Dubinse <[EMAIL PROTECTED]> wrote in sci.stat.edu:
>The story is about six
> students who go off on a trip and get drunk the weekend before
>their statistics final.
...
>caught in a storm and their car blew a tire and ended up
>in a ditch and they needed brief hospitalization etc.
...
> The one question was, "Which tire?" I remember that
>the likelihood of all four pickng the same tire was quite small, but I
>forgot how to calculate it explicitly
>
>I would particularly appreciate a general solution (N students, M tires).
(1/M)^(N-1) is the probability that N students responding randomly
on M choices will all make the same response.
In your scenario, N=6, M=4, p=.25^5 = 1/1024 = about .098%.
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://oakroadsystems.com
My reply address is correct as is. The courtesy of providing a correct
reply address is more important to me than time spent deleting spam.
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