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Q5E

Expert-verifiedFound in: Page 475

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

**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?**

**Answer**

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

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

**Bayes’ Probability:**

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

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

It is given that,

$P\left(E\right)=0.08\phantom{\rule{0ex}{0ex}}P\left(E\right|F)=0.96\phantom{\rule{0ex}{0ex}}P(E|\overline{)F})=0.09$

Using the complement rule, we can have

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

= 0.92

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 \( = \frac{{0.0768}}{{0.1596}} \approx 0.4812\)

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