On Tue, 20 Apr 2004 16:59:30 -0400, Reef Fish
<[log in to unmask]> wrote:

>On Tue, 20 Apr 2004 15:46:19 -0400, Reef Fish
><[log in to unmask]> wrote:
>
>>The conditional probabilities are NOT multiplicative.
>>
>>Here's a textbook example (in a textbook written by me, coauthored
>>with a professor at the University of Chicago, and used in MANY
>>graduate schools :-)) that will give readers here some IDEA how
>>deceptive the conditional probability concepts are.
>>
>>Let me pose this question first:
>>
>>There is a VERY reliable screening text for AIDS, with 95%
>>reliability, in the sense:
>>
>>P(positive diagnosis | AIDS ) = 0.95   or 5% false negatives.
>>P(negative diagnosis | AIDS ) = 0.95   or 5% false positives.


Ooops!! the 5% false positive is correct, but the statement should read

>>P(negative diagnosis | no AIDS ) = 0.95   or 5% false positives.

The remaining statements and the solution is orrect,
subject to this typo correction.


>>Give it a shot -- to test your knowledge or intuition, or lack thereof.
>>Those familiar with Bayes Theorem should know what the answer is.  :-)
>
>For those few who are actually trying out ... DON'T read the solution
>below until you've tried ...
>
>                        space inserted
>                             for
>                        readers to skip
>                              :
>                              :
>                              :
>                              :
>                              :
>                              :
>               Here's the Bayes Theorem Solution
>
>using an actual figure of P( AIDS ) = 17.2/100,000 = 0.000172,
>
>from http://www.bamc.amedd.army.mil/DCI/ProbTheo/ppframe.htm
>     Click "OBSERVED AIDS PER 100,000" for the 17.2 figure.
>
>Given:               P(AIDS) = 0.000172
>>P(positive diagnosis | AIDS ) = 0.95   or 5% false negatives.
>>P(negative diagnosis |no AIDS ) = 0.95   or 5% false positives.
>
>To abbreviate, Let P(positive diagnosis | AIDS ) be denoted by P(+|A).
>
>
>     P( A | + ) = P (+ AND A)/P(A) = P(+|A)P(A)/P(+)
>                = P(+|A)P(A)/(P(+|A)P(A) + P(+|notA)P(notA)
>
>                = 0.95 * 0.000172 /(.95*.000172) + .05*.999828)
>                = .0001634/.0501548 = 0.00032579
>
>In other words, if the reliable AIDS test is given to ALL divers,
>(assume same unconditional probability as overall population),
>and you are one of those getting a positive diagnosis for AIDS,
>
>......... don't jump to the wrong conclusion and PANIC, because
>the probability you ACTUALLY have AIDS is only 0.0003 given the
>positive diagnosis!
>
>                 P( AIDS | positive diagnosis ) = 0.00032579
>
>That's how SURPRISING conditional probabilities can be!
>For the actual problem given, with P(AIDS) = .0001.
>
>                 P( AIDS | positive diagnosis ) = 0.0018966,
>
>miniscule, about 0.002, as opposed to 0.95 as most layman would think.
>
>It PAYS (in more ways than one :-)) to know some statistics!
>
>-- Bob.
>
>P.S.  I don't have the corresponding PFO figures, but have good
>      reason to believe that the "surprise factor" or the actual
>      conditional probability figure is not far from the example
>      above, for the same statistical reason.