phil@wubios.wustl.edu (J. Philip Miller) (06/01/90)
[the following was sent to me by:
From: Jimmy Dean <CSVCJLD%NNOMED.BITNET@VM.TCS.Tulane.EDU>
for posting - jpm]
From elementary probability theory, the probability that one is
HIV+ given that one is a male homosexual is the probability that one is
a male homosexual given that one is HIV+, times the probability that one
has AIDS, divided by the probability that one is a male homosexual.
According to the CDC, in the United States, the probability of
being a male homosexual given that one has AIDS is 2/3. Let's assume
that the probability of being a male homosexual given that one is HIV+
is also 2/3. Assuming 1,000,000 HIV+ people in the United States, the
probability of being HIV+ is then 1,000,000/260,000,000. Assume that
the probability of being a male homosexual is 1/20 (half the population
is male). Then, in the United States, the probability of being HIV+
given that one is a male homosexual must be
(2/3)*(1,000,000/260,000,000)/(1/20) = 2/39 or roughly 1/20.
If this number is too low, then more than two thirds on the HIV+
are homosexual men, or more people are HIV+, or fewer people are male
homosexuals. Any comments?