Question

Enzyme immunoassay (EIA) tests are used to screen blood specimens for the presence of antibodies to HIV, the virus that causes AIDS. Antibodies indicate the presence of the virus. The test is quite accurate but is not always correct. Suppose that 1% of a large population carries antibodies to HIV in their blood. Of those that carry the HIV antibodies in their blood, 99.85% will have a positive test result and 0.15% will have a false-negative test result. Of those that do not carry the HIV antibodies in their blood, 99.4% will have a negative test result and 0.60% will have a false-positive test result. a) Draw a tree diagram for selecting a person from this population and testing his or her blood. b) Construct a probability table that shows the probabilities for individuals in this population with respect to the presence of antibodies and test results.

Answer 1

We are given here that:
P(HIV) = 0.01, therefore P(no HIV) = 1 - 0.01 = 0.99

Also, for HIV antibodies people, we have here:
P( + | HIV) = 0.9985 and therefore P(- | HIV) = 1 - 0.9985 = 0.0015
P( + | no HIV) = 0.006 and therefore P(- | HIV) = 1 - 0.006 = 0.994

a) From the above computed probabilities, the decision tree here is obtained as:

Positive Test 0.9985 001 0.01 Carrying HIV Garing tov 0.0015 s pacate tort Negative Test 0.99 0.006 Positive Test Not Carrying HIV 0.994 Negative Test

b) The joint probability distribution table here is given as:

	Positive Result	Negative Result
HIV antibodies	0.01*0.9985 = 0.009985	0.01*0.0015 = 0.000015
No HIV	0.99*0.006 = 0.00594	0.99*0.994 = 0.98406