Question

Steroids A company has developed a drug test to detect steroid use by athletes. The test is accurate $95\%$ of the time when an athlete has taken steroids. It is $97\%$ accurate when an athlete hasn’t taken steroids. Suppose that the drug test will be used in a population of athletes in which $10\%$ have actually taken steroids. Let’s choose an athlete at random and administer the drug test.

(a) Make a tree diagram showing the sample space of this chance process.

(b) What’s the probability that the randomly selected athlete tests positive? Show your work.

(c) Suppose that the chosen athlete tests positive. What’s the probability that he or she actually used steroids? Show your work.

• Determine whether two events are independent.

Short answer

Part (b) The probability is $0.122.$

Part (c) The probability is $0.7787$

Part (a) The tree diagram is

Step-by-step solution

Part (a) Step 1. Given Information

Percentage of athletes that use steroids $10\%=0.10$ Percentage of tests that are accurate when steroids are used $95\%=0.95$ Percentage of tests that are accurate when steroids are not used $97\%=0.97$

Part (a) Step 2. Concept

Formula used:$P\left(A\right)+P\left(B\right)=1$

$P\left(A\right)=1-P\left(B\right)$

Part (a) Step 3. Calculation

Let S stand for the taking of a steroid. The letter NS denotes the absence of a steroid. The letter P stands for a positive outcome. The letter N stands for a negative outcome. Based on the information provided:$$\begin{gathered} P\left(S\right)=0.10 \\ P\left(P\right|S)=0.95 \\ P(N\left|NS\right)=0.97 \end{gathered}$$

Part (b) Step 1. Calculation

Here,

$$\begin{gathered} P\left(P\right)=P(S\cap P)+P(NS\cap P) \\ P(S\cap P)=P\left(S\right)\times P\left(P\right|S) \\ P(NS\cap P)=P(NS)\times P(P\left|NS\right) \end{gathered}$$

The probabilities

$$\begin{gathered} P(S\cap P)=0.10\times 0.95=0.095 \\ P(NS\cap P)=0.90\times 0.03=0.027 \end{gathered}$$

Adding the probabilities together to arrive at the required probability

$=0.095+0.027=0.122$

As a result, $0.122$is the required probability.

Part (c) Step 1. Explanation

From the part b

$$\begin{gathered} P\left(P\right)=0.122 \\ P(S\cap P)=0.095 \end{gathered}$$

In this case,

$P\left(S\right|P)=\frac{P(S\cap P)}{P\left(P\right)}$

Using the probabilities shown above,

$P\left(S\right|P)=\frac{0.095}{0.122}=0.7787$

As a result, $0.7787$ is the required probability.