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suppose,H be the event for HIV virus,H' for not HIV.pos for postive,neg for negative. How low would P(H) have to be so that the conclusion should be "no HIV" even if the result test is positive? How can i calculate it? answer :0.0206 i have solved a problem,which tries to find P(H|pos) using these values P(H)=0.15,P(H')=0.85, P(Pos|H)=0.95,P(Pos|H')=0.2

2 Answers 2

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It seems like you are asking: "how low does P(H) have to be in order for one to be certain that the patient does not have HIV?"

Well, just using common sense, that would have to be 0, because if there was any non-zero chance of there being HIV, then the test being positive could have come from the patient having HIV virus, so you can't say with certainty that the patient does not have HIV.

OK, but given that the answer you are looking for is not 0, you are asking a different question, apparently.

Maybe you are simply asking: what is $P(H|pos)$?

OK, use Bayes' theorem:

$$P(H|pos) = \frac{P(pos|H)*P(H)}{P(pos)}$$

And you have:

$$P(pos) = P(pos|H)*P(H) + P(pos|H')*P(H')$$

OK .... that's $\frac{0.1425}{0.3125} = 0.456$ ... which isn't 0.026 either, so apparently that is not the question either ...

OK, maybe the question is: For what value of $P(H)$ will $P(H|pos)

OK, so let's solve

$$P(H|pos) = P(H'|Pos)$$

Where we assume that $P(H) = p$ and thus $P(H') = 1-p$

So:

$$P(H|pos) = \frac{P(pos|H)*P(H)}{P(pos)} = \frac{0.95*p}{P(pos)}$$

and

$$P(H'|pos) = \frac{P(pos|H')*P(H')}{P(pos)} = \frac{0.2*(1-p)}{P(pos)}$$

So these are equal when:

$$0.95*p = 0.2 *(1-p)$$

i.e. when:

$$1.15*p = 0.2$$

and thus:

$$p = 0.1739..$$

No, that's not it either ...

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    answer is:0.0206 – 2017-02-24
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    @MohammadMohsin OK, you'll need to rephrase your question then, since it looks like you asked "how low does P(H) have to be in order for one to be certain that the patient does not have HIV" – 2017-02-24
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    This is one confused question - Should of stayed away from the "light" ;). – 2017-02-24
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    sorry for the typing mistakes, i type ever instead of even. – 2017-02-24
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    when i saw the problem first,at first it seems to me 0,but how answer is 0.0206 i cannot figured out – 2017-02-24
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I don't think your question is at all clear, but I will assume that you mean to ask "assuming the data, as in the post, compute $P(H\,|\,pos)$". I'll answer that question.

There are two ways to get a positive result. Either the person has the virus (prob $=.15$) and tests positive (prob $=.95$) or the person hasn't got it (prob $=.85$) but tests positive anyway (prob $=.2$). Thus the probability of getting a positive result on a random subject is $$.15\times .95+.85\times .2= .3125$$

Of that, the portion which is explained by the person actually having the virus is $.15\times .95=0.1425$ so the answer you want is $$\frac {0.1425}{.3125}=\boxed {0.456}$$

Note this does not agree with your answer, but your answer doesn't seem sensible. Getting a positive result should not lower the estimate of the subject's risk exposure.