hw1sln

# hw1sln - 4 5 1 5 3 4 5 2 1 5 4 4 5 3 1 5 0(5 6 4 5 4 1 11 E...

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IEOR 161 Operations Research II University of California, Berkeley Spring 2008 Homework 1 Suggested Solution Chapter 2. 12. P { X 4 } = P { X = 4 } + P { X = 5 } = ± 5 4 ² ( 1 3 ) 4 ( 2 3 ) + ± 5 5 ² ( 1 3 ) 5 = 10 + 1 243 = 11 243 . 20. ± 5 2 ² ( 1 5 ) 2 ± 3 1 ² ( 1 2 ) 1 ± 2 2 ² ( 3 10 ) 2 = 5! 2!2! ( 1 5 ) 2 ( 1 2 ) 1 ( 3 10 ) 2 = 0 . 054 28. (a). Consider the ﬁrst time that the two coins give diﬀerent results. Then P { X = 0 } = P { ( t, h ) | ( t, h ) or ( h, t ) } = p (1 - p ) 2 p (1 - p ) = 1 2 . (b). No, with this procedure P ( X = 0) = P (ﬁrst ﬂip is a tail) = 1 - p. Chapter 3. 5 (a) Let Z= number of red balls selected P { X = i | Y = 3 } = X = i, Y = 3 , Z = 3 - i Y = 3 = ( 3 i )( 5 3 )( 6 3 - i ) ( 9 3 )( 5 3 ) = ( 3 i )( 6 3 - i ) ( 9 3 ) , i = 0 , 1 , 2 , 3 . (b) Given Y=1, we know that the other 5 are either white or red. For each ball selected, there is a 3 9 probability that it is a white ball. E [ X | Y = 1] = 5 × 3 9 = 5 3 8 (a) X Geo( 1 6 ), E[X]=6. (b) E[X | Y=1]=1+E[X]=7 (c) By conditioning E [ X | Y = 5] = E [ X | Y = 5 , X = 1] P ( X = 1 | Y = 5) + E [ X | Y = 5 , X = 2] P ( X = 2 | Y = 5) + E [ X | Y = 5 , X = 3] P ( X = 3 | Y = 5) + E [ X | Y = 5 , X = 4] P ( X = 4 | Y = 5) + E [ X | Y = 5 , X = 5] P ( X = 5 | Y = 5) + E [ X | Y = 5 , X 6] P ( X 6 | Y = 5) = 1( 1 5 ) + 2(

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Unformatted text preview: 4 5 )( 1 5 ) + 3( 4 5 ) 2 ( 1 5 ) + 4( 4 5 ) 3 ( 1 5 ) + 0 + (5 + 6)( 4 5 ) 4 1 11 E [ X | Y = y ] = Z y-y xf X | Y ( x | y ) dx = Z y-y x f ( x, y ) f Y ( y ) dx = C ( y ) Z y-y x ( y 2-x 2 ) dx = 0 21 (a) X = ∑ N i =1 T i . (b) E ( N ) = 3 . (c) E [ T N ] = 2 (d) Given that the miner selects the ﬁrst door on the nth attempt, we know all the previous n-1 attempts are either the second door or the third door wither equal probability. E [ N X i =1 T i | N = n ] = E [ N-1 X i =1 T i | N = n ]+ E [ T N | N = n ] = ( 3 2 + 5 2 )( n-1)+2 = 4 n-2 (e) E [ X ] = E [ E [ X | N ]] = E [ E [ N X i =1 T i | N ]] = E [4 N-2] = 4 E [ N ]-2 = 4 × 3-2 = 10 2...
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hw1sln - 4 5 1 5 3 4 5 2 1 5 4 4 5 3 1 5 0(5 6 4 5 4 1 11 E...

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