2 Bin ( n 2 ,p ), where X 1 and X 2 are independent. Note that X 1 + X 2 Bin ( n 1 + n 2 ,p ). We would like to compute the conditional distribution of X 1 given X 1 + X 2 = k , for k = 1 ,...,n 1 + n 2 . In order to do so, we apply the definition of conditional probability: P ( X 1 = x | X 1 + X 2 = k ) = P ( X 1 = x,X 1 + X 2 = k ) P ( X 1 + X 2 = k ) (Definition) = P ( X 1 = x,X 2 = k - x ) P ( X 1 + X 2 = k ) = P ( X 1 = x ) P
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This note was uploaded on 01/21/2010 for the course STATS 37850 taught by Professor Phelan during the Spring '09 term at UC Irvine.

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hw6vqn - Vinh Nguyen Stat 120A(Fall 2007): HW5 Solutions...

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