Answer:
a) Pr(drug user| positive test) = 0.3333
b) The probability that he will failed his first test = 0.9703
c) the probability that he is a drug user since failed his second drug test
= 0.961165
Step-by-step explanation:
From the given information:
Suppose that 1% of the employees of a certain company use illegal drugs.
Probability of illegal drug user = 0.01
Probability of user that do not use drug = 1 - 0.01 = 0.99
From the person that is a illegal drug user, the company performs random drug tests that return positive results = 0.99
Therefore, the negative result for illegal drug user = 1 - 0.99 = 0.01
However, it also has a 2% false positive rate.
i.e the probability of the user that do not use drug has a positive result of 2% = 0.02
Thus, the probability of the user that do not use drug has a negative result of = 1 - 0.02
= 0.98
We are tasked to answer the following questions.
a) Steve, an employee at the company, has a positive test. What is the probability that he is a drug user?
i.e This employee we are taking about is a drug user and he has a positive test.
Thus;
Pr(drug user| positive test) = [tex]\dfrac{0.99 \times 0.01}{0.99 \times 0.01+ 0.02 \times 0.99}[/tex]
Pr(drug user| positive test) = [tex]\dfrac{0.0099}{0.0099+0.0198}[/tex]
Pr(drug user| positive test) = [tex]\dfrac{0.0099}{0.0297}[/tex]
Pr(drug user| positive test) = 0.3333
b) Knowing he failed his first test, what is the probability that Steve will fail his next drug test?
The probability that he will failed his first test = ((0.01 × 0.01) + (0.99×0.98))
The probability that he will failed his first test = ( 1 × 10⁻⁴ + 0.9702)
The probability that he will failed his first test = 0.9703
c) Steve just failed his second drug test. Now, what is the probability that he is a drug user?
the probability that he is a drug user since he failed his second drug test using Bayes theorem can be expressed as:
= [tex]\dfrac{0.01 \times(0.99\times 0.99)}{0.01 \times (0.99 \times0.99)+ 0.99(0.02 \times0.02)}[/tex]
the probability that he is a drug user since failed his second drug test
= [tex]\dfrac{0.01 \times(0.9801)}{0.01 \times (0.9801)+ 0.99(4 \times 10^{-4})}[/tex]
the probability that he is a drug user since failed his second drug test
= [tex]\dfrac{0.009801}{0.009801+ 3.96 \times 10^{-4}}[/tex]
the probability that he is a drug user since failed his second drug test
= 0.961165
