Avi Lev wrote:
>
> well let's express this probablity that all will meet in plain english
> and the formulae should then be more clearly understood and will make
> perfect sense.
>
> P = sum of 1 to meet all n-1 already there - sum of all of the rest (n-1) to
> come at the same time as 1 individual.
>
> this translates into: P = SIG[i=1..n](PI[j=1,j!=i..n] (dj)) -
Hi. What is j! ? I know what the PI operator is, but what does
PI[j=1,j!=1..n] do?
Can you expand the P_3 formula that I gave into P_4? Thanks in advance.
> SIG[i=1..n](PI[j=1..n] (dj)).
>
> SIG is the normal summation operator and PI is the multiplication operator.
>
> bye.
>
> <[EMAIL PROTECTED]> wrote in message news:83cd5m$jo$[EMAIL PROTECTED]...
> > In article <[EMAIL PROTECTED]>,
> > [EMAIL PROTECTED] wrote:
> > > On Tue, 14 Dec 1999 "Patrick D. Rockwell" wrote:
> > > > Let's say that n people agree to meet between 2:00 P.M. and
> > > > 3:00 P.M... Now, what if each person is willing to wait a
> > > > different amount of time d1, d2, d3, etc,... [expressed as
> > > > fractions of an hour] What is the probability that ALL of
> > > > them meet?
> > >
> > > Obviously with n=2 the probability is simply
> > >
> > > P_2 = d1 + d2 - (d1^2 + d2^2)/2
> > >
> > > The subtractive components result from the truncation of the waiting
> > > intervals at the boundaries. It's quite a bit more work to show
> > > that with n=3 the probability, with d1 < d2 < d3, is given by
> > >
> > > P_3 = d1 d2 + d1 d3 + d2 d3
> > >
> > > - (1/2)(d1 d2^2 + d1 d3^2 + d2 d3^3)
> > >
> > > - (1/3)d1^3 - (1/6)d2^3
> >
> > Thanks. I think that I understand. If it's not too much trouble, could
> > you please write the P_4 case of the above formula? I want to see if I
> > can discern a pattern, or at least satisfy myself that one can't be
> > discerened (at least by me).
> >
> >
> >
> > Sent via Deja.com http://www.deja.com/
> > Before you buy.
--
Patrick D. Rockwell
mailto:[EMAIL PROTECTED]
mailto:[EMAIL PROTECTED]
mailto:[EMAIL PROTECTED]
