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rijaalu
Junior Member

Taiwan
3 Posts

Posted - 09/04/2010 :  03:23:07  Show Profile  Reply with Quote
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 virus? What is P[H|+], the conditional probability that a random chosen person has the HIV virus given that the person tests positive?

I started working it, but i cannot continue further . Please i will be Glad if someone help me ! Some of my sloution is below.

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


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someguy
Advanced Member

Canada
143 Posts

Posted - 09/06/2010 :  14:05:01  Show Profile  Reply with Quote
P(HIV) = 1/5000
P(no HIV) = 4999/5000


You state "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."
Lets interpret this in the easiest way.

P(+|HIV) = 0.99
P(-|no HIV) = 0.99


What do you think the following probabilities are?
P(-|HIV)
P(+|no HIV)

Hint:
P(-|HIV) + P(+|HIV) = ?
P(-|no HIV) + P(+|no HIV) = ?


To find P(HIV|+), use Bayes rule.

P(HIV|+)
= P(+ and HIV) / P(+)
= P(+ and HIV) / ( P(+ and HIV) + P(+ and no HIV) )
= P(HIV)P(+|HIV) / ( P(HIV)P(+|HIV) + P(no HIV)P(+|no HIV) )

Plug in the values into the last line to find P(HIV|+).
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