Midterm Solutions

Midterm Solutions - stage 2 give each girl 2 jellybeans...

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CSE 21: Solutions to Midterm (May 10, 2007) 1. By direct calculation, (a) A B B = (1234)(5). (b) B A A = (1324)(5). (c) A B A = (1524)(3). (d) B A B = (1534)(2). So the answer is (c). 2. Without any restriction, we have ( 3+4+5 3 ) = 220 ways to choose 3 balls. To get three balls with different colors, we have ( 3 1 )( 4 1 )( 5 1 ) = 60 choices. Then the probability is 60 / 220 = 3 / 11. 3. By the linearity of expected values, E ( X + Y ) = E ( X ) + E ( Y ). To compute E ( X ), Pr ( X = 1) = ± 4 1 ²± 48 1 ² / ± 52 2 ² = 32 221 Pr ( X = 2) = ± 4 2 ² / ± 52 2 ² = 1 221 Then E ( X ) = 1 · Pr ( X = 1) + 2 · Pr ( X = 2) = 32 221 + 2 · 1 221 = 2 13 To compute E ( Y ), Pr ( Y = 1) = ± 13 1 ²± 39 1 ² / ± 52 2 ² = 13 34 Pr ( Y = 2) = ± 13 2 ² / ± 52 2 ² = 1 17 Then E ( Y ) = 1 · Pr ( Y = 1) + 2 · Pr ( Y = 2) = 13 34 + 2 · 1 17 = 1 2 1
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Finally, E ( X + Y ) = E ( X ) + E ( Y ) = 2 13 + 1 2 = 17 26 . 4. We can distribute the cookies and jellybeans in two stages: stage 1: give each boy 2 cookies, each girl 1 cookie. Then distribute the remaining 20 - 2 · 3 - 1 · 4 = 10 cookies with no restriction.
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Unformatted text preview: stage 2: give each girl 2 jellybeans. Then distribute the remaining 25-2 · 4 = 17 jellybeans with no restriction. Using the trick of Stars and Bars , we have ( 10+7-1 7-1 ) = ( 16 6 ) choices for stage 1, and ( 17+7-1 7-1 ) = ( 23 6 ) choices for stage 2. Then the total number of ways is ± 16 6 ²± 23 6 ² = 808 , 383 , 576 . 5. To count the choices with two defective bulbs, we need two stages: stage 1: choose 2 defective bulbs. Number of choices: ( 5 2 ) stage 2: choose 4 effective bulbs. Number of choices: ( 15 4 ) . The number of choices with no restriction is ( 20 6 ) . Then our probability is ± 5 2 ²± 15 4 ² ± 20 6 ² = 455 1292 . 2...
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This note was uploaded on 12/03/2009 for the course CSE 21 taught by Professor Graham during the Spring '07 term at UCSD.

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Midterm Solutions - stage 2 give each girl 2 jellybeans...

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