Question

Suppose that 8% of all bicycle racers use steroids, that a bicyclist who uses steroids tests positive for steroids 96% of the time, and that a bicyclist who does not use steroids tests positive for steroids 9% of the time. What is the probability that a randomly selected bicyclist who tests positive for steroids actually uses steroids?

Short answer

Answer

The probability that a randomly selected bicyclist who tests positive for steroids actually uses steroids is$=\frac{0.0768}{0.1596}\approx 0.4812$

Step-by-step solution

Step-1: Given Information

1. 8 % of all bicycle racers use steroids.
2. Bicyclistswhousesteroidstestpositive for steroids 96 % of the time.
3. Bicyclists who do not use steroids test positive for steroids 9 % of the time.

Step-2: Definition and Formula

Bayes’ Probability:

$\mathit{P}\mathbf{\left(}\mathit{F}\mathbf{\right|}\mathit{E}\mathbf{)}\mathbf{=}\frac{\mathbf{P}\mathbf{\left(}\mathbf{E}\mathbf{\right|}\mathbf{F}\mathbf{\left)}\mathbf{P}\mathbf{\right(}\mathbf{F}\mathbf{)}}{\mathbf{P}\mathbf{\left(}\mathbf{E}\mathbf{\right|}\mathbf{F}\mathbf{\left)}\mathbf{P}\mathbf{\right(}\mathbf{F}\mathbf{)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{P}\mathbf{(}\mathbf{E}\mathbf{|}\overline{\mathbf{F}}\mathbf{}\mathbf{)}\mathbf{P}\mathbf{(}\mathbf{}\overline{\mathbf{F}}\mathbf{}\mathbf{)}}$

Step-3: Assumption and Finding Steroids Probability

Let E be the event of test positive for steroids and F be the event of racers use steroids.

It is given that,

$$\begin{gathered} P\left(E\right)=0.08 \\ P\left(E\right|F)=0.96 \\ P(E|\overline{F})=0.09 \end{gathered}$$

Using the complement rule, we can have

$P\left(\overline{F}\right)=1-0.08$

= 0.92

Step-4: Finding P (E|F)

The probability that a randomly selected bicyclist who tests positive for steroids actually uses steroids is as follows:

Bayes’ probability

\(P(F|E) = \frac{{P(E|F)P(F)}}{{P(E|F)P(F) + P(E|\overline F )P(\overline F )}}\)

\( = \frac{{0.96(0.08)}}{{0.96(0.08) + 0.92(0.09)}}\)

\(\begin{aligned}{l} &= \frac{{0.0768}}{{0.1596}}\\ \approx 0.4812\end{aligned}\)

Hence,

The probability that a randomly selected bicyclist who tests positive for steroids actually uses steroids is \( = \frac{{0.0768}}{{0.1596}} \approx 0.4812\)