Help with a Statistics Homework Problem?

A new technology has been developed to test for HIV. If a patient has HIV, there is a probability of .96 the test will detect it. If the person does not have HIV, there is a .05 probability the test will conclude that they do (a false positive). Assume that in reality, only 2 out of 100 people actually have HIV.

a) What is the probability a randomly selected person will have a positive test?
b)Given a randomly selected person has a positive test, what is the probability he actually has HIV?
c)Given a randomly selected person has a positive test, what is the probability he does not have HIV?

If someone could please show me how to work these (any or all) it would be greatly appreciated!

Here is what may be an unusual way to do it. Consider 10,000 people. Of these 2% or 200 will actually have HIV. Of that 200, 96% or 192 will test positive. Of the other 9,800 people who do not have HIV there will still be 5% or 490 that wrongly test positive.

Does this help you answer the question yourself?

Let
H= person has HIV
H' = person does not have HIV
D = detects correctly
D' = does not detect correctly
Let P denote probability
P(D / H)=0.96
P(D' /H') =0.05
P(H)=2/100=0.02
P(H')+98/100 =0.98
DP=Positive diagnosis
DP'= negative diagnosis
a)
P(DP)=P(H)P(D/H)+P(H')P(D'/H') =(0.02) (0.96)+(0.98)(0.05) =0.0602
P(DP')=1-P(DP)=0.9398
b)
P(H/DP) = P(H)P(DP/H) / P(DP) = (0.02)(0.96) / 0.0602 =0.2815
The denominator is from (a)

c)P(H'/DP) =P(H')P(DP'/H') / P(DP')= (0.98)(0.95) / 0.9398=0.9906
The denominator is from 1-(a)