Math 301 2008

# Math 301 2008 - 10(14 2 HW2 2.1 Problem 5(a P[0 1 = R 1 e-t...

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HW1 and HW2 Solution 1 HW1 1.1 Problem 16 Assume A j ,j = 1 , 2 , 3 , 4 , stands for the event that a soldier lost one eye, lost one ear, lost one hand and lost one leg respectively. Then, P ( A 1 ) . 70 (1) P ( A 2 ) . 75 (2) P ( A 3 ) . 80 (3) P ( A 4 ) . 85 . (4) Goal : Minimize P ( A 1 A 2 A 3 A 4 ). Method I: P ( j =4 \ j =1 A j ) P ( j =3 \ j =1 A j ) + P ( A 4 ) - 1 (5) P ( j =2 \ j =1 A j ) + P ( A 3 ) + P ( A 4 ) - 2 (6) j =4 X j =1 A j - 3 (7) . 70 + . 75 + . 80 + . 85 - 3 (8) . 10 (9) 1

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Method II: P ( j =4 \ j =1 A j ) = 1 - P (( j =4 \ j =1 A j ) c ) (10) = 1 - P ( j =4 [ j =1 A c j ) (11) 1 - j =4 X j =1 P ( A c j ) (12) 1 - (1 - . 70) - (1 - . 75) - (1 - . 80) - (1 - . 85) (13)
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Unformatted text preview: . 10 (14) 2 HW2 2.1 Problem 5 (a) P ([0 , 1]) = R 1 e-t dt = 0 . 63212 (b) P ([0 , 3]) = R 3 e-t df = 0 . 95021 (c) Since the particle is not necessarily the ﬁrst one, the probability here is the same as (a), which is 0.63212. (d) The same reason as above, the probality here is P ([0 , ∞ ]), which is 1. 2...
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Math 301 2008 - 10(14 2 HW2 2.1 Problem 5(a P[0 1 = R 1 e-t...

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