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Q5E

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Discrete Mathematics and its Applications
Found in: Page 475
Discrete Mathematics and its Applications

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

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Short Answer

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 =0.07680.1596≈0.4812

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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:

P(F|E)=P(E|F)P(F)P(E|F)P(F) + P(E|F¯ )P( F )

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(E) = 0.08P(E|F) = 0.96P(E|F ) = 0.09

Using the complement rule, we can have

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

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