Question

Suppose that the reliability of a HIV test is specified as follows Of people having HIV, $90\%$ of the test detect the disease but $10\%$ go undetected. Of people free of HIV, $99 \%$ of the test are judged HIV-ive but $1 \%$ are diagnosed as showing HIV+ive. From a large population of which only $0.1 \%$ have HIV, one person is selected at random, given the HIV test, and the pathologist reports him/her as HIV+ive. Then the probability that the person actually has HIV is

• A. $0.081$
• B. $0.082$
• C. $0.084$
• D. $0.083$

Answer 1

Answer: (D)

Let $E$ denote the event that the person selected in actually having HIV and $A$ the event that the person's HIV test is diagnosed as positive. We need to find $P(E / A)$. Also, $E^{\prime}$ denotes the event that the person selected is actually not having HIV.
Clearly, $\left\{E, E^{\prime}\right\}$ is a partition of the sample space of all people in the population.
$ P(E)=0.1 \%=\frac{0.1}{100}=0.001$
$ P\left(E^{\prime}\right)=1-P(E)=0.999$
$P(A / E)=P$ (person tested as HIV positive given that he/she is actually having HIV)
$=90 \%=\frac{90}{100}=0.9$
and $P\left(A / E^{\prime}\right)=P($ person tested as HIV +ive given that he/she is actually not having HIV)
$=1 \%=\frac{1}{100}=0.01$
Now, by Bayes' theorem
$P(E / A) =\frac{P(E) P(A / E)}{P(F) P(A / F)+P\left(F^{\prime}\right) P\left(A / F^{\prime}\right)}$
$ =\frac{0.001 \times 0.9}{0.001 \times 0.9+0.999 \times 0.01}=\frac{90}{1089}$
$ =0.083 \text { approx. }$
Thus, the probability that a person selected at random is actually having HIV given that he/she is tested HIV positive is $0.083$.