5.12: Base Rates
- Page ID
- 2368
Learning Objectives
- Compute the probability of a condition from hits, false alarms, and base rates using a tree diagram
- Compute the probability of a condition from hits, false alarms, and base rates using Bayes' Theorem
Suppose that at your regular physical exam you test positive for Disease \(X\). Although Disease \(X\) has only mild symptoms, you are concerned and ask your doctor about the accuracy of the test. It turns out that the test is \(95\%\) accurate. It would appear that the probability that you have Disease \(X\) is therefore \(0.95\). However, the situation is not that simple.
For one thing, more information about the accuracy of the test is needed because there are two kinds of errors the test can make: misses and false positives. If you actually have Disease \(X\) and the test failed to detect it, that would be a miss. If you did not have Disease \(X\) and the test indicated you did, that would be a false positive. The miss and false positive rates are not necessarily the same. For example, suppose that the test accurately indicates the disease in \(99\%\) of the people who have it and accurately indicates no disease in \(91\%\) of the people who do not have it. In other words, the test has a miss rate of \(0.01\) and a false positive rate of \(0.09\). This might lead you to revise your judgment and conclude that your chance of having the disease is \(0.91\). This would not be correct since the probability depends on the proportion of people having the disease. This proportion is called the base rate.
Assume that Disease \(X\) is a rare disease, and only \(2\%\) of people in your situation have it. How does that affect the probability that you have it? Or, more generally, what is the probability that someone who tests positive actually has the disease? Let's consider what would happen if one million people were tested. Out of these one million people, \(2\%\) or \(20,000\) people would have the disease. Of these \(20,000\) with the disease, the test would accurately detect it in \(99\%\) of them. This means that \(19,800\) cases would be accurately identified. Now let's consider the \(98\%\) of the one million people (\(980,000\)) who do not have the disease. Since the false positive rate is \(0.09\), \(9\%\) of these \(980,000\) people will test positive for the disease. This is a total of \(88,200\) people incorrectly diagnosed.
To sum up, \(19,800\) people who tested positive would actually have the disease and \(88,200\) people who tested positive would not have the disease. This means that of all those who tested positive, only
\[\dfrac{19,800}{19,800 + 88,200} = 0.1833\]
of them would actually have the disease. So the probability that you have the disease is not \(0.95\), or \(0.91\), but only \(0.1833\).
These results are summarized in Table \(\PageIndex{1}\). The numbers of people diagnosed with the disease are shown in red. Of the one million people tested, the test was correct for \(891,800\) of those without the disease and for \(19,800\) with the disease; the test was correct \(91\%\) of the time. However, if you look only at the people testing positive (shown in red), only \(19,800 (0.1833)\) of the \(88,200 + 19,800 = 108,000\) testing positive actually have the disease.
True Condition | |||
---|---|---|---|
No Disease 980,000 |
Disease 20,000 |
||
Test Result |
Test Result |
||
Positive 88,200 |
Negative 891,800 |
Positive 19,800 |
Negative 200 |
Bayes' Theorem
This same result can be obtained using Bayes' theorem. Bayes' theorem considers both the prior probability of an event and the diagnostic value of a test to determine the posterior probability of the event. For the current example, the event is that you have Disease \(X\). Let's call this Event \(D\). Since only \(2\%\) of people in your situation have Disease \(X\), the prior probability of Event \(D\) is \(0.02\). Or, more formally, \(P(D) = 0.02\). If \(P(D')\) represents the probability that Event \(D\) is false, then \(P(D') = 1 - P(D) = 0.98\).
To define the diagnostic value of the test, we need to define another event: that you test positive for Disease \(X\). Let's call this Event \(T\). The diagnostic value of the test depends on the probability you will test positive given that you actually have the disease, written as \(P(T|D)\), and the probability you test positive given that you do not have the disease, written as \(P(T|D')\). Bayes' theorem shown below allows you to calculate \(P(D|T)\), the probability that you have the disease given that you test positive for it.
\[P(D|T)=\frac{P(T|D)P(D)}{P(T|D)P(D)+P(T|D')P(D')}\]
The various terms are:
\(P(T|D) = 0.99\)
\(P(T|D') = 0.09\)
\(P(D) = 0.02\)
\(P(D') = 0.98\)
Therefore,
\[P(D|T)=\frac{(0.99)(0.02)}{(0.99)(0.02)+(0.09)(0.98)}=0.1833\]
which is the same value computed previously.