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. What is the probability that the person actually has HIV?

$0.083$

The correct answer is Option (2) → $0.083$ ##

Let $E$ denote the event that the person selected is actually having HIV and $A$ the event that the person's HIV test is diagnosed as +ive. We need to find $P(E|A)$.

Also $E'$ denotes the event that the person selected is actually not having HIV.

Clearly, $\{E, E'\}$ is a partition of the sample space of all people in the population.

We are given that

$P(E) = 0.1\% = \frac{0.1}{100} = 0.001$

$P(E') = 1 - P(E) = 0.999$

$P(A|E) = P(\text{Person tested as HIV+ive given that he/she is actually having HIV}) = 90\% = \frac{90}{100} = 0.9$

$P(A|E') = P(\text{Person tested as HIV+ive given that he/she is actually not having HIV}) = 1\% = \frac{1}{100} = 0.01$

Now, by Bayes' theorem

$ P(E|A) = \frac{P(E)P(A|E)}{P(E)P(A|E) + P(E')P(A|E')}$

$= \frac{0.001 \times 0.9}{0.001 \times 0.9 + 0.999 \times 0.01} = \frac{90}{1089}$

$= 0.083 \text{ approx.}$

Thus, the probability that a person selected at random is actually having HIV given that he/she is tested HIV+ive is $0.083$.