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 $0.083$.