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rijaalu
Junior Member

Taiwan
3 Posts

 Posted - 09/04/2010 :  03:23:07 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(+) = 1/5000P(-) = 4999/5000P(H|+) = 1/5000 x 99/100P(H|-) = 4999/5000 x 1/100P(no H|+) = 1/5000 x 1/100P(no H|-) = 4999/5000 x 99/100

someguy

143 Posts

 Posted - 09/06/2010 :  14:05:01 P(HIV) = 1/5000P(no HIV) = 4999/5000You 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) = 0.99P(-|no HIV) = 0.99What 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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