The probability of having HIV for someone who tests positive

Question

Assume one person out of 10,000 is infected with HIV, and there is a test in which 2.5% of all people test positive for the virus although they do not really have it. If you test negative on this test, then you definitely do not have HIV. Let H be the event of having HIV and T be the event of testing positive. Find the following.

(a) Pr(T|H), the probability of testing positive for someone with HIV.

(b) Pr(H ∩ T), the probability of having HIV and testing positive.

(c) Pr(T|H'), the probability of testing positive for someone without HIV.

(d) Pr(H'∩ T), the probability of not having HIV and testing positive.

(e) Pr(T), the probability of testing positive.

(f) Pr(H|T), the probability of having HIV for someone who tests positive.

I already get Pr(H) = 0.0001, Pr(H') = 0.9999 ,Pr(T|H')= 0.025 , Pr(H|T)= 0.0395 -Calcaluted with bayes theorem Now the question is how to do part (b) , part (d), and part (e)

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Nov 20 '21 at 09:40
what does "if you test negative on the test then you definitely don't have HIV" translate to in terms of a probability? — Henno Brandsma, Nov 20 '21 at 09:45
The question seems to be poorly worded: Clarification requested: Are we to assume that in a group of exactly $(10000)$ people, exactly $(1)$ of whom has HIV, that if all $(10000)$ are tested, then the $(1)$ who has HIV will test positive, and $(249)$ of the other $(9999)$ will also test positive? — user2661923, Nov 20 '21 at 09:45
@HennoBrandsma You and I just approached the same question from different directions. Apparently, we are both waiting for the OP (i.e. original poster) to clarify the problem. — user2661923, Nov 20 '21 at 09:47
@HennoBrandsma Good catch, thanks - I edited my comment/question. — user2661923, Nov 20 '21 at 09:49
$P(T) = P(T|H)P(H) + P(T|H')P(H')$ or other such facts are options. — Henno Brandsma, Nov 20 '21 at 09:51
This only works if you test people at random. You don’t. You test people who are at higher risk than average. — gnasher729, Nov 21 '21 at 07:50

ryang · Accepted Answer · 2021-11-21T04:18:06.560

p: HIV prevalance (ignoring prior risk factors)
v: test sensitivity
f: test specificity
H: HIV-positive
H': HIV-negative
+: Positive test result
-: Negative test result

We are given that \begin{align}p&=0.0001,\\1-f=0.025,\text{ i.e., }f&=0.975\\v&=1.\end{align}

Referring to the above probability tree and given values,

(a) Pr(T|H)

$P(T|H)=v$

(b) Pr(H ∩ T)

$P(H ∩ T)=pv$

(d) Pr(H'T)

$P(H'∩ T)=(1-p)(1-f)$

(e) P(T)

$P(T)=P(H∩T)+P(H'∩T)$

(f) Pr(H|T)

Pr(H|T)= 0.0395

\begin{aligned}P(H|T)&=\frac{P(H∩T)}{P(H∩T)+P(H'∩T)}\\&=\frac{pv}{pv+(1-p)(1-f)}\\&=0.398\%.\end{aligned} (How did you obtain your derived value? It differs from mine.)

More information here: The accuracy of a medical test.

The probability of having HIV for someone who tests positive

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