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.

Bicyclists who do not use steroids test positive for steroids 9 % of the time

Step-2: Definition and Formula

Bayes’ Probability:

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

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,

\[\begin{array}{l}P(E) = 0.08\\P(E|F) = 0.96\\P(E|\overline F ) = 0.09\end{array}\]

Using the complement rule, we can have

\[P(\overline F ) = 1 - 0.08\]

\[ = 0.92\]

Step-4: Finding \(P(F|E)\)

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

Bayes’ probability

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

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

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

Hence,

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