sol_prac_midterm-1

sol_prac_midterm-1 - CSE 21: Discussion 5, Solutions to...

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CSE 21: Discussion 5, Solutions to Practice Midterm (May 7, 2007) 1. The number of onto functions is 4! S (5 , 4) = 24 · 10 = 240. The total number of functions is 4 5 . Then the probability is 240 / 4 5 = 60 / 4 4 = 15 / 64. 2. Without any restriction, we have ( 2+3+4 3 ) = 84 choices. To get three balls with different colors, we have ( 2 1 )( 3 1 )( 4 1 ) = 24 choices. Then the probability is 24 / 84 = 2 / 7. 3. We could have k = 0 , 1 , 2 , 3 or 4 Aces in the four cards. For each k ( k Aces in the four cards), the number of possibilities can can calculated as follows: stage 1: choose k Aces. Number of choices: ( 4 k ) . stage 2: choose 4 - k more cards from the 48 cards which are not Aces. Number of choices: ( 48 4 - k ) . Then for each k , the number of possibilities is ( 4 k )( 48 4 - k ) . Then the corresponding probability is ( 4 k )( 48 4 - k ) ( 52 4 ) . Finally, the expected value is E ( X ) = 4 X k =0 k · ( 4 k )( 48 4 - k ) ( 52 4 ) = ( 4 1 )( 48 3 ) ( 52 4 ) + 2 ( 4 2 )( 48 2
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