2. Baye’s Theorem Working with Conditional Probabilities

3. Drug Testing Suppose it is assumed that about 5% of the population uses drugs. A drug test that is given to employees of a certain company is 95% accurate; that is, it correctly labels a drug user 95% of the time and correctly labels a nonuser 95% of the time. If a random employee takes the drug test and tests positive, what is the probability that he/she is a drug user?

4. Baye’s Theorem 1) P(A and B) = P(A) * P(B | A) 2) P(B and A) = P(B) * P(A | B) Since the P(A and B) is the same as the P(B and A), we can equate statements 1 and 2 3) P(A) * P(B | A) = P(B) * P(A | B) Divide both sides by P(A) to get

5. Solution to Drug Problem Let A = event that a person tests positive on the drug test Let B = event that a person is a drug user We are looking for the probability that a person is a drug user given that their test turned out positive. P(drug user | positive test result) =

6. P(drug user) = 0.05given in problem P(positive test result | drug user) = 0.95given in problem P(positive test result) = P(positive test result and drug user) + 0.95*0.05 P(positive test result and nonuser) 0.05*0.95 = 0.095 P(drug user | positive test result) =

7. Another Method Drug Status Drug User Non User Test result Positive Negative 10000 5009500 0.95*500 = 475 0.95*9500 = 9025 475 25 950 9050 P(drug user | Test positive) =

8. 0.05 0.95 Drug user Non-Drug user Positive Test Negative Test Positive Test Negative Test 0.05 0.95 0.05 0.95 0.05*0.095 = 0.0475 0.95*0.05 = 0.0475 0.05*0.05 = 0.0025 0.95*0.95 = 0.9025