If the applicant has used illegal drugs the test has a 90 pe

If the applicant has used illegal drugs, the test has a 90 percent chance of a positive result. If the applicant has not used illegal drugs, the test has an 85 percent chance of a 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?

Solution

let A be the event denoting the test has a positive result.

B be the event that a randomly selected applicant has used illegal drugs.

given that P[B]=0.04

P[A|B]=0.90

P[A^c|B^c]=0.85

so P[B^c]=1-0.04=0.96

P[A|B^c]=1-0.85=0.15

we are to find P[the applicant has used illegal drug given that the test is positive]=P[B|A]

using Bayes\' theorem

P[B|A]=P[A|B]*P[B]/P[A]=P[A|B]*P[B]/(P[A|B]*P[B]+P[A|B^c]*P[B^c])

          =0.90*0.04/(0.90*0.04+0.15*0.96)=0.2 [answer]