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RE: Odds & Chances
Posted:
Aug 1, 2006 2:09 PM
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Just to clarify.
The odds of laying off the oldest person 3 times in a row is
1/(39 x 38 x 37) or 1 in 54,834. The problem is a mathematical permutation
written as 39P3 or 39! /36! .In this type of problem the oddest must be laid
off before the second oldest, etc.
The odds are different for having the 3 oldest people laid off period where
the order of layoffs is immaterial as long as the end result is that the 3
oldest are laid off (like the Pick-Six lottery). This is a combinatorial
problem, mathematically written as 39C3 or 39! /(36! X 3!). Again the odds
are 1 in 9,139.
> -----Original Message-----
> From: numeracy-approval@world.std.com [mailto:numeracy-
> approval@world.std.com] On Behalf Of richard boone
> Sent: Monday, July 31, 2006 1:14 PM
> To: numeracy@europe.std.com
> Subject: RE: Odds & Chances
>
> The odds depend on exactly how you word the question.
>
> A permutation problem: 1 in 54,834 laying off 3 people, 1 at a time and
> happening to lay off the oldest each time.
>
> Combinatorial problem: 1 in 9,139 laying off the three oldest at the same
> time. This is the probability of choosing 3 people out of 39 at random.
>
> > -----Original Message-----
> > From: numeracy-approval@world.std.com [mailto:numeracy-
> > approval@world.std.com] On Behalf Of Earl Sellie
> > Sent: Thursday, July 27, 2006 5:19 PM
> > To: numeracy@world.std.com
> > Subject: Re: Odds & Chances
> >
> > I need to amend my previous question by adding in chronilogial order
> from
> > oldest to youngest.
>
> If a Company had 39 employees and they laid off the 3 oldest employees,
> does
> anybody know what the odds of that occuring by chance would be?
> >
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