On Thu, 6 Nov 2003, jim clark wrote:

> Hi
>
> On 6 Nov 2003, Curious George wrote:
>
> > [EMAIL PROTECTED] (Curious George) wrote in message
> > news:<[EMAIL PROTECTED]>...
> > > I am an instructor, using 'Elementary Statistics: A Brief
> > > Version, 2/e'  by
> > > Allan G. Bluman.  A questions in the database states:
> > >
> > > A family has three children. How many combinations of boys and
> > > girls will be in the sample space?
> > >
> > > The answer given is: 8.
> > > _______________________________
> > >
> > > Even if you allow for 'birth order' I can not identify more than 6
> > > possible outcomes.  Am I missing something?

Yes, evidently;  though it's hard to say what, since you didn't
enumerate the six.  The following may help, if you haven't thought of it
this way already:  Either there are 3 boys, or 2 boys and a girl, or a
boy and 2 girls, or 3 girls.  If there are 2 boys, the girl may be the
youngest, the middle, or the eldest child;  and similarly if there is
one boy he maybe youngest, middle, or eldest.  That brings you to 8:
1 (all boys) + 3 (2 boys) + 3 (2 girls) + 1 (all girls).  Enumerated in
birth order, the 8 possibilities are:  BBB, BBG, BGB, BGG, GBB, GBG,
GGB, GGG.  This is all easily related to Pascal's triangle if you get
into binomial computations, as Jim suggests.  -- DFB.

> > > Thanks,
> > > George
> >
> > "I see, said the blind man....."  I needed to reverse the birth order.
> > George
>
> Depending on what you plan to do later (e.g., binomial), you
> might want to teach this as a general principle based on the
> multiplication rule.  That is, 2 outcomes (BG) on each of 3
> trials gives
>
> 2 x 2 x 2 = 2^3 = 8
>
> and then relate this to the enumeration and/or a tree diagram of
> the outcomes.
>
> Best wishes
> Jim

 -----------------------------------------------------------------------
 Donald F. Burrill                                         [EMAIL PROTECTED]
 56 Sebbins Pond Drive, Bedford, NH 03110                 (603) 626-0816
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