55010Solution3 - STAT 550 HW #3 Solution 1.5-1 sample space...

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STAT 550 HW #3 Solution 1.5-1 {( , ), , 1 6} {(3,4),(4,3)} { 7} {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} 26 1 2 () , () , ( ) 36 36 6 36 2 / 3 6 1 (|) 1 / 6 3 s am p l es p a c eS ijij A B the sum of face is PA PB PA B PAB == = = = = K I I 1.5-4 (a) 554 111 3 ⎛⎞ ⎜⎟ ⎝⎠ ++ (b) Let A, B, C be events that there is a person selected in Democrats, Republicans and Independents respectively: 11 1 55 4 14 13 12 ( ) ()( | )( | ) PABC PAPB APC AB CC C CCC ⋅⋅ = =⋅⋅ 1.5-5 (a) 0.15 1 (4 | ) 0.6 4 P more than hours student is in a tech area = = (b) 0.15 3 (| 4 ) 0.15 0.05 4 P student is in a tech area more than hours = = + (c) 0.20 0.15 7 4 ) 0.25 0.55 16 P student is in a nontech area less than or equal to hours + = = + 1.5-6 Let E be an event that red marble is selected in BOX I; A be an event that the marble is red in BOX II. Then 22 4 , ( ) , (| ) 44 6 PE PAE = 3 6 = By the total probability, we have 2423 7 ()( | ) ( )( | ) 46461 2 PEPAE PE PAE =+ = × + × = 1.5-8 Let A be the event that a student will graduate, E be the event that a student will study 1
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(a) ( ) ( ) ( | ) ( ) ( | ) 75% 85% 25% 35% 0.725 CC PA PEPAE PE PAE =+ = + = (b) ( ) ( ) ( | ) 75% 85% ( | ) 0.879 ( ) ( ) 0.725 PE A PEPAE == = = I 1.5-9 (a) ( ) ( 1) ( | 1) ( 2) ( | 2) 16 17 0.59 21 0 21 2 P cookie is broken P Box P broken Box P Box P broken Box =⋅ +⋅ = (b) 22 67 10 12 ( ) ( | ( 2) ( | 2) 11 0.325 P both broken P Box P both broken Box P Box P both broken Box 1.6-3 (a) (i) ( ) ( ) ( | ) 0.25 0.5 0.125 PA B PBPAB × = I (ii) ( ) ( ) ( ) ( ) 0.75 0.125 0.25 0.625 PA B PA B PB =+− = + = UI (iii) (
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This note was uploaded on 09/25/2010 for the course STAT 550 taught by Professor Staff during the Spring '08 term at San Diego State.

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55010Solution3 - STAT 550 HW #3 Solution 1.5-1 sample space...

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