Question

All interns for Comcast are required to be tested for drug use. The imperfect test gives positive results (indicating drug use) for 80% of all interns who actually use drugs but also (incorrectly) for 3% of those who do not use drugs. Suppose that only 10% of all of these interns actually use drugs.
If an intern is tested negative, what is the probability that he/she uses drugs?

Answer 1

Answer:

0.0224 or 2.24%

Step-by-step explanation:

The probability that an intern uses drugs, given that the test is negative, is determined by  the probability of false negative divided by the probability of a negative (true or false).

A false negative occurs when an intern who uses drugs (10%) takes the test and gets a negative. Since the test gets it right 80% of the time, the probability of this happening is:

$P(FN) = 0.10*(1-0.80)=0.02$

A true negative occurs when an intern who does not use drugs (90%) takes the test and gets a negative. Since the test gets it wrong 3% of the time, the probability of this happening is:

$P(TN) = 0.90*(1-0.03)=0.873$

Therefore, the probability that an intern uses drugs, given that the test is negative is:

$P(D|N) = \frac{P(FN)}{P(FN)+P(TN)} \\P(D|N) = \frac{0.02}{0.02+0.873} \\P(D|N) = 0.0224 = 2.24\%$

The probability is 0.0224 or 2.24%.

Answer 2

Answer:

don't do drugs or you'll end up like macauley culkin

Step-by-step explanation: