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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?

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P[-|H]= P[- * H] / P[H] .01/(1/5000)
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Other answers:

this is a baye's formula problem, right?
easiest way to see the answer is to do it with actual numbers. then we can use the formula, although i have to go soon.
suppose 1,000,000 people are tested. of those 1,000,000, exactly 200 will have the disease (assuming the probabilities are exact) then of those 200, 99% will test positive, that is, 198 will test positive, and 2 will test negative. out of the remaining 999,800 1% or 99,980 will also test positive
that was wrong, sorry. last line should be out of the remaining 999,800 1% or 9,998 will test positive
now you can just go ahead and compute
so that 1% accuracy is shared among those wrongly tested positive and wrongly tested negative or it is each 1%?
1% inaccuracy
that is what i am assuming from the wording, yes
ok thanks
usually false positive and false negative are different percents, but in this case you were just given one number
9998+200 test positive, but of those with positive tests, only 198 actually have the disease, so the probability that you have the disease given that the test is positive is only \(\frac{198}{9998+200}\) now if you redo the problem using only decimals instead of 1,000,000 you will get bays formula

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