This might seem somewhat counterintuitive as we know the test is quite accurate. The point is that the disease is also very rare. Thus, there are two competing forces here, and since the rareness of the disease (1 out of 10,000) is stronger than the accuracy of the test (98 or 99 percent), there is still good chance that the person does not have the disease.
Another way to think about this problem is illustrated in the tree diagram in Figure . Suppose 1 million people get tested for the disease. Out of the one million people, about 100 of them have the disease, while the other 999,900 do not have the disease. Out of the 100 people who have the disease 100×=99 people will have positive test results. However, out of the people who do not have the disease 999,900×.02=19998
999,900 ×=19998 people will have positive test results. Thus in total there are 19998 people with positive test results, and only 99 of them actually have the disease. Therefore, the probability that a person from the “positive test result“ group actually have the disease is
P(D|T)=99/9919998 99=
The Problem:
A certain disease affects about 1 out of 10,000 people. There is a test to check whether the person has the disease. The test is quite accurate. In particular, we know that
the probability that the test result is positive (suggesting the person has the disease), given that the person does not have the disease, is only 2 percent;
the probability that the test result is negative (suggesting the person does not have the disease), given that the person has the disease, is only 1 percent.
A random person gets tested for the disease and the result comes back positive. What is the probability that the person has the disease?
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