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Odds of Left-Handedness in a Group


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.

Thanks for your response!


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 ).
That's about 9 percent.

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).

For more information about the "choose formula", have a look through 
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/   
    
Associated Topics:
High School Probability

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