Question

Drug usage in the workplace costs employers incredible amounts of money each year. Drug testing potential employees has become so prevalent that drug users are finding it extremely hard to find jobs. Drug tests, however, are not completely reliable. The most common test used to detect drugs is approximately 95% accurate. To decrease the likelihood of making an error, all potential employees are screened through two tests, which are independent, and each has 97% accuracy. If a potential employee were a drug user, what is the probability that the person would fail the first test and pass the second test?

Answer 1

We are given :

Probability that a drug user fails the test = 97% = 0.97

So,

Probability that a drug user pass the test = 100-97 = 3% = 0.03

So,

Probability that the person would fail the first test and pass the second test =

= Probability that person would fail the first test * Probability that person would pass the second test

= 0.97 * 0.03 = 0.0291 = 2.91%

Answer 2

Let's use Bayes' theorem to solve this problem. We want to find the probability that a potential employee is a drug user given that they fail the first test and pass the second test.

Let D be the event that the potential employee is a drug user, and let F1 and F2 be the events that they fail the first test and pass the second test, respectively.

From the problem statement, we know that:

P(F1 | D) = 0.05 (the probability of failing the first test given that the person is a drug user) P(F1 | not D) = 0.01 (the probability of failing the first test given that the person is not a drug user) P(F2 | D) = 0.03 (the probability of passing the second test given that the person is a drug user) P(F2 | not D) = 0.97 (the probability of passing the second test given that the person is not a drug user)

We want to find P(D | F1 and F2), the probability that the person is a drug user given that they failed the first test and passed the second test. We can use Bayes' theorem to express this probability in terms of the conditional probabilities above:

P(D | F1 and F2) = P(F2 | D and F1) * P(F1 | D) * P(D) / P(F2 | F1)

We can calculate the denominator P(F2 | F1) using the law of total probability:

P(F2 | F1) = P(F2 | D and F1) * P(D) + P(F2 | not D and F1) * P(not D)

We know that P(not D) = 1 - P(D), and we can calculate P(F2 | not D and F1) using the complement rule:

P(F2 | not D and F1) = 1 - P(not F2 | not D and F1)

where not F2 is the event that the person fails the second test. We can use the conditional probabilities given in the problem statement to calculate P(not F2 | not D and F1):

P(not F2 | not D and F1) = P(not F2 | not D) * P(not D | F1) / P(not F2 | F1)

We know that P(not F2 | not D) = 0.03 and P(not F2 | F1) = 0.01 * P(not D | F1) + 0.97 * P(not F1 | not D). We can calculate P(not F1 | not D) using the complement rule:

P(not F1 | not D) = 1 - P(F1 | not D) = 1 - 0.01 = 0.99

Now we have all the conditional probabilities we need to calculate P(D | F1 and F2):

P(F2 | F1) = P(F2 | D and F1) * P(D) + P(F2 | not D and F1) * P(not D) = 0.03 * P(D) + (1 - 0.01 * P(not D | F1) - 0.97 * 0.99) * (1 - P(D)) P(not F2 | not D and F1) = 1 - P(not F2 | F1) * P(not F2 | not D) / P(not F1 | not D) = 1 - 0.01 * P(not D | F1) / 0.99 P(D | F1 and F2) = P(F2 | D and F