3. One of the things graduating engineers fear most about their first job assign

Question

3. One of the things graduating engineers fear most about their first job assignment is the required urinalysis test to screen for the use of illicit substances (THC, cocaine, opiates). USA Today reports that such tests have "false positive" rates (that is, they identify non-users as users) of about 15%. The rate for "false negatives" (where users are identified as non-users, is lower, perhaps around 10%. Assume that 1% of students are users of illicit substances. Use the following event notation: U = the event that a student uses an illicit substance NU = the event that a student does not use an illicit substance +the event that the urinalysis test identifies a student as a user -= the event that the urinalysis test identifies a student as a non-user (10) From the above description, determine the following 6 probabilities: P(U) P(NU) = PI+/NU) (5) Determine the unconditional probability that a student selected at random will test positive for illicit substances. (5) Suppose an individual selected at random tests positive for illicit substances. Determine the probability that this person is actually not a user. (5) An individual selected at random tests negative for illicit substances. Determine the probability that he/she actually is a user. not use drugs will get a positive reading on two consecutive tests?

Explanation / Answer

QUestion 1

P(U) = 0.01

P(NU) = 0.99

P(+ l U) = 1 - Pr(False NEgative) = 1- 0.10 = 0.90

P(+ l NU) = Pr(False Positive) = 0.15

P(- l U) = Pr (False Negative) = 0.10

P(- l NU) = 1 - Pr(False Positive) = 1 - 0.15 = 0.85

QUestion 2

Pr(Tested Positve) = Pr(+ l NU) * Pr(NU) + Pr(+ l U) * Pr(U) = 0.15 * 0.99 + 0.90*0.01 = 0.1575

QUestion 3

Pr(NOt a user l Tested positve) = Pr(+ l NU) * Pr(NU)/ [Pr(+ l NU) * Pr(NU) + Pr(+ l U) * Pr(U)]

= 0.15 * 0.99/ (0.15 * 0.99 + 0.90*0.01) = 0.9429

Question 4

Pr(User l Tested negative) = Pr(- l U) * Pr(U) / [Pr(- l U) * Pr(U) + Pr(- l NU) * Pr(NU)]

= (0.10 * 0.01)/ (0.10 * 0.01 + 0.85 * 0.99) = 0.0012

QUestion 5

Pr(Two false positive test in different tests) = [Pr(False positive test)] 2 = Pr(+ l NU) 2 = 0.152 = 0.0225

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