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Suppose there a medical test is administered to test if a person has a particular disease:

  • If they have the disease, there is 10% probability that the test says they don't have the disease. This is called false negatives.
  • If they don't have the disease, there is 30% probability that the test says they have the disease. This is called false positives.

Suppose that a random patient is given this test. If the test result is positive, what is the probability they have the disease?

Logically, it goes like this Pr(Positive,Positive) = 100% - False Positive = 100% - 30% = 70%.

Suppose that now it is known that the disease only occurs in 10% of the population. Using posterior probability, the probability is Pr(Positive,Positive) = 25%

Why does knowing that the disease only occurs in 10% of population change the probability that the patient has the disease? I'm confused; can someone please help me clear my confusion?

ryang
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i'm ashamed with what i asked
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1 Answers1

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The way to frame and interpret medical tests in general is to understand them as updating one's level of certainty that the patient has the disease:

  • without a medical-test result, the disease prevalence (a measure of disease frequency) can be taken as the patient's probability of having the disease;
  • however, in the context of a medical-test result, the aforementioned probability has changed: its updated value depends not just on the disease prevalence (as before), but now also on the test's sensitivity (true positive rate) and specificity (true negative rate).

Given a positive test result, the (updated) probability $P(D|+)$ that the patient is indeed diseased can be derived from the following probability tree:

enter image description here

p:  disease prevalance and other (prior) risk factors
v:  test sensitivity
f:  test specificity
D:  Diseased
H:  Healthy

\begin{aligned}P(D|+)&=\frac{P(D+)}{P(D+)+P(H+)}\\&=\frac{pv}{pv+(1-p)(1-f)}.\end{aligned} This formula makes clear that $P(D|+)$ is a function of disease prevalence $p,$ test sensitivity $v,$ and test specificity $f.$

It makes sense that information about the test's technical characteristics ($v$ and $f$), as well as the disease prevalence and the patient's prior health ($d$), should refine our knowledge of the probability that the patient has the disease.

Addendum

OP: Does fewer people having the disease increase the probability that the test result is just a false positive?

Yes. From the above probability tree, the probability that the test result is a false positive is $$P(H+)=(1-p)(1-f).$$ So, the lower the disease prevalence $p,$ the greater this probability; in fact, unless the test has 100% specificity, the number of false-positive results is directly proportional to $(1-p).$

ryang
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  • I gave Jmoravitz the chance to write this answer first , but he doesnt seems interested . Yes , from your answer and from the help of Jmoravitz , i finally understand that the disease prevalent must be taken into account to calculate the whole probability a patient have disease . Calculating only the test sensitivity means that i assume either all patient have disease or all patient doesnt have the disease (thanks Jean-Claude Arbaut). Correct me if i'm wrong. – i'm ashamed with what i asked Aug 30 '21 at 00:49
  • Before i accept this answer , can you please add explanation what does the "D" and "H" mean ? – i'm ashamed with what i asked Aug 30 '21 at 01:02
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    @i'mashamedwithwhatiasked 2. “Calculating only the test sensitivity means that i assume either all patient have disease or all patient doesn't have the disease” If you test a batch of samples that are all actually from diseased patients, the percentage of these samples that return from the test with a positive result is the ‘test sensitivity’. – ryang Aug 30 '21 at 07:20