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Binomial Experiments

Date: 09/05/2003 at 19:21:26
From: Confused
Subject: What is the Binomial Distribution?

How can I explain the binomial distribution to my six grade math class?  

Date: 09/05/2003 at 20:50:36
From: Doctor Tom
Subject: Re: What is the Binomial Distribution?

Well, I'm not sure how far you can hope to get with sixth graders,
but there are some things you can do.

A binomial distribution arises any time an event with exactly two
outcomes is repeated a bunch of times.

The simplest example is the case where the two outcomes are
equally likely -- flipping a coin, for example.

For the class, what I'd do is pick a fixed number of flips, say 10.
And rather than flip one coin ten times, it's easier if they flip 10
coins once and count the number of times "heads" appears.

Have everybody in the class do this 10 times (or more) and
record the number of "heads" for each attempt.  There are
11 possible outcomes, from 0 heads to 10 heads.  Obviously, it'll
be unlikely to get any zeros or tens, but just collect the data
and draw it as a bar chart and notice that the most likely
outcome seems to be 5, and that it does seem to form a
bell-like distribution.

With more and more repeats of the experiement, you'll find
that the proportions of times each outcome occurs matches
better and better the theoretical results (which you can obtain
from the tenth row of Pascal's triangle, where I count the
top row as row zero).

In other words, if you add the eleven numbers in that tenth
row, they will add to 2^10 = 1024, and so as the number
of experiments (where an experiment consists of flipping
a coin ten times) with results 0 through 10 heads should
occur with the following likelihoods:

     1    10    45   120   210   252   210   120    45    10     1
  ----, ----, ----, ----, ----, ----, ----, ----, ----, ----, ----
  1024  1024  1024  1024  1024  1024  1024  1024  1024  1024  1024

In other words, about 1/4 (252/1024, to be precise) of the
times that you do the experiment, there will be exactly 5
heads.  Only 1 time in about a thousand will all ten be
heads, and so on.

You can compare these theoretical results with those
generated by your experiments.

That may be good for the whole lesson, but a binomial
distribution does not require that the likelihoods of the
results be the same.  For example, suppose your
experiment is to roll a die with six sides, and you
consider that you "win" if you obtain a 5 or a 6, and
that you "lose" if you obtain 1, 2, 3 or 4.

If you run that experiment ten times (ten rolls of a die or
1 roll of ten dice), there are again eleven possible outcomes:
zero through ten wins.  But it stands to reason (since you
usually lose) that it will be FAR more likely to get ten
losses than ten wins.

With some dice and time, you can do the same thing here
as you did with the coins.  You'll see that you also get a
bell-shaped curve, but it'll be lopsided, with the bump
nearer zero than 10.  In fact, since you win 1/3 of the
time, the bump will be near the 1/3 point -- near 3 wins
in 10.

The expected results can be calculated, but it's quite a
bit messier to do so.  You can use the same fraction
numerators that we worked out with Pascal's triangle
(1, 10, 45, 120, 210, 252, 210, 120, 45, 10, and 1), but they
need to be multiplied by (1/3)^W(2/3)^L to get the
probability of W wins and L losses, where W+L = 10.

For example, the probability of zero wins is:

  1 x (1/3)^0 x (2/3)^10 = .01734...

The probability of ten wins is:

  1 x (1/3)^10 x (2/3)^0 = .000016935...

The probability of exactly 3 wins (most likely result) is:

  120 x (1/3)^3 x (2/3)^7 = .26012...

I'll bet this is beyond a sixth grade class, however.

You might look at it not with 10 rolls, but rather with 2.
Then there's a hope of seeing what's going on by drawing
a tree for the possible outcomes: start at a root and go
left (win) with 1/3 probability, and go right (lose) with 2/3
probability.  From those nodes repeat for the probabilites
after two events.  You can then sum up the probabilities
to obtain:

              1/3 x 1/3 = 1/9 chance of two wins

  1/3 x 2/3 + 2/3 x 1/3 = 4/9 chance of one win 
                              (could be first or second)

              2/3 x 2/3 = 4/9 chance of zero wins

Maybe you could extend this to three experiments and
notice that there is:

  1 way to get 3 losses
  3 ways to get 2 losses and 1 win
  3 ways to get 1 loss and 2 wins
  1 way to get 3 wins.

Point out that this looks like Pascal's triangle's third row and
that the only difference is that as there are more and
more wins, the likelihood gets smaller and smaller.

Look in any introductory book on statistics for more material,
but I'm pretty sure it will be beyond what you can use in a
sixth grade class.

Good luck!

- Doctor Tom, The Math Forum 
Associated Topics:
High School Probability
High School Statistics
Middle School Probability
Middle School Statistics

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