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- Path: sparky!uunet!pipex!unipalm!uknet!pavo.csi.cam.ac.uk!camcus!cet1
- From: cet1@cus.cam.ac.uk (C.E. Thompson)
- Newsgroups: sci.math
- Subject: Re: Probably you know the probability ;-)
- Message-ID: <1992Oct15.224325.12471@infodev.cam.ac.uk>
- Date: 15 Oct 92 22:43:25 GMT
- References: <1992Oct14.193802.25129@noose.ecn.purdue.edu> <1992Oct15.025957.11871@galois.mit.edu> <1992Oct15.045308.3552@noose.ecn.purdue.edu> <stephen.719167970@mont>
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-
- In article <stephen.719167970@mont>, stephen@mont.cs.missouri.edu
- (Stephen Montgomery-Smith) writes:
- |>
- |> 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.
-
- Well, with any plausible interpretation of the original problem this will
- of course be true: they will all have probability 0. You will need to be
- more precise than that!
-
- |> I
- |> think that if you pick the break points uniformly along the rod that
- |> you get a different distribution.
-
- It is fairly clear to me that the intended distribution is the one in
- which the two breakpoints are uniformly *and independantly* distributed
- along the rod.
-
- Assuming this, the probability of the pieces being capable of forming
- a triangle is 1/4 (not 1/8, as stated by Terry Tao in article
- <1992Oct15.010724.6653@Princeton.EDU>, who had the right idea but
- must have drawn his diagram wrong). Given that they can form a triangle
- the conditional probability that it is acute is 12 log 2 - 8 = 0.31776...
- for a total probability of 3 log 2 - 2 = 0.07944...
-
- The probability of the triangle being isosceles (or right angled) is,
- of course, zero.
-
- Chris Thompson
- JANET: cet1@uk.ac.cam.phx
- Internet: cet1@phx.cam.ac.uk
-