Question

The sensitivity is about 0.993. That is, if someone has HIV, there is a probability of 0.993 that they will test positive. • The specificity is about 0.9999. This means that if someone doesn’t have HIV, there is probability of 0.9999 that they will test negative. In the general population, incidence of HIV is reasonably rare. It is estimated that the chance that a randomly chosen person has HIV is 0.000025. To investigate the possibility of implementing a random HIV-testing policy with the Elisa test, calculate the following: a. The probability that someone will test positive and have HIV. b. The probability that someone will test positive and not have HIV. c. The probability that someone will test positive. d. Suppose someone tests positive. What is the probability that they have HIV? In light of the last calculation, do you envision any problems in implementing a random testing policy?

Answer 1

(a) The probability that someone will test positive and have HIV is 0.000025.

(b) The probability that someone will test positive and not have HIV is 0.0001.

(c) The probability that someone will test positive is 0.000125.

(d) The probability a person has HIV given that he/she was tested positive is 0.1986.

Explanation:

Denote the events as follows:

X = a person has HIV

Y = a person is tested positive for HIV.

The information provided is:

`P(Y|X)=0.993\\P(Y^(c)|X^(c))=0.9999\\P(X)=0.000025`

Compute the probability of a person not having HIV as follows:

`P(X^(c))=1-P(X)=1-0.000025=0.999975`

Compute the probability of
`(Y^(c)|X)` as follows:

`P(Y^(c)|X)=1-P(Y|X)=1-0.993=0.007`

Compute the probability of as
`(Y|X^(c))` follows:

`P(Y|X^(c))=1-P(Y^(c)|X^(c))=1-0.9999=0.0001`

(a)

Compute the probability that someone will test positive and have HIV as follows:

`P(Y\cap X)=P(Y|X)P(X)\\=0.993*0.000025\\=0.000024825\\\approx0.000025`

Thus, the probability that someone will test positive and have HIV is 0.000025.

(b)

Compute the probability that someone will test positive and not have HIV as follows:

`P(Y\cap X^(c))=P(Y|X^(c))P(X^(c))\\=0.0001*0.999975\\=0.0000999975\\\approx0.0001`

Thus, the probability that someone will test positive and not have HIV is 0.0001.

(c)

Compute the probability that someone will test positive as follows:

`P(Y)=P(Y\cap X)+P(Y\cap X^(c))=0.000025+0.0001=0.000125`

Thus, the probability that someone will test positive is 0.000125.

(d)

Compute the probability a person has HIV given that he/she was tested positive as follows:

`P(X|Y)=(P(Y|X)P(X))/(P(Y)) \\=(0.993*0.000025)/(0.000125)\\ =0.1986`

Thus, the probability a person has HIV given that he/she was tested positive is 0.1986.

As the probability of a person having HIV given that he was tested positive is not very large, it would not be wise to implement a random testing policy.