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- From: kavuri@lips.ecn.purdue.edu (Surya N Kavuri )
- Subject: Re: Probably you know the probability ;-)
- Message-ID: <1992Oct15.053443.4372@noose.ecn.purdue.edu>
- Sender: news@noose.ecn.purdue.edu (USENET news)
- Organization: Purdue University Engineering Computer Network
- References: <1992Oct14.193802.25129@noose.ecn.purdue.edu> <1992Oct15.045308.3552@noose.ecn.purdue.edu>
- Date: Thu, 15 Oct 1992 05:34:43 GMT
- Lines: 44
-
- In article <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.
- >
- >
- > There is a part (c) to this problem that I later added.
- >
- > (c) what is the probability that they form an acute-angled triangle ?
- >
- >
- > Surya Kavuri
- >
- >
-
- It appears that the above "breakage procedure" is still ambiguous.
-
- Let us redefine the problem this way.
-
-
- " I have a rod of length l. I take TWO hammers and hit the rod
- simultaneously to break the rod into three parts. All points
- on the rod have equal probability of being the breakage points
-
-
-
- Now consider parts (a), (b) and (c) above.
-
-
- Surya Kavuri
-
-
-