Bayes theorem eg2 S13 - Rhea

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↳ Bayes' Theorem

↳ Example 1: Quality Control

↳ Example 2: False Positives

↳ Example 3: Monty Hall Problem

Example 2: The False Positive Paradox

by Maliha Hossain, proud Member of the Math Squad.

keyword: probability, false positive, Bayes' Theorem, Bayes' Rule 

The false positive paradox occurs when false positive tests are more probable than true positive tests. The fewer the number of incidents in the overall population, the higher the likelihood of a false positive test.

The following example employs Bayes' Theorem to illustrate how a high false positive rate can affect the results of a drug test.

A manufacturer claims that its product can detect drug use among athletes 97% of the time (i.e. the test will show a positive 97% of the time given that the athletes used drugs). However, there is a 10% chance of a false alarm (i.e. non drug users will show positive results 10% of the time). Given that only 5% of the team actually use drugs, what is the probability that an athlete who tested positive is a non user?

Let $ D $ be the event that the athlete used drugs.

Therefore, $ D' $ is the event that the athlete did not use drugs.

Let $ Y $ be the event that the test result was positive

Therefore a negative result is described by the event $ Y' $

From the problem statement, we can infer the following.

$ P[D] = 0.05 $

$ P[D'] = 1 - P[D'] = 0.95 $

$ P[Y|D] = 0.97 $ (i.e. the probability of a positive test given the athlete used drugs)

$ P[Y'|D] = 1 - P[Y|D] = 0.03 $ (i.e. the probability of a negative test given the athlete used drugs)

$ P[Y|D'] = 0.1 $ (i.e. the probability that the test was positive given the athlete did not take drugs)

$ P[Y'|D'] = 1 - P[Y|D'] = 0.9 $ (i.e. the probability of a negative test given the athlete did not use drugs)

Now we need to find $ P[D'|Y] $, in other words, the probability that the athlete did not use drugs given the test was positive.

By Bayes' Theorem, we have that

$ \begin{align} P[D'|Y] &= \frac{P[Y|D']P[D']}{P[Y|D]P[D] + P[Y|D']P[D']} \\ &= \frac{0.01\times 0.95}{0.97 \times 0.05 + 0.1 \times 0.95} \\ &= 0.662 \end{align} $

So we see that a positive test is more likely to be a false alarm since the false positive rate is too high for this low incidence population.

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