Question

Suppose that a new Internet company Mumble.com requires all employees to take a drug test. Mumble.com can afford only the inexpensive drug test - the one with a \(5 \%\) false-positive rate and a \(10 \%\) false-negative rate. (That means that \(5 \%\) of those who are not using drugs will incorrectly test positive and that \(10 \%\) of those who are actually using drugs will test negative.) Suppose that \(10 \%\) of those who work for Mumble.com are using the drugs for which Mumble is checking. (Hint: It may be helpful to draw a tree diagram to answer the questions that follow.) a. If one employee is chosen at random, what is the probability that the employee both uses drugs and tests positive? b. If one employee is chosen at random, what is the probability that the employee does not use drugs but tests positive anyway? c. If one employee is chosen at random, what is the probability that the employee tests positive? d. If we know that a randomly chosen employee has tested positive, what is the probability that he or she uses drugs?

Short answer

a. The probability that a randomly chosen employee uses drugs and tests positive is 0.09 \n b. The probability that a randomly chosen employee does not use drugs but tests positive anyway is 0.045 \n c. The probability that a randomly chosen employee tests positive is 0.135 \n d. The probability that a randomly chosen employee uses drugs given a positive test is 0.6667

Step-by-step solution

Understanding the Problem

We are given: The probability that an employee uses drugs (P(D)) = 0.10, The probability that an employee doesn’t use drugs (P(ND)) = 1 – P(D) = 0.90, The probability that the test is positive given the person is a drug user (P(T|D)) = 1 – false negative rate = 1 - 0.10 = 0.90, The probability that the test is positive given the person is not a drug user (P(T|ND)) = false positive rate = 0.05.

Probability of a Drug User Testing Positive

To find P(D and T), we multiply P(D) and P(T|D), thus P(D and T) = P(D) * P(T|D) = (0.10) * (0.90) = 0.09.

Probability of a Non-Drug User Testing Positive

To find P(ND and T), we multiply P(ND) and P(T|ND), thus P(ND and T) = P(ND) * P(T|ND) = (0.90) * (0.05) = 0.045.

Probability that any Employee will Test Positive

To find this, we add all the ways an employee can test positive. Thus, P(T) = P(D and T) + P(ND and T) = 0.09 + 0.045 = 0.135.

Probability that an Employee Uses Drugs Given a Positive Test

Use the formula for conditional probability P(D|T) = P(D and T) / P(T) = 0.09 / 0.135 = 0.6667.