Question

Assume that among the general population, 13 people out of every 1000 have HIV/AIDS. A pharmaceutical company designs a screening test for HIV/AIDS that has a 99% sensitivity and an 89% specificity. Given that someone has a positive test result, what is the probability they have HIV/AIDS?

Answer 1

The probability that a person has HIV\AIDS given that the person was tested positive for HOV|AIDS is 0.1060.

Explanation:

Let a set be events that have occurred be denoted as:

S = {A₁, A₂, A₃,..., Aₙ}

The Bayes' theorem states that the conditional probability of an event, say Aₙ given that another event, say X has already occurred is given by:

`P(A_(n)|X)=\fracP(X{\sum\limits^(n)_(i=1)P(X}`

The screening test for HIV/AIDS designed by a pharmaceutical company is being analysed.

Denote the events as follows:

A = a person has HIV/AIDS

X = the test turns out as positive

The information provided is:

`P(A)=0.013\\P(X|A)=0.99\\P(X^(c)|A^(c))=0.89`

The probability that the test result is positive given that the person does not have HIV\AIDS is:

`P(X|A^(c))=1-P(X^(c)|A^(c))=1-0.89=0.11`

Compute the value of P (A|X) using the Bayes' theorem as follows:

`P(A|X)=(P(X|A)P(A))/(P(X|A)P(A)+P(X|A^(c))P(A^(c)))`

`=((0.99* 0.013))/((0.99* 0.013)+(0.11* (1-0.013)))`

`=(0.01287)/(0.01287+0.10857)`

`=0.105978\\\approx0.1060`

Thus, the probability that a person has HIV\AIDS given that the person was tested positive for HOV|AIDS is 0.1060.