Assignment 5 solutions Final

# Assignment 5 solutions Final - CE93-Engineering Data...

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Unformatted text preview: CE93-Engineering Data Analysis Mark Hansen, Fall 2009 Assignment 5 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) = 1− − 1− − 1− 1− 1− 1− 1− 1− 1− 1− − 1− − 1− 1− 1− 1− 1− 1− Can also do P(2 out of 4 system works)= P(2 work)+P(3 work)+P(4 work) 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) =1− 1− 1− 1− 1− 1− − 1− 1− 1− 1− − 1− 1− 1− − 1− 1− 1− 1− − 1− 1− 1− − 1− 1− 1− 1− − 1− 1− 1− − 1− 1− 1− − 1− 1− 1− − 1− 1− 1− − 1− 1− 1− − 1− 1− 1− − 1− 1− 1− − 1− 1− 1− − 1− 1− 1− − 1− 1− 1− 1− 1− Can also do P(3 out of 5 system works)= P(3 work)+P(4 work)+P(5 work) 2. Problem 3.41 in Ross C1=component 1 works Use Bayes’ Theorem 1 ≥1 = 1 ≥1 1 =1 = ≥1 =1− = 1 − 0.5 1 = 0.5 1 − 0.5 3. Problem 3.47 in Ross For parts a and b find union of A and B, P(A or B) a. Mutually exclusive: ∪ ≥1 if C1 works then you know that system works 1 = 0.5 ≥1 1∗ = + = 0.2 + 0.3 = 0.5 1 CE93-Engineering Data Analysis ∪ b. Independent: Where ∩ Mark Hansen, Fall 2009 = = + − ∩ = 0.2 + 0.3 − 0.2 ∗ 0.3 = 0.44 ∗ For parts c and d find intersection of A, B, and C: P(A&B&C) ∩ c. Independent: ∩ = ∩ ∩ d. Mutually exclusive: ∗ ∗ = 0.2 ∗ 0.3 ∗ 0.4 = 0.024 =∅ 4. Given are: D: detect G: gross polluter System A: = 0.70, = 0, = 0.30 =1 System B: = 0.90, = 0.01, = 0.10 = 0.99 = 0.05 a. Find P(G|Dc) for A only. Use Bayes’ Theorem ∗ = ∗ = 0.0155 + = ∗ 0.3 ∗ 0.05 0.3 ∗ 0.05 + 1 ∗ 1 − 0.05 b. Find P(G|D) for B only. Use Bayes’ Theorem ∗ = = 0.8257 ∗ + ∗ = 0.9 ∗ 0.05 0.9 ∗ 0.05 + 0.01 ∗ 1 − 0.05 c. P(DA|G)⊥P(DB|G) i. Find ∪ ∪ ii. Find = + − ∩ = + − ∗ = 0.7 + 0.9 − 0.7 ∗ 0.9 = 0.97 ∩ ∩ = 0.7 ∗ 0.9 = 0.63 d. Grossly polluting if either sensor indicates this. = + = = 0.7 ∗ 0.05 + 0 ∗ 0.95 = 0.035 + ∩ = = 0.9 ∗ 0.05 + 0.01 ∗ 0.95 = 0.0545 ∩ + ∩ = 0.63 ∗ 0.05 + 0 ∗ 0.95 = 0.0315 CE93-Engineering Data Analysis Mark Hansen, Fall 2009 ∪ = + − ∩ = 0.035 + 0.0545 − 0.0315 = 0.058 ∪ i. Given 1. Sensor A only ∩ ∪ − = ∩ = = ∪ 0.035 − 0.0315 = 0.060 0.058 0.0545 − 0.0315 = 0.397 0.058 2. Sensor B only ∩ ∪ − = ∩ ∪ 3. Both sensors ∩ ∪ = ∩ ∪ ii. False alarm rate – find = 0.0315 = 0.543 0.058 ∪ ∪ = = 0.01 ...
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