## Re: [probability] Re: Binomial/Hypergeometric probability function??

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• I see what you mean, however... Couldn t we look at the problem as follows? Wouldn t you agree that the bags are independent, and that there is no
Message 1 of 6 , Dec 31, 2003
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I see what you mean, however...

Couldn't we look at the problem as follows?
Wouldn't you agree that the bags are independent,
and that there is no "with/without replacment" issues
to worry about here? So couldn't we write out the
probability of choosing 0 red balls as:

Prob(X=0) = 0.5 * 0.7 * 0.9 * 0.2 = 0.063

And there is only one way to do this: 4C0.

Also, one might take the position that the special
cases you noted about the one bag having all the reds,
or one particular bag having no reds, would be
pathelogical examples and so wouldn't normally be

For Prob(X=1) you have 4 bags and so there are 4C1=4
ways to choose one marble, but now I think you've got
to find the individual probabilities of each of the 4
ways of choosing only one marble and then sum them up.
Hope this track is more fruitful.

Happy New Year!

--- lk_maxwell <lk_maxwell@...> wrote:
> I appreciate the answer, but the bags do make a
> differnce. If one of
> the bags had zero red marbles, then the chances of
> pulling four out
> would be reduced to zero and conversely if one of
> red marbles the chances of pulling none out would be
> zero.
>
> LKM
>
>
>
> --- In probability@yahoogroups.com, The Webmaster
> <maintainer_wiz@y...> wrote:
> > On second thought, how does this sound?
> >
> > I don't think the fact that the marbles are in 4
> > separate bags affects the way the choices can be
> > (there are no replacement/with replacement issues
> to
> > worry about here) so I think this is simply a
> > hypergeometric probelm...much like choosing 5
> cards
> > from a deck of 52 (i.e. ways of choosing face
> cards,
> > non face cards for each hand of 5 as in poker).
> >
> >
>
P(#reds=0)=>{=(COMBIN(17,0)*COMBIN(23,4))/COMBIN(40,4)}
> >
>
P(#reds=1)=>{=(COMBIN(17,1)*COMBIN(23,3))/COMBIN(40,4)}
> >
>
P(#reds=2)=>{=(COMBIN(17,2)*COMBIN(23,2))/COMBIN(40,4)}
> >
>
P(#reds=3)=>{=(COMBIN(17,3)*COMBIN(23,1))/COMBIN(40,4)}
> >
>
P(#reds=4)=>{=(COMBIN(17,4)*COMBIN(23,0))/COMBIN(40,4)}
> >
> > Paste the above functions found between the
> brackets
> > into an Excel spreadsheet to see the probability
> > calculation.
> >
> > Hope this is correct. ;-)
> >
> >
> > --- lk_maxwell <lk_maxwell@h...> wrote:
> > > Can you walk me through this? I'm not following
> you
> > > on how it should
> > > be done.
> > >
> > > Thanks,
> > >
> > > LKM
> > >
> > > --- In probability@yahoogroups.com, The
> Webmaster
> > > <maintainer_wiz@y...> wrote:
> > > > I think in this case p is dist hypergeo and
> once
> > > you
> > > > figured this out apply the binomial
> probability
> > > > function to sum up the number of ways of
> chosing
> > > > k = 0,1,...4 reds from n total marbles. This
> is
> > > my
> > > > basic sketch.
> > > >
> > > >
> > > > --- lk_maxwell <lk_maxwell@h...> wrote:
> > > > > Hello all, excellent group!
> > > > >
> > > > > My question is:
> > > > >
> > > > > I have four bags of marbles. The first bag
> has
> > > 5
> > > > > red and 5 blue, the
> > > > > second bag has 3 red/7 blue, the third 1
> red/9
> > > blue,
> > > > > and the fourth
> > > > > bag 8 red/2 blue. If I draw one marble out
> of
> > > each
> > > > > bag, what is the
> > > > > probability of the red marbles totaling: 0,
> 1,
> > > 2, 3,
> > > > > or 4?
> > > > >
> > > > > I would prefer a step by step explaination,
> but
> > > can
> > > > > someone at least
> > > > > point me in the right direction as to what
> type
> > > of
> > > > > formula I would
> > > > > use?
> > > > >
> > > > > Thanks,
> > > > >
> > > > > LKM
> > > > >
> > > > >
> > > > >
> > > >
> > > >
> > > > __________________________________
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> > >
> > >
> > >
> >
> >
> > __________________________________
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