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I'm refreshing on bayes theorem and conditional probability and I ran across these practice problems. I was trucking along until problem 9, which states:

HIV The New York State Health Department reports a 10% rate of the HIV virus
for the “at-risk” population. Under certain conditions, a preliminary screening test
for the HIV virus is correct 95% of the time. (Subjects are not told that they are
HIV infected until additional tests verify the results.) If someone is randomly
selected from the at-risk population, what is the probability that they have the
HIV virus if it is known that they have tested positive in the initial screening?

Seems pretty straight forward. Let P(H) denote the probability of having HIV.

Then,

P(H) = .1
P(~H) = .9
P(+ | H) = .95

and we need to find

P(H | +) = P(H)P(+ | H)/[(P(H)P(+ | H)) + (P(~H)P(+ | ~H))

After running the numbers, they somehow conclude that

P(+ | ~H) = 1 - P(+ | H) = .05 

I figured this out by working backwards from their answer.

The problem is I have no idea how they reached this conclusion. Doing what they did above they should be including both P(- | H) and P(+ | ~H) together since these are the cases that are the opposite of P(+ | H).

The reason I'm confused is this doesn't seem like the proper way to refer to the probability that they tested positive given that they dont have HIV, strictly because what I suppose they did above also includes the case where they tested negative even though they have HIV.

Can someone walk me through their logic here? Thanks!

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  • $\begingroup$ +1. Because this is a self-study question, you should add the self-study tag to your question. $\endgroup$ – Patrick Coulombe Jul 26 '14 at 21:01
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Let $H$ denote the event "has HIV". Let $+$ denote the event "tests positive in initial screening".

HIV The New York State Health Department reports a 10% rate of the HIV virus for the “at-risk” population.

$P(H)=0.1$

Under certain conditions, a preliminary screening test for the HIV virus is correct 95% of the time.

This statement implies both that $P(+|H)=0.95$ and $P(-|\overline{H})=0.95$; if only one of them were 0.95, they'd have to talk about either sensitivity or specificity, rather than accuracy (the intended meaning of 'accuracy' should probably have been made explicit, but as a plain-English phrase, for it to be accurate it must have some level of performance under both $H$ and $\overline{H}$) -- the implication is then that a single number for accuracy would imply it holds for both; while related to the technical meaning in binary classification, I don't think they intended exactly that, but rather the more general sense of 'measuring what it's supposed to measure').

In short, the question could be less ambiguous than it is.

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So I think the question is actually quite poorly worded. First, when they say that the preliminary screening test is correct 95% of the time, I believe that it correctly should be interpreted as the following: P(+|H) + P(- | ~H) = .95. Basically, it's correct for true positives and true negatives 95% of the time.

In the set up of this problem however, the author wasn't too clear and confused what "correct diagnoses" actually means. You picked up on that during your opposite of P(+|H) :-).

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  • $\begingroup$ $P(+|H)+P(−|¬H)$ seems like a ridiculous quantity to consider. I would interpret "95% correct" as either $P(+|H)=.95$,$P(-|\lnot H) = .95$ (as @Glen_b does in his answer), or possibly $P(+|H) P(H) + P(-|\lnot H) P(\lnot H) = .95$ (though not on this kind of question). $\endgroup$ – Dougal Jul 27 '14 at 6:02
  • $\begingroup$ I completely agree, but stating that P(+|H) and P(-|~H) are both .95 as determined correct is also quite odd -- especially given the specificity and sensitivity tradeoffs of these tests. I completely agree that the wordings are quite off. $\endgroup$ – user1357015 Jul 27 '14 at 16:44

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