hwk3 - 4.5 We need to assume each outcome is equally likely...

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4.5 We need to assume each outcome is equally likely for each person to choose each door. Each person has four choices, thus there are 3 46 4 possible outcomes for three people to enter the building. If x people choose door I, there are 3 x    ways to specify the x people, and for the remaining 3 x people, each of them has 3 choices to enter the building. Thus there are 3 3 3 x x possible outcomes if x people choose entrance one, the probability of X x is 3 3 3 ( ) , 0,1,2,3 64 x x PX x x  Then 27 27 9 1 ( 0 ), ( 1 ( 2 ( 3 ) 64 64 64 64 pp 4.6 (a) For a given structure fire, the probability that it is caused by cooking is 29706 0.5712 52006 p Now if x out of four are caused by cooking, there are 4 x ways to specify the x fires that are caused by cooking, and each outcome is equally likely, with probability 4 (1 ) x x , thus the probability of X x is 4 4 () ( 1 ) , 0 , 1 , 2 , 3 xx p p x x , 4 Then we have x 01234 p(x) 0.0338 0.1801 0.3599 0.3197 0.1065 (b) The probability of interest is ( 1) 1 ( 0) 1 0.0338 0.9662 PX Px   
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4.7 (a) X can take value 0,1,2,3,4,5,6. Let Y denote the number of sales in the first day, Z denotes the number of sales in the second day, Y and Z are independent. Then ( 0) ( 0, 0) ( 0) ( 0) 0.5 0.5 0.25 PX PY Z PZ   ( 1) ( 1) ( 0) ( 0) ( 0.3 0.5 0.5 0.3 0.3       ( 2) ( 2) ( 0) ( 0) ( 2) ( 1) ( 0.15 0.5 0.5 0.15 0.3 0.3 0.24  ( 3 )(3 ) (0 )(0 ) (3 )(1 ) ( 2 )(2 ) (1 0.05 0.5 0.5 0.05 0.3 0.15 0.15 0.3 0.14   ( 4 ) ) ) (2 ) 0.3 0.05 0.05 0.3 0.15 0.15 0.0525 ) ) ) 1 ) ( 5 ) ) 0.05 0.15 0.15 0.05 0.015      ( 6 ) 0.05 0.05 0.0025 (b) The probability of interest is ) 1 ) ( 1 0.25 0.3 0.45   4.8 a. there are ways to choose two chips out of four in total. 4 6 2    0: X only one possible outcome, we choose the two non-defective ones. Then 1 ) 6 1: X we have chosen 1 defective and 1 non-defective, there are two ways to specify the defective one and there are two ways to specify the non-defective one, thus there are 4 possible outcomes. Then
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42 (1 ) 63 PX  2: X only one possible outcome, we choose the two defective ones. Then 1 (2 ) 6
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This note was uploaded on 02/14/2012 for the course STA 4321 taught by Professor Staff during the Spring '08 term at University of Florida.

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hwk3 - 4.5 We need to assume each outcome is equally likely...

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