Question

Consider a test for using a particular drug. The test will produce 99% true positive results for drug users and 99% true negative results for non-drug users. Suppose that 0.5% of people are users of the drug. What is the probability that a randomly selected individual with a positive test is a drug user?

Answer 1

Answer:

Step-by-step explanation:

Using  Bayes' theorem, we have:

$P(A|B)= \frac{P(B|A)P(A)}{P(B)}$

$P(A|B)$ is a conditional probability: the likelihood of event A occurring, given that B is true.

$P(B|A)$ is also a conditional probability: the likelihood of event B occurring, given that A is true.

P(A) and P(B) are the marginal probabilities of observing A and B, independently of each other.

We solve thus:

$P(User|+) = \frac{P(+|User)P(User)}{P(+)}$

                 = $\frac{P(+|User)P(User)}{P(+|User)P(User) + P(+|Non-user)P(Non-user)}$

                 = $\frac{0.99 X 0.005}{0.99 X 0.005 + 0.01 X 0.995}$

                 = $\frac{0.00495}{0.00495 + 0.00995}$

                 = $\frac{0.00495}{0.0149}$

                 = $0.3322$ or $33.22%$%

Therefore, if an individual tests positive, it is more likely than not (1 - 33.2% = 66.8%) that they do not use the drug.