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- Path: sparky!uunet!portal!lll-winken!iggy.GW.Vitalink.COM!cs.widener.edu!ukma!mont!mont!stephen
- From: stephen@mont.cs.missouri.edu (Stephen Montgomery-Smith)
- Newsgroups: sci.math
- Subject: Re: Probably you know the probability ;-)
- Message-ID: <stephen.719167970@mont>
- Date: 15 Oct 92 16:52:50 GMT
- References: <1992Oct14.193802.25129@noose.ecn.purdue.edu> <1992Oct15.025957.11871@galois.mit.edu> <1992Oct15.045308.3552@noose.ecn.purdue.edu>
- Organization: University of Missouri
- Lines: 30
-
- In <1992Oct15.045308.3552@noose.ecn.purdue.edu> kavuri@lips.ecn.purdue.edu (Surya N Kavuri ) writes:
-
- >In article <1992Oct15.025957.11871@galois.mit.edu>, jbaez@riesz.mit.edu (John C. Baez) writes:
- >> In article <1992Oct14.193802.25129@noose.ecn.purdue.edu> kavuri@lips1.ecn.purdue.edu (Surya N Kavuri ) writes:
- >> > I am given a "rod" of fixed length l. I took a hammer and
- >> > broke it into three parts.
- >> >
- >> > (a) what is the probability that they form a triangle ?
- >> > (b) what is the probability that they form an isoceles triangle ?
- >>
- >> There is not enough information to answer unless you tell us the
- >> probability distribution of the lengths of the 3 parts. In real
- >> life this would depend in a complicated way on how you broke the
- >> rod.
- >>
-
- > All the points on the rod are equally likely to be breakage points.
- >
-
- It seems to me that one would break the rod into lengths (a,b,c) such
- that all admissible triplets (i.e. a+b+c = l) are equally likely. I
- think that if you pick the break points uniformly along the rod that
- you get a different distribution.
-
- Well, I haven't done any calculations, so I might be wrong. But thats the
- beauty of nn --- write first, think later !!!!
-
- Am I right?
-
- Stephen
-