Does the new hire use drugs? Many employers require prospective employees to

take a drug test. A positive result on this test suggests that the prospective employee uses

illegal drugs. However, not all people who test positive use illegal drugs. The test result

could be a false positive. A negative test result could be a false negative if the person

really does use illegal drugs. Suppose that $4\%$of prospective employees use drugs and

that the drug test has a false positive rate of $5\%$and a false negative rate of$10\%$.

Imagine choosing a prospective employee at random.

a. Draw a tree diagram to model this chance process.

b. Find the probability that the drug test result is positive.

c. If the prospective employee’s drug test result is positive, find the probability that she

or he uses illegal drugs.

We have been given that $4\%$of prospective employees use drugs and that the drug test has a false positive rate of $5\%$and a false negative rate of $10\%$.

We need to draw a tree diagram for this process.

The tree diagram for the given process is shown below.

We have been given that $4\%$of prospective employees use drugs and that the drug test has a false positive rate of $5\%$and a false negative rate of $10\%$.

We need to find out the probability that the drug test result is positive.

Let D$=$Drug use, Dn$=$no drug use ,P$=$positive test result ,Pn$=$negative test result

$P\left(D\right)$$=\frac{4}{100}=0.04$

$P$$\left(P\overline{)D}\right)$$=$$1-P\left(Pn\overline{)D}\right)=1-0.10=0.90$

$P\left(Dn\right)=1-P\left(D\right)=1-0.04=0.96$

$P\left(P\overline{)Dn}\right)=\frac{5}{100}=0.05$

$P\left(PandD\right)=P\left(D\right)\times P\left(P\overline{)D}\right)=0.04\times 0.90=0.036$

$P\left(PandDn\right)=P\left(Dn\right)\times P\left(P\overline{)Dn}\right)=0.96\times 0.05=0.048$

Now,

$$\begin{gathered} P\left(P\right)=P\left(PandD\right)+P\left(PandDn\right) \\ =0.036+0.048 \\ =0.084 \\ =8.4\% \end{gathered}$$

We have been given that the prospective employee’s drug test result is positive.

We need to find out the probability that she or he uses illegal drugs.

Using the concept of conditional probability

$$\begin{gathered} P\left(D\overline{)P}\right)=\frac{P\left(DandP\right)}{P\left(P\right)} \\ =\frac{0.036}{0.084} \\ =\frac{36}{84}\approx 0.4286 \\ =42.86\% \end{gathered}$$