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anonymous
 4 years ago
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?
anonymous
 4 years ago
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0P[H]= P[ * H] / P[H] .01/(1/5000)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is a baye's formula problem, right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0easiest way to see the answer is to do it with actual numbers. then we can use the formula, although i have to go soon.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0suppose 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that was wrong, sorry. last line should be out of the remaining 999,800 1% or 9,998 will test positive

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now you can just go ahead and compute

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so that 1% accuracy is shared among those wrongly tested positive and wrongly tested negative or it is each 1%?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that is what i am assuming from the wording, yes

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0usually false positive and false negative are different percents, but in this case you were just given one number

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.09998+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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