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### Odds of Left-Handedness in a Group

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Date: 9/2/96 at 17:34:1
From: Anonymous
Subject: Odds of Left-Handedness in a group

I teach a class of 2nd and 3rd graders. It turns out that out of 23
children, 7 are left-handed!

My question is, assuming that 10 percent of the population is
left-handed, what are the chances of having 7 lefties in the room?

I assume this is akin to the 2 identical birth dates in a group
problem.

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Date: 03/09/97 at 15:47:26
From: Doctor Ken
Subject: Re: Odds of Left-Handedness in a group

Hi there -

We were going through some old questions, and found this one - I hope
it's not too late and that you haven't lost interest in the problem!

Actually, this problem isn't quite like the birthday problem. It's
more like this: if there were only 10 days in every year (and thus
only 10 possible birthdates), what are the odds that 7 people in a
class of 23 were born on the first day of the year? The big difference
is that you specify the birthday you're looking for.

In your problem, you specify the characteristic you're looking for,
left-handedness.  If you want to know more about the birthday problem,
you can look in our archives. It's a deceptively difficult problem.

So here's how you'd solve your problem. It's actually easier than the
birthday problem - good news!

Instead of thinking of this problem as about handedness, think of it
as a bunch of bags of marbles. Specifically, there are 23 bags of
marbles, and each bag has 9 red marbles in it and one lavender marble.
The red marbles represent a right-handed kid, the lavender marbles
represent a lefty. So we could draw one marble from each bag, and that
would determine whether a child is a lefty or a righty.

When you're doing probability, the basic rule to keep in mind is that
you want to take the number of favorable possibilities and divide that
by the total number of possibilities.

Let's first find the total number of possibilities.  From each bag,
you could choose any of 10 marbles, right? That means that there are
10*10*10*...*10 (23 times) = 10^23 ways of choosing your 23 marbles.
So that's going to be the denominator of all of our equations.  Now we
just have to find some numerators!

Let's start simple: what's the probability of drawing 23 marbles, one
from each bag, such that all of them are red?  Well, there are 9 ways
we could draw a white marble from 1 bag, so there are 9^23 ways to
draw a red marble from all 23 bags. So the probability here is
(9^23)/(10^23) = 0.0886294  ( you can evaluate it by taking 0.9^23 ).

Now what is the probability of drawing 23 lavender marbles?  There's
only one way to do it, so the probability is 1/(10^23), which is a
decimal point followed by 22 zeroes and a one.  That's not very
likely!

Let's move on to something a little harder: what's the probability of
drawing exactly 22 red marbles and 1 lavender marbles?  Well, the
lavender marble could come from any of the 23 bags, so there are 23
ways of drawing it. And once we know which bag the lavender marble
came from, then there are 9^22 ways to choose the 22 red marbles.  So
the total number of possibilities is 23*(9^22), and when we divide
that by 10^23 we get 0.2265, which is about 23 percent.

What's the probability of drawing 22 red marbles and 2 lavender ones?
First let's figure out how many ways to draw the lavender ones, then
it's just like the previous case.  Here I'll introduce the "choose
formula," which we can use to figure out the number of ways of drawing
k items from a set of n items.  I'll write that as [n:k], and in this
case it's [23:2]; 23 choose 2.

The formula for [n:k] is n!/(k! * (n-k)!).  The !'s are factorial
symbols, a la 5! = 5*4*3*2*1.  So 23 choose 2 is 23!/(2! * 21!).
After some cancellation, that becomes 23*22/2, which is 253.  So the
number of ways of drawing 21 red marbles and 2 lavender ones is 253*
(9^21). Dividing that by 10^23, we get 0.27683 as our probability.

And now, (ta dah!) for our probabilty of having 7 lefties in your
class.

Using the choose formula, the number of ways of drawing 7 lavender
marbles is 23!/(7! * 16!), which is 245157.  Multiply this by the
number of ways of choosing the remaining 16 marbles, which is 9^16,
and the total number of ways of choosing exactly 7 lavender marbles is
454280870438350784037.

Divide this by 10^23, and we get ........drum roll.........

0.00454281, i.e. 0.5 percent.  That's quite unlikely.

Of course, a more useful number would probably be the probability of
having _at least_ 7 lefties in your class.  Do you know how you would
find that? I'll leave it to you (there's a medium-easy way and a hard
way).

our archives - it's in there a lot, and it's also in the FAQ.

-Doctor Ken,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
High School Probability

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