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Title: WILLYWUTANG HAS STDS? Post by rugga on Sep 9th, 2002, 3:34am Quote:
Trick question! Not to offend Willywutang, but anyone who would attempt to satisfy three women (one really skanky looking) at one sitting :o -- improperly using comdoms to boot -- has a higher chance of having an STD than a "random person". Without getting graphic, Willy's willy may have been exposed to 1, 2, or all 3 of the women depending on when and where the tear occurred:-[ This is not even considering what this event implies about willy's previous (probably impressive) social life. More seriously, this is one reason why doctors/nurses take a patient's history along with their blood when doing tests. The patient's history gives context for interpreting the test results by way of adjusting the a priori probability for that individual. Of course in real life it's impossible to do this adjustment quantitatively, but it's more information on which to base the necessary human judgement. Now, if the question concerned billybutane about whom I know nothing, then I could reason as follows... Using Bayes' rule (see any probablity textbook): p(STD | positive) = p(STD) p(positive | STD) / p(positive) = p(STD) p(positive | STD) / [ p(positive | STD) p(STD) + p(positive | no STD) p(no STD) ] p(STD) = 0.001 because billy is a random person p(no STD) = 0.999 p(positive | STD) = 0.93 = success rate of test p(positive | no STD) = 0.07 = error rate of test Plugging in the numbers, billybutane finds out .... only 1.3% chance of STD. He (she?) can relax. Or more accurately, I can relax because I don't know anything about him that might affect the 0.001. Of course he may know something about himself that raises his risk like our poor friend willy. I suppose another way to approach the question is to consider the chance that willywutang caught an STD during this one encounter, but we really need more information to approach it that way, like the chance of transmitting an STD in one encounter, how many of the three women he was exposed to, whether skanky looking people are more likely to have STDs, etc. |
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Title: Re: WILLYWUTANG HAS STDS? Post by Chronos on Sep 11th, 2002, 7:57pm I see four or five ways to interpret this problem: 1: The figure of .001 is the a priori probability of Willy having an STD, with all relevant factors considered. In this case, enough information is presented in the problem. 2: The skanky girl had a .001 chance of having an STD, and the other two girls (and, initially Willy) are presumed clean. In this (and following) case(s), we need to know what the probability of disease transmission is. 3: All three girls have a .001 chance of having an STD, but Willy is initially presumed clean. Again, we need to know the transmission probability. 4: Before the encounter, all four people have an a priori .001 chance of having an STD. After the encounter, due to being exposed to each other, they all have probabilities higher than that. So, Willy, which is it? |
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Title: Re: WILLYWUTANG HAS STDS? Post by william wu on Sep 28th, 2002, 1:46pm guess i goofed. hehe, sorry guys. ok, to salvage the riddle, i have added the following line: Quote:
i think this fixes it, but let me know otherwise. love the sense of humor btw, you guys are the best :) |
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Title: Re: WILLYWUTANG HAS STDS? Post by luke's new shoes on Nov 12th, 2002, 3:50pm there are two possibilities 1. he has stds and the kit is correct 2. he doesnt have stds and the kit is wrong probability of 1. is 0.001*0.93 = 0.00093 probability of 2. is 0.999*0.07 = 0.06993 because they are the only two possibilites, their sum must add up to one (well the sum of a common ration of the two must add up to one) i.e. 0.00093*x + 0.06993*x = 1 (0.00093 + 0.06993)*x = 1 0.07086*x = 1 x = 14.1 therefore probabiltiy of willy having an std is 0.00093*14.1 which equals 0.013 |
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