Information Technology Reference
When doing an annual check, assume that 85% of the people with lung cancer will
show positive for the chest x-ray test. About 6% of the people without lung cancer will
also show positive for the chest x-ray test.
The second test called CT scan is done independently. It returns positive for 85% of the
people with lung cancer; its false rate is 0.1%.
Answer: For first part of the question, we can use the result in the previous section. Here is
the repeat: For x-ray test, we have the following:
strength(x-ray test) = P(positive x-ray | cancer) / P(positive x-ray | ~cancer)
= 0.85 / 0.06 = 14.17
For CT scan, we have:
strength(CT scan test) = P(positive CT scan | cancer) / P(positive CT scan | ~cancer)
= 0.85 / 0.001 = 850
For the second part of the problem (the distance each test sways our beliefs), we will
proceed as follows:
We started in the initial world with following probabilities:
P(cancer) = 0.2%, P(healthy) = 99.8%
For a person in this initial world, the probability of having lung cancer is 0.2% (pretty low).
If we use the x-ray as a membership test, then the probability become following (already
calculated in previous sections):
P(cancer | positive x-ray) = 2.8%, P(healthy | positive x-ray) = 97.2%
The x-ray test shifted our view from P(cancer) = 0.2% to P(cancer | positive x-ray) = 2.8%. It
is a positive evidence. The percentage increase is 2.6%.
Now, let's see how much the CT scan test will shift our view. Starting from the initial world,
if we use the CT scan as a membership test, then the probability can be calculated as
We use the Bayes' theorem:
And plug in the following data:
P(cancer) = 0.2% (20 out of 10,000 have cancer)
P(~cancer) = 99.8% (9980 out of 10,000 have no cancer)
P(positive CT scan | cancer) = 85%