Dolon Web Security Consultants requries all job applicants to submit to a test for illegal drugs. If the applicant has used illegal drugs, the test has 90 percent chance of a positive result. If the applicant has not used illegal drugs, the test has an 85 percent chance of negative result. Actually, 4 percent of the job applicants have used illegal drugs. If an applicant has a positive test, what is the probability that he or she has actually used illegal drugs

20 percent

Step-by-step explanation:

P(Used illegal drugs) = 4%

P(+ test | Used illegal drugs) = 90%        

P(- test | Has not Used illegal drugs) = 85%      

P(+ test ) =

P(+ test and Used illegal drugs) + P(+ test and Has not used Illegal drugs) =

[ P(+ test | Used illegal drugs) * P(Used illegal drugs) ] + [ P(+ test | Has not Used illegal drugs) * P(Has not Used illegal drugs) ],

since

P(+ test and Used illegal drugs) = P(+ test | Used illegal drugs) * P(Used illegal drugs)

P(+ test and Has not used illegal drugs) =P(+ test | Has not Used illegal drugs) * P(Has not Used illegal drugs)  

Thus,

P(+ test ) =  

[ (90%)(4%) ]  +  [ (100% - 85%)(100% - 4%) ] =

0.9*.04 + .15*.96 =

0.18

and,

we want  

P(Used illegal drugs | + test),

and

P(Used illegal drugs and + test) = P(+ test | Used illegal drugs) * P(Used illegal drugs) = 0.9*0.04 = 0.036

and we have that

P(Used illegal drugs and + test) = P(Used illegal drugs | + test) * P(+test)

so that

0.036 = P(Used illegal drugs | + test) * 0.18

so that

P(Used illegal drugs |+ test) =

0.036 / 0.18 =

0.2 =    

20 percent

answer: idk

step-by-step explanation: