A patient takes a special cancer test which has the accuracy test_accuracy=99.9%: if the result is positive, then 99.9% of the patients tested will suffer from the special type of cancer. 99.9% of the patients with a negative result do not suffer from the cancer.

Suppose that a patient is tested and scores positive on the test. What is the probability that a patient suffers from the special type of cancer?

Analysis:

We will use Bayes' theorem to find out the probability of the patient having the cancer:

To know the prior probability that a patient has the cancer, we have to find out how frequently the cancer occurs among people. Say that we find out that 1 person in 100,000 suffers from this kind of cancer. Then P(cancer)=1/100,000. So, P(test_positive|cancer) = test_accuracy=99.9%=0.999 given by the accuracy of the test.

P(test_positive) has to be computed:

P(test_positive)=P(test_positive|cancer)*P(cancer)+P(test_positive|no_cancer)*P(no_cancer)

= test_accuracy*P(cancer)+(1-test_accuracy)*(1-P(cancer))

= 2*test_accuracy*P(cancer)+1-test_accuracy-P(cancer)

Therefore, we can compute the following:

P(cancer|test_positive) = (test_accuracy * P(cancer))/(2 * test_accuracy * P(cancer)+1-test_accuracy-P(cancer))

= 0.999 * 0.00001 / (2 * 0.999 * 0.00001 + 1 - 0.999-0.00001)

= 0.00989128497 which is approximately 1%

So, even if the result of the test is positive and the test has accuracy is 99.9%, the probability of the patient having the tested type of cancer is only approximately 1%. This probability of having the cancer after taking the test is relatively low when compared to the high accuracy of the test, but is much higher than the probability of 1 in 100,000 (0.001%), as known prior to taking the test based on its occurrence in the population.