131a3sol - -1 2 > 1 4 . Yes, A 1 is...

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OLEE 131A Homework #3 Fall 2010 Solution K. Yao 1. a. b. P(X=0)=0.07; P(X=1)=.23; P(X=2)=0.57; P(X=3)=0.13. c. d. 2. a. i =1 A i = { x : 0 x < 1 } .
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b. Since all the A i ’s are mutually exclusive, then i =1 A i = . 3. Solve for C 2 0 C n - 2 4 C n 4 = 1 2 × C 2 2 C n - 2 2 C n 4 C n - 2 4 = C n - 2 2 / 2 ( n - 2)! 4!( n - 6)! = ( n - 2)! 2 × 2!( n - 4)! , n 6 , 1 4 × 3 = 1 2 × ( n - 4)( n - 5) ( n - 4)( n - 5) = 6 n 2 - 9 n + 14 = 0 = ( n - 2)( n - 7) . Thus, n = 7 . 4. a. 4/499 = 0.008 b. (5/500)(4/499)=0.00008 c. (495/500)(494/499)=0.98 5. a. To check for independence between A and B, verify whether P ( B | A ) = P ( B )? P ( B | A ) = 4 / 499 = 0 . 008; P ( B ) = (495 / 500)(5 / 499) + (5 / 500)(4 / 499) = 0 . 01 . Thus, A and B are not independent. b. P ( B ) = 5 / 500 = 0 . 01 = P ( B | A ) . Thus, A and B are independent. 6. a. P [ B 0 ] = P [ B 0 | A 0 ] P [ A 0 ] + P [ B 0 | A 1 ] P [ A 1 ] = (1 - ± 1 ) 1 2 + ± 2 · 1 2 . b. P [ A 0 | B 1 ] = P [ B 1 | A 0 ] P [ A 0 ] P [ B 1 ] = ± 1 · 1 2 1 - 1 2 (1 - ± 1 ) - 1 2 ± 2 = ± 1 1 - ± 2 + ± 1 . P [ A 1 | B 1 ] = 1 - P [ A 0 | B 1 ] = 1 - ± 2 1 - ± 2 + ± 1 . Given B 1 , input A 1 is more likely if P [ A 1 | B 1 ] > P [ A 0 | B 1 ] 1 - ± 2 1 - ± 2 + ± 1 > ± 1 1 - ± 2 + ± 1 1 - ± 2 > ± 1 1
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Unformatted text preview: -1 2 &amp;gt; 1 4 . Yes, A 1 is probable. 2 7. a. F (0-) = lim x -F ( x ) = lim x 1 2 e x = 1 2 . Therefore, P ( X = 0) = F (0)-F (0-) = (1-1 4 e-)-1 2 = 1 4 . b. P (0 &amp;lt; X 2) = F (2)-F (0) = (1-1 4 e-2 )-(1-1 4 e-) = 1 4 (1-e-2 ). c. P (0 X 2) = F (2)-F (0)+ P ( X = 0) = 1 4 (1-e-2 )+ 1 4 = 1 2-1 4 e-2 . d. P (-3 &amp;lt; X &amp;lt; 0) = F (0)-F (-3)-P ( X = 0) = 3 4-( 1 2 e-3 )-1 4 = 1 2-1 2 e-3 . e. P (-3 X 0) = F (0)-F (-3) + P ( X =-3) = 3 4-1 2 e-3 + 0 = 3 4-1 2 e-3 . 3...
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131a3sol - -1 2 &amp;amp;amp;gt; 1 4 . Yes, A 1 is...

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