Question

Suppose that for the general population, 1 in 5000 people carries the human immunodeficiency virus (HIV). A test for the presence of 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 tests negative given that the person does have the HIV virus? What is P[H|+], the conditional probability that a randomly chosen person has the HIV virus given that the person tests positive?

Answer 1

The answer is "0.019".

Step-by-step explanation:

For HIV positive:
`P[HIV{+}] =(1)/(5000)`

For HIV negative:
`P[HIV{-}] =(4999)/(5000)`

calculating the proability for test gives right reslut:
`P[TV] =(99)/(100)`

calculating the proability for test gives wrong reslut:
`P[TX] =(1)/(100)`

For HIV negative:
`P[HIV{-}] = p[TX]= (1)/(100)`

calculating proability to have HIV:

`P[HIV{+}] = \frac{P[HIV{-} * HIV{-}]}{P{+}}`

`= ((1)/(5000) * (99)/(100))/((1)/(5000) * (99)/(100)+(4999)/(5000) *(1)/(100))\\\\= ((1)/(5000) * 0.99)/((1)/(5000) * 0.99+(4999)/(5000) *0.01)\\\\= ((0.99)/(5000) )/((0.99)/(5000) +(49.99)/(5000))\\\\= (0.000198)/(0.000198 +0.009998)\\\\= (0.000198)/(0.010196)\\\\=0.019`