Q&A #3784

Tree diagram: Four Children (2B's/2G's any order)

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From: Pat Ballew (for Teacher2Teacher Service)
Date: Apr 29, 2000 at 02:45:12
Subject: Re: Tree diagram: Four Children (2B's/2G's any order)

Yes, and so simple it makes you laugh out loud.... The solution uses
something called the binomial theorem if you want to explore this some more,
and Pascal's triangle would be another related idea that would help to
explain this.....

  To find the probability of getting b boys and g girls in any order, find
out the probability of getting them in one particular order, then multiply by
how many orders there are of getting b and g.  For this case it is so easy it
masks how well the solution works, so I will do a harder one at the bottom,
but first to answer the question you asked...
  The probability of getting bbgg is 1/16  Now all we need to do is count how
many ways there are to get two of each... this is pretty simple to do by hand
when the numbers are this small.:
So there are six orders, and all of them have a probability of 1/16, so the
total probability of this event is 6/16 or 3/8.

To find how many orders there are when there are only two possible outcomes
(h,t or b,g etc)  we use the notation of combinations.  The combinations of n
things taken r at a time is given by
comb = ----------
         r! (n-r)!

the exclamation point is read "factorial".  It means to multiply that number
times every integer smaller down to one.... 5! = 5*4*3*2*1  etc...
  That is not a very complete explanation, but you can find more by checking
the links at Dr. Math FAQ   here are two I think you may find helpful...

To show you how effective this is, suppose you wanted to know how the
probability that, of five radom children selected three of the children had
blue eyes and two had brown.  The probability of blue eyes is  1/4, brown
eyes is 3/4.. to find the probability of 3 blue and 2 brown in ANY order, we
first find the probability in ONE particular order,, say Blue blue blue brown
brown  which would have a probability of 1/4*1/4*1/4*3/4*3/4 = 9/1024..

Now we just count all the ways to pick two of the five to have brown eyes..
Well we could pick 12, 13, 14, 15, 23, 24, 25, 34, 35, 45  which is ten
different ways... so our probability of three blue & two brown is 90/1024...

Using combination of 5 taken 2 at a time we would write
            5!          5*4*3*2*1
  comb = --------- = ------------- =  10 also....
          2! * 3!       2*1 * 3*2*1

This is a very interesting area, and kids can love it if they get a chance to
explore... have fun

 -Pat Ballew, for the Teacher2Teacher service

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