Question

A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (that is, if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Answer 1

Let P( A ), P( B ) and P( C ) be the probabilities that are defined below,

The probability of person has disease is P( A ).

The probability of person does not have the disease is P( B ).

The probability of test result is positive is P( C ).

The probability that a person has the disease give that his test result is positive P( A C ),

P( A C )= P( A )P( C A ) P( B )⋅P( C B )+P( A )⋅P( C A ) (1)

The probability of a person that the person has disease is,

P( A )=0.1% =0.001

The probability that the test is positive, given that the person has the disease,

P( C A )=99% =0.99

The probability of a person that the person does not has disease is,

P( B )=99.9% =0.999

The probability that the test result is positive, given that the person does not have the disease,

P( C B )=0.005

Put these values in equation (1),

P( A C )= ( 0.001×0.99 ) ( 0.999×0.005 )+( 0.001×0.99 ) = 990 5985 = 22 133

Thus, the required probability is 22 133 .