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exam1v2olutions - STAT 350 Probability and Random Processes...

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STAT 350: Probability and Random Processes for Engineers Spring, 2008 ANSWERS TO EXAM #1, VERSION #2 You will need to follow along with your exam. 1. (a) Combinations Rule: C (32 , 4) = 35,960 (b) Partitions Rule: 32! 4!3!25! = 117,804,960, or C (32 , 4) × C (28 , 3) = 117,804,960 (c) Order matters, so Permutations Rule: P (32 , 4) = 863,040 (d) The chance of getting exactly 3 EEs and 1 CPE is C (23 , 3) × C (9 , 1) C (32 , 4) = 1771 × 9 35960 = . 4432 2. If you hit black 0 times, you lose 5(\$10) = \$50. If you hit black 1 time, you lose \$30 (four losses, one win). Otherwise, you lose less than \$30. So, you want the chance of 0 or 1 black spins in 5 tries. This requires the binomial model, with n = 5 and p = 18 / 38: P ( k 1) = 5 0 (18 / 38) 0 (20 / 38) 5 + 5 1 (18 / 38) 1 (20 / 38) 4 = . 2221 3. For each component, let A i means that component # i is working properly; P ( A i ) = . 88 for each i . The reliability of the 1-2 subsystem is R 12 = P ( A 1 A 2 ); by deMorgan’s Law and independence, P ( A 1 A 2 ) = 1 - P ( ¯ A 1 ¯ A 2 ) = 1 - P ( ¯ A 1 ) × P ( ¯ A 2 ) - 1 - ( . 12)( . 12) = . 9856. Likewise, R 34 = . 9856, and of course R 5 = . 88. Since these subsystems are connected in series, by independence R sys = R 12 × R 34 × R 5 = ( .

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exam1v2olutions - STAT 350 Probability and Random Processes...

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