exam1v1solutions

exam1v1solutions - STAT 350 Probability and Random...

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Unformatted text preview: STAT 350: Probability and Random Processes for Engineers Spring, 2008 ANSWERS TO EXAM #1, VERSION #1 You will need to follow along with your exam. 1. (a) Combinations Rule: C (30 , 4) = 27,405 (b) Partitions Rule: 30! 4!3!23! = 71,253,000, or C (30 , 4) × C (26 , 3) = 71,253,000 (c) Order matters, so Permutations Rule: P (30 , 4) = 657,720 (d) The chance of getting exactly 3 EEs and 1 CPE is C (23 , 3) × C (7 , 1) C (30 , 4) = 1771 × 7 27405 = . 4524 2. If you hit black 5 times, you win 5(\$10) = \$50. If you hit black 4 times, you win 4(\$10)- 1(\$10) = \$30. Otherwise, you win less than \$30. So, you want the chance of at least 4 black spins in 5 tries. This requires the binomial model, with n = 5 and p = 18 / 38: P ( k ≥ 4) = 5 4 ¶ (18 / 38) 4 (20 / 38) 1 + 5 5 ¶ (18 / 38) 5 (20 / 38) = . 1563 3. For each component, let A i means that component # i is working properly; P ( A i ) = . 84 for each i . The reliability of the 1-2 subsystem is R 12 = P ( A 1 ∪ A 2 ); by deMorgan’s Law and independence,...
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This note was uploaded on 04/18/2008 for the course STAT 350 taught by Professor Carlton during the Spring '07 term at Cal Poly.

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exam1v1solutions - STAT 350 Probability and Random...

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