What is the conditional probability of this case?

Question

A diagnosis for a test is such that it (correctly) detects the disease in 85% of the individuals who actually have the disease. Also, if a person does not have the disease, the test will report that he or she does not have it with probability 0.80. Only half of one percent or 0.5% of the population has the disease in question. If a person is chosen at random from the population and the diagnostic test indicates that he has the disease, what is the conditional probability that he does, in fact, have the disease?

Answer 1

Abood,

 

You can either solve this formally defining variables, etc, or just use a 2 X 2 probability matrix to handle this problem. I prefer the latter being a visual person.

 

TEST RESULT     |         Among the 0.005 of     |     Among the 0.995 of

                        |         Population that has      |     Population that does

                        |          this disease                |     NOT have this disease

-------------------------------------------------------------------------------------------

                        |                                           |    False Positive                                   

Positive             |        p = 0.00425                 |         p = 0.1990                                   

                        |                                           |                                     

-------------------------------------------------------------------------------------------

                        |   False Negative                   |                                         
Negative           |                                           |
                        |       p = 0.00075                  |        p = 0.796 
-------------------------------------------------------------------------------------------

 

You get these p table entries by applying Test accuracy statistics to the incidence of the disease in the population.

 

Your answer = Prob [ has the disease/ given a pos Test result]  = 0.00425/(0.0025 + 0.199) = 0.00425/0.20325 = 0.0209

 

Here's the problem with screening tests that have significant inaccuracy when applied to rare diseases. Without any test, your chance of having the disease equals its incidence in the population = 0.005 decimal or 1/2 of 1%.  Taking this test and getting a positive reading ups this probability by a factor of about 4 to 0.0209 decimal or 2.09%.  That is still very small, and your gain in assurance is basically meaningless, medically speaking in the real world.

 

But now look at the FALSE Positives from taking the test = 0.199 decimal or about 20%. Now these people are going to undergo various appropriate medical procedures that could involve blood tests, x-rays, MRIs, even surgery, biopsies, etc., with their attendant risks that could be in the 2% range (not counting the time, cost, and anxiety involved).  So the medical profession benefits in providing these services, but you really don't since your chance of having the disease is miniscule because its incidence is rare.

 

BTW, another way of looking at that 0.0209 answer is 1/0.0209 = 47.8.  So you have to test 48 people to find one real case of the disease on average, and 9.8 or about 9-10 people are going to get false positives that will require followups.  That test is just not accurate enough to be very useful in my judgment as an impartial observer, being not the patient or in the medical profession.