Medical Testing & Conditional Probability

Question

A certain genetic condition affects 5% of the population in a city of 10,000. Suppose there is a test for the condition that has an error rate of 1% (i.e., 1% false negatives and 1% false positives). 

Consider the values that would complete the table below.

	Has condition	Does not have condition	Totals
Test positive
Test negative
Totals

What is the probability (as a percentage) that a person does not have the condition if he or she tests negative? (Round your answer to once decimal place.) 

Answer 1

Hi Catherine,

 

                           Has condition   Does not have condition      Total

 

Test Positive            0.83898            0.16100                      0.99998                                                               

Test Negative           0.00053            0.99947                      1.00000                                       

Total                       0.83951            1.16047                                 

 

G: a person has the genetic disorder; P(G) = 0.05

G': a person does not have the genetic disorder; P(G') = 0.95

p : an event of a positive test; P(p|G) = 0.99, P(p|G') = 0.01

n : an event of a negative test; P(n|G') = 0.99, P(n|G) = 0.01

 

P(G|p)= P(p|G)P(G)/(P(p|G)P(G) + P(p|G')P(G'))

          = (0.99*0.05)/(0.99*0.05 + 0.01*0.95)

          = 0.0495/0.0590

          =0.83898

P(G'|p) = P(p|G')P(G')/(P(p|G')P(G') + P(p|G)P(G))

           =(0.01*0.95)/(0.01*0.95 + 0.99*0.05)

           =0.0095/0.0590

           =0.16100  

P(G'|n) = P(n|G')P(G')/(P(n|G')P(G') + P(n|G)P(G))

            =(0.99*0.95)/(0.99*0.95 + 0.01*0.05)

            =0.9405/0.941

            =0.99947

P(G|n) = P(n|G)P(G)/(P(n|G)P(G) + P(n|G')P(G'))

           =(0.01*0.05)/(0.01*0.05 + 0.99*0.95)

           =0.0005/0.941

           =0.00053

 

Probability that a person does not have the genetic condition and that the test was negative is 99.9%

Answer 2

.............................