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Unformatted text preview: STAT/MATH 418 Solution to Exam 1 Exam 1
1. (a) 8!3! (b) 3!3!4!3! 2. (a) P ( suit U shirt U tie ) = 0.22 + 0.28 + 0.30  0.14  0.11  0.10 + 0.06 = 0.51 P ( None of these items ) = 1  P ( suit U shirt U tie ) = 1  0.51 = 0.49 (b) 0.03 + 0.1 + 0.15 = 0.28 3 4 4 3 1 1 1 1 3. (a) 14 4 4 4 2 2 (b) 14 4 8 4 (c) 14 4 2 1 3 1 1 1 1 1 1 1 = 2 , P letter V = 1 = 1 , P letter E = 1 = 3 ( ) 7 8 ( ) 7 8 4. P ( letter R ) = 7 56 8 56 56 1 1 1 1 1 1 Therefore, the probability the same letters is chosen is P ( letter R ) + P ( letter V ) + P ( letter E ) = 5. (a) 2 . 3 4 = 24 (b) 2 . 3 1 = 6 (c) 1 . 3 4 = 12 6 56 (d) {(chicken, pasta, Ice cream), (chicken, rice, Ice cream), (chicken, potatoes, Ice cream)} 6. (a) 7! = 21 2!3! n + 2 (b) 2 ...
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This note was uploaded on 06/28/2008 for the course MATH 418 taught by Professor Arnold,stevenfdong,yuexiao during the Summer '08 term at Pennsylvania State University, University Park.
 Summer '08
 ARNOLD,STEVENFDONG,YUEXIAO
 Math, Probability

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