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Expert-verified Found in: Page 475 ### Discrete Mathematics and its Applications

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?

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

See the step by step solution

## Step-1: Given Information

1. 8 % of all bicycle racers use steroids.
2. Bicyclists who use steroids test positive 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:

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

## 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,

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

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 $$= \frac{{0.0768}}{{0.1596}} \approx 0.4812$$ ### Want to see more solutions like these? 