#### Am I Ill or Not? (After a Test)

Back to the original question: You have three independent, negative test results. What is the probability that you are now healthy? To answer this question, it is not enough to know only the sensitivity, i.e. how reliably the test detects diseased persons.

So as the question changes direction, we jump from the concept of sensitivity to that of** segregancy**. This is another term we introduced in our mini-series. It is the probability that someone who tests negative is actually healthy. However, before we can extend this concept for multiple testing, we must first master the simple test. Without explaining the formula in detail here, we need three probabilities for this:

- The probability of testing negative if the person is not ill.
- The probability of not having the disease.
- The probability for a negative test.

The first probability is just the specificity of the test. In our reader question, a bad test was given as an example with 60 percent sensitivity. For the sake of simplicity, we adopt this assumption for the specificity. Fortunately, real specificities are much higher. So, with 60 percent probability, I will get a negative test if I do not have the disease. Please note the order:

»{negative test} if {not diseased}« = 60 percent.

To get to the reverse probability, that is, »{not ill} if {negative test}«, we must quantify the other two probabilities.

The second probability – not being diseased – depends heavily on prior knowledge. If we randomly select a person from the population of Germany, then we are probably talking about about 99 percent. However, if this person had contact with infected persons or even had symptoms himself, the probability naturally jumps down uncontrollably. Let us therefore assume that there is no cause or symptoms and set the second probability to 99 percent.

This leaves the probability of a negative test. The data on this are not robust. Not all laboratories report negative tests, especially if they are rapid tests. We can only estimate here that we are above 90 percent. For our sought probability we get thus: