Assignment 4 solutions

Assignment 4 solutions - CE 93 – Engineering Data...

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Unformatted text preview: CE 93 – Engineering Data Analysis Prof. Joan Walker, Spring 2011 Assignment 4 Solutions 1. Problem 3.38 in Ross a. 2 out of 4 system P(2 out of 4 system works)=P(>=2 work)=1-P(only 1 works)-P(none work) Can also do: P(2 out of 4 system works)= P(only 2 work)+P(only 3 work)+P(all 4 work) The point is that you always need to consider all the components. When you are saying only two of them work, then it means 2 work and the other two do not. One possible outcome is 1&2 work but 3&4 do not work. If I denote it as W1W2W3cW4c, then the probability is P(W1W2W3cW4c)= P(W1) P(W2) P(W3c) P(W4c) (By independence)=P1 P2 (1-P3)(1-P4) b. 3 out of 5 system P(3 out of 5 system works)=P(>=3 components work) =1-P(none work)-P(only 1 works)- P(only 2 work) Can also do: P(3 out of 5 system works)= P(only 3 work)+P(only 4 work)+P(all 5 work) 2. Problem 3.41 in Ross C1=component 1 works Use conditional probability: if C1 works then you know that system works CE 93 – Engineering Data Analysis Prof. Joan Walker, Spring 2011 3. Problem 3.47 in Ross For parts a and b find union of A and B, P(A or B) a. Mutually exclusive: b. Independent: Where For parts c and d find intersection of A, B, and C: P(A&B&C) c. Independent: d. Mutually exclusive: 4. a. b. 5. Given are: D: detect DA: detect by A DB: detect by B G: gross polluter System A: , , System B: , , , a. Find P(G|Dc) for A only. Use Bayes’ Theorem b. Find P(G|D) for B only. Use Bayes’ Theorem c. The question is saying that DA|G and DB|G are independent but NOT DA and DB. i. Find CE 93 – Engineering Data Analysis Prof. Joan Walker, Spring 2011 ii. Find d. Grossly polluting if either sensor indicates this. i. Given 1. Sensor A only Since 2. Sensor B only 3. Both sensors ii. False alarm rate – find , we can get CE 93 – Engineering Data Analysis 6. Prof. Joan Walker, Spring 2011 Bay Bridge repairs: Total days (T) days for repair A + Probabilities days for repair B (TA+TB) 13 6+7 14 6+8 7+7 15 6+9 7+8 8+7 7+9 8+8 16 17 Total probability 0.06 0.06 0.06 0.15 0.08 0.15 0.09 0.32 0.2 0.09 0.29 0.12 0.12 8+9 PMF of total duration T to finish both repairs 0.21 ...
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