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From: prodeng@fluke.UUCP (Jim Hirning)
Newsgroups: net.math
Subject: Re: A statistics paradox
Message-ID: <711@vax2.fluke.UUCP>
Date: Mon, 19-Sep-83 16:29:12 EDT
Article-I.D.: vax2.711
Posted: Mon Sep 19 16:29:12 1983
Date-Received: Tue, 20-Sep-83 21:55:34 EDT
References: <205@ecn-ed.UUCP>
Organization: John Fluke Mfg. Co., Everett, Wash
Lines: 32


	>Let p = .5 be the probability that a phenomenum will happen to
   >an item. Thus the probability that it won't happen is 1-p = .5
   >So with a population of 2, the probability that the phenomenum
   >happens to at least one of the items will be p+p = 1 , while the probability
   >that the phenomenum will not happen to either is (1-p)(1-p) = .25
   >hence making a total of 1.25 ?!?!
         	>Where did I go wrong ?

      	   >Hao-Nhien Vu (pur-ee!norris)

The problem lies in your calculation of the first probability to be 1.
Obviously this cannot be true (you cannot guarantee to get a "heads"
by flipping two coins).

The way to calculate this probability:  for the phenomenum to happen
to "at least one" of the items, it must 1) happen to exactly one of
them or 2) happen to both of them.

Calling the two items A and B, and the phenomenum happening T, and not
happening F, there are four possible outcomes.  1) A T, B T; 
2) A T, B F; 3) A F, B T; 4) A F, B F.

There are two ways (cases 2 & 3) of the phenomenum happening to exactly
one of the items, and one way (case 1) of it happening to both.  Since
there are four outcomes, the probability of the phenomenum happening to
at least one of the items is 2/4 + 1/4 = 3/4.  Then the other case,
the phenomenum happening to neither, (case 4) has a probability of 1/4.

Thus, 3/4 + 1/4 = 1  as required.

	Debbie Smit