Probability of HIV

Question

There is a new home test kit for HIV detection. This test kit is known to be 99% accurate (ie. its specificity and sensitivity are both 99%.) Which of the following statements are correct?

(I) Among those with HIV, roughly 1% tests negative.

(II) Among those who test negative, roughly 1% has HIV.

My guess is that only (I) is correct. (Sensitivity is the probability of positive result given that you have HIV. Specificity is probability of negative result given no HIV). Besides, for (I) is there a difference between "roughly 1% tests negative" and "roughly 1% chance of testing negative" ?

If the test's sensitivity and specificity are both 99%, then its overall accuracy is indeed 99%; however, the converse is not true. — ryang, Dec 09 '21 at 05:17

S.C.B. · Answer 1 · 2017-02-16T04:51:53.927

3

You are correct. Option I follows from the definition of accuracy.

For Option II, there is a counterexample, assume that $99.99\%$ of the population has HIV. Then $1 \%$ of those with HIV will test negative , and $99 \%$ of those without HIV will test negative.

Thus, $0.9999 \%$ of the population will have HIV and test negative, while $0.0099 \%$ of the population will not have HIV and test negative. So among those who test negative, more than half will have HIV. Thus Option II is incorrect. Indeed, for Option II to be true the population of people who have HIV must be the same as those that dont.

Only Option I is correct.

edited Feb 16 '17 at 04:51

answered Feb 15 '17 at 16:01

S.C.B.

22,768

do u mean option I is true? – Loki123 Feb 16 '17 at 04:17
1

@Loki123 Sorry, I misread something. I could swear it was different 12 hours ago. Yes, you're right. Only option I is true. – S.C.B. Feb 16 '17 at 04:20

score 1 · Answer 2 · answered Feb 16 '17 at 04:50

This test kit is known to be 99% accurate (ie. its specificity and sensitivity are both 99%.)

Then we have $\mathsf P(T\mid H)=0.99$, and $\mathsf P(\neg T\mid \neg H)=0.99$

Directly: $\mathsf P(\neg T\mid H)=0.01$, so the first claim is okay.

Bayes' Rule: $\mathsf P(H\mid \neg T) = \dfrac{p~\mathsf P(\neg T\mid H)}{p~\mathsf P(\neg T\mid H)+(1-p)~\mathsf P(\neg T\mid \neg H)} = \dfrac{p}{99-98p}$ where $p$ is the probability of having HIV. For the second claim to be true we would require half the population to have HIV. So the second claim is unlikely. $$p=\frac{99~\mathsf P(H\mid \neg T)}{1+98~\mathsf P(H\mid\neg T)} =\dfrac{99\cdot 0.01}{1+98\cdot 0.01} = \dfrac{1}{2}$$

Probability of HIV

2 Answers2