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

@satellite73

Answer 2

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

Answer 3

is it right?

Answer 4

this is a baye's formula problem, right?

Answer 5

yes

Answer 6

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.

Answer 7

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

Answer 8

that was wrong, sorry. last line should be out of the remaining 999,800 1% or 9,998 will test positive

Answer 9

now you can just go ahead and compute

Answer 10

so that 1% accuracy is shared among those wrongly tested positive and wrongly tested negative or it is each 1%?

Answer 11

1% inaccuracy

Answer 12

that is what i am assuming from the wording, yes

Answer 13

ok thanks

Answer 14

usually false positive and false negative are different percents, but in this case you were just given one number

Answer 15

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