Suppose that for the general population, 1 in 5000 people carries the human immunodeficiency virus (HIV). A test for the presence oh HIV yields either a positive (+) or negative (-) response. Suppose the test gives the correct answer 99% of the time. What is P[-|H], the conditional probability that a person test negative given the person does have the HIV? What is P[H|+], the conditional probability that a random chosen person has the HIV given that the person tests positive?

You need to draw a probability tree to help you. Having done this, you'll get:

P(+) = 1/5000
P(-) = 4999/5000
P(H|+) = 1/5000 x 99/100
P(H|-) = 4999/5000 x 1/100
P(no H|+) = 1/5000 x 1/100
P(no H|-) = 4999/5000 x 99/100

From those, you should be able to work out the rest.

Do you know about probability trees?