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Calculation of the probability of HIV transmission between sexual partners
Also see: http://bmj.com/cgi/content/full/324/7344/1034#resp3 This is a letter published in the BMJ arguing there is no proof of heterosexual transmission either in the West or in Africa.
Livio Mina. Mathematician and Statistician, Department of Medical Physics, Royal Perth Hospital, Perth, Western Australia
Suppose that for each contact episode there is a constant probability, p of being infected which is independent of any previous contact history.
The number of contacts needed to first contract the disease follows the geometric distribution with probability function
P(n) = p (1p)^{n1}
where P(n) is the probability of first contracting the disease at the nth contact.
If we are interested in knowing the probability of having caught the disease after a given number of contacts (say n) we must sum all the probabilities of first catching the disease at the first, second, third, etc. contact up to n. This is somewhat tedious, and for this question we can turn instead to the Binomial distribution which gives us the distribution of the number of times we would catch the disease (at least notionally) in n contact episodes.
The probability function here is
P(x) = n!/[x!(nx)!] p^{x} (1p)^{nx}
where P(x) is the probability of being infected x times (sic) in n contacts.
The idea of multiple infection may not make a great deal of sense biologically but we can legitimately ask what is the probability that there be no infection at all ( ie. x = 0 ) after n contacts. From the formula we see that this will be (1p)^{n} so that the probability of contracting the disease (regardless of the notion of multiple infections) is 1  (1p)^{n}.
This formula can be put into an EXCEL file (download here) in which these parameters can be varied to make the calculations.
