One website gives the risk of infection by HIV for a woman during unprotected vaginal intercourse with an infected male partner as "between 1/1000 and 1/100,000." The CDC estimated the number of undiagnosed persons in the greater U.S. with HIV in 2009 at a conference as 230,000. Using a very rough estimate of the number of males over the age of 18 in the U.S. (150 M--it is of course smaller), one might give a ballpark estimate of the risk of unprotected sex (per event) for a woman as [using 1/1000 times 230,000/150M] about 2/million.(adsbygoogle = window.adsbygoogle || []).push({});

Finding the annual risk for such behavior, with the many simplifications above, one could calculate Probability(one or more encounters that lead to infection) as 1-Probability(no infectious encounters). The probability of no infectious encounters is

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N = (1 - 2/1000000)^365.

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The probability of at least one adverse event in a year is then 1 - N = 0.00073 or roughly 7/10,000. I notice that one non-scholarly article gives the annual probability of being murdered as about 1/16,500.

I don't think statistics help much in this area, but as a back-of-envelope calculation, it seems low. Can anyone suggest a factor that would change the order of magnitude of the risk? Thanks.

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# Medical HIV question: risk of infection

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