Question

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The President of a large company with 10,000 employees is considering mandatory cocaine testing for every employee. The test that would be used is 90% accurate, meaning that it will detect 90% of the cocaine users who are tested and that 90% of the nonusers will test negative. This also means that the test gives 10% false positive. Suppose that 1% of the employees actually use cocaine. Find the probability that someone who tests positive for cocaine is, indeed, a user. Hint: The probability fraction is The number who test+and are users The total number who test+ SHOW WORK and FLOW CHART or TREE. (hand in) Answer What is the probability in % of those employees who test + but are not actually drug users? Do the percentages seem to favor a case for or against mandatory drug testing? Explain.

Answer 1

let probability of using Cocaine and not using Cocaine are represented by P(C) and P(NC) . Also being tested positive is P(P). and tested negative P(N)

P/CnP)=0.009 PIP|C)=0.9 PINC)=0.1 Ple=0.01 PICN)=0.001 PINCHP)=0.099 PINC)-0.99 PIPINC)=0.1 PINNC)=0.9 PINCnN)=0.891

from above:

P(User| Positive) =P(C|P) =P(C n P)/P(P) =0.009/(0.009+0.099)=0.0833

P(who test + but are not actually drug users) =P(NC n P) =0.99*0.1 =0.099 ~ 9.9 %

SInce there is a high probability a person who tested positive is not a user and inccorrectly identified as a drug user this favor a case against mandatory drug testing