Editing Base rate fallacy (section)

===Example 2: Drunk drivers===
Imagine that a group of police officers have [[breathalyzer]]s displaying false drunkenness in 5% of the cases in which the driver is sober. However, the breathalyzers never fail to detect a truly drunk person. One in a thousand drivers is driving drunk. Suppose the police officers then stop a driver at random to administer a breathalyzer test. It indicates that the driver is drunk. No other information is known about them.

Many would estimate the probability that the driver is drunk as high as 95%, but the correct probability is about 2%.

An explanation for this is as follows: on average, for every 1,000 drivers tested,
* 1 driver is drunk, and it is 100% certain that for that driver there is a ''true'' positive test result, so there is 1 ''true'' positive test result
* 999 drivers are not drunk, and among those drivers there are 5% ''false'' positive test results, so there are 49.95 ''false'' positive test results
Therefore, the probability that any given driver among the 1 + 49.95 = 50.95 positive test results really is drunk is <math>1/50.95 \approx 0.019627</math>.

The validity of this result does, however, hinge on the validity of the initial assumption that the police officer stopped the driver truly at random, and not because of bad driving. If that or another non-arbitrary reason for stopping the driver was present, then the calculation also involves the probability of a drunk driver driving competently and a non-drunk driver driving (in-)competently.

More formally, the same probability of roughly 0.02 can be established using [[Bayes' theorem]]. The goal is to find the probability that the driver is drunk given that the breathalyzer indicated they are drunk, which can be represented as

<math display="block">p(\mathrm{drunk}\mid D)</math>

where ''D'' means that the breathalyzer indicates that the driver is drunk. Using Bayes's theorem,

<math display="block">p(\mathrm{drunk}\mid D) = \frac{p(D \mid \mathrm{drunk})\, p(\mathrm{drunk})}{p(D)}.</math>

The following information is known in this scenario:

<math display="block">\begin{align}
p(\mathrm{drunk}) &= 0.001,\\
p(\mathrm{sober}) &= 0.999,\\
p(D\mid\mathrm{drunk}) &= 1.00,\\
p(D\mid\mathrm{sober}) &= 0.05.
\end{align}</math>

As can be seen from the formula, one needs ''p''(''D'') for Bayes' theorem, which can be computed from the preceding values using the [[law of total probability]]:

<math display="block">p(D) = p(D \mid \mathrm{drunk})\,p(\mathrm{drunk})+p(D\mid\mathrm{sober})\,p(\mathrm{sober})</math>

which gives

<math display="block">p(D)= (1.00 \times 0.001) + (0.05 \times 0.999) = 0.05095.</math>

Plugging these numbers into Bayes' theorem, one finds that

<math display="block">p(\mathrm{drunk}\mid D) = \frac{1.00 \times 0.001}{0.05095} \approx 0.019627,</math>

which is the precision of the test.