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Math Help - this is a diffrent one

  1. #1
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    this is a diffrent one

    Sorry but i am not good at math, can someone help me ,what is the probality of a person contacting hiv in a city with a population of 190,000.with 6,000 people carrying the virus in the city.
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  2. #2
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    Quote Originally Posted by canyouhelpme View Post
    Sorry but i am not good at math, can someone help me ,what is the probality of a person contacting hiv in a city with a population of 190,000.with 6,000 people carrying the virus in the city.
    Not enough information. Over what period? how many times will they
    have unprotected sex? What is the transmission probability per encounter?
    Does this depend on sexual practice? Gaye straight? ...

    RonL
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    sorry

    thanks for you reply,im talking a one time encounter with a female using a Condom,what the transmission probality in this one encounter,and lets just say she has been with 20 men 18 protected 2 unprotected
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    Quote Originally Posted by canyouhelpme View Post
    thanks for you reply,im talking a one time encounter with a female using a Condom,what the transmission probality in this one encounter,and lets just say she has been with 20 men 18 protected 2 unprotected
    You are supposed to tell us the rate, this is not an epidemiology site
    we help with the maths not gathering data.

    (don't worry: one encounter, hetro, protected in this case probably <1/100
    chance of transmission other things being equal)

    RonL
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    Quote Originally Posted by canyouhelpme View Post
    thanks for you reply,im talking a one time encounter with a female using a Condom,what the transmission probality in this one encounter,and lets just say she has been with 20 men 18 protected 2 unprotected
    Your professor gave you such a question?

    The probability that a person have HIV is:
    p=\frac{6000}{190000}\approx .0315
    (Problem is, we do not know the male/female population. I will assume it is equal then what I said is true).

    I will assume when it is protected the probability is zero. Thus, the harlot did it twice with two different men. The 18 make no difference it was protected.
    The probability she did contract HIV is (assume unprotected leads to HIV):
    2(.0315)+(.0315)^2\approx .064

    Thus the probability the harlot is safe is about 94%.
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    Quote Originally Posted by ThePerfectHacker View Post
    Your professor gave you such a question?

    The probability that a person have HIV is:
    p=\frac{6000}{190000}\approx .0315
    (Problem is, we do not know the male/female population. I will assume it is equal then what I said is true).

    I will assume when it is protected the probability is zero. Thus, the harlot did it twice with two different men. The 18 make no difference it was protected.
    The probability she did contract HIV is (assume unprotected leads to HIV):
    2(.0315)+(.0315)^2\approx .064
    Even here we have a problem, as we don't know the infection prob. per
    encounter, but it is definitly less than 1

    Thus the probability the harlot is safe is about 94%.
    There are lots of other things we don't know which make this an impossible
    sum, we need to know the hetro infected population, and ...

    Also it does not look like homework to me...

    All we can say is the use of a condom should make the transmission via
    the obvious route unlikley, but there are other transmission mechanisms
    about which very little is known.

    RonL

    RonL
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    And, unfortunately, a condom is not a perfect guard against the disease either, as they can break, slip off due to improper use, etc. This factor also should be included.

    -Dan
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    thanks

    thank you for your help everyone
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