View single post by HankC
 Posted: Tue May 12th, 2009 06:55 pm
 PM  Quote  Reply  Full Topic 

Joined: Tue Sep 6th, 2005
Posts: 517

  back to top

I need simple… If not, I’d probably say something like :

Fill a bowl with 31 marbles. Color 17 of them, say, red, and 14 of them, say, gray.

Stir the bowl well, close your eyes and choose 11.

What do you think the chances are of all 11 being gray?

This is a problem of probability, specifically combinations where order is not important and repetition is not allowed (e.g. the marble named 'Virginia' can be chosen, at most, once).

First we need to know how many possible combinations of 11 items out of the 31.

The equation for this is 31!/(11! * 20!) where n! = n*n-1*n-2*n-3*…*2*1 (called n factorial, so 6! = 5*4*3*2*1 = 120). We see that these numbers can get big in a hurry.

Anyway, 31!/(11! * 20!) = 84,672,315 . That’s a lot of possibilities!

Then we need to figure how many combinations of 11 we can get from just the 14 gray marbles. This equation is 14!/(11! * 3!) or 364. Here, try it yourself (14*13*12*11*10*9*8*7*6*5*4*3*2*1)/((11*10*9*8*7*6*5*4*3*2*1)*(3*2*1)) = 364. Okay, well, take my word for it ;)

So if we predict the chances of picking 11 gray marbles out of our bowl of 31 total marbles, we’ll divide 364 by 84,672,315 and get .0000043 or .00043 % (4 chances in a million) chance of picking 11 gray marbles purely by chance.

Given such a small probability, we’d then have to find some dependent variable among the gray marbles that causes them to, more or less, increase the chance of being picked once one or more is picked. Finding this dependency is left to the reader ;)

 Close Window