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Unformatted text preview: coin ǀ head) = Now, P (head ǀ weighted coin) = 0.75; P (weighted coin) = 1/3; P (head) = 1/3 * 0.5 + 1/3* 0.5 + 1/3* 0.75 = 0.5833 So, P (weighted coin ǀ head) = 0.75*(1/3)/0.5833 = 0.4286 1/9 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) Homework 3 Bayes’ Theorem and Probability Dist. 2) Police plan to enforce speed limits by using radar traps near A) Mattis and Windsor; B) Lincoln and Florida; and C) Lincoln and University. These locations are operated 40%, 45%, and 30% of the time, respectively. Students living on campus won’t be caught speeding to class, of course; but a random student living off
campus has a probability of 0.1, 0.33, and 0.2 (respectively) of passing through these locations. Three locations: Mattis and Windsor (1) , Lincoln and Florida (2), Lincoln and University(3) Assume, S1, S2 and S3 = the events that a person is speeding as he passes through the respective locations R = represent the event that the radar traps is operating resulting in a speeding ticket Given, P (R ǀ S1)= 0.4 ; P (S1) = 0.1; P (R ǀ S2)= 0.45 ; P (S2) = 0.33; P (R ǀ S3)= 0.3 ; P (S3) = 0.2 a) What is the probability that a random student living off
campus will receive a speeding ticket? a) P (R) = ǀ Si) * P (Si) = 0.4 * 0.1 + 0.45 * 0.33 + 0.3*0.2 = 0.2485 b) If the student does receive a speeding ticket, what is the probability that it came from the Lincoln and Florida speed trap? b) Using Bayes’ Theorem, P (S2 ǀ R) = P (R ǀ S2) *...
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This document was uploaded on 03/06/2014 for the course CEE 202 at University of Illinois, Urbana Champaign.
 Spring '08
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