Probability Question from a Math Test

Date: 3/8/96 at 21:9:23
From: Anonymous
Subject: Question

Dear Dr. Math,

I have a question about a math problem that came up on a math test 
I took recently.  

There are 15 homerooms in the school. There are 20 students in 
each homeroom.  If the principal selects 5 of the homerooms for a 
pizza party, what is the probability of Mr. Smith's homeroom being 
selected?

a. 1/3  b. 1/5  c. 15/20 d. 1/20

Can you help me? I feel the answer is a. 1/3, but why is it not 
1/15 * 14 * 1/13 * 1/12 * 1/11?  And is this a way to find the 
probability of anything, and if so what would the question be?

Date: 3/21/96 at 21:4:52
From: Doctor Aaron
Subject: Re: Question

Your feeling is right, if we pick 5 out of 15 rooms the 
probability is 1/3. You could be thinking about several things by 
the multiplication 1/15 * 1/14 * 1/13 * 1/12 * 1/11. One 
probability question that this would answer is:

What is the probability of the principal picking 5 specific rooms 
in a specific order? The first time (s)he must pick one room out 
of 15, then out of 14, and so on, so the probability of the 
principal picking all 5 rooms in order is the product of the 
probability of picking each room individually, i.e., 
1/15 * 1/14 * 1/13 * 1/12 * 1/11.

A more common question would ask: what is the chance of the 
principal picking 5 specific rooms in no particular order. In this 
case, the probability is that of picking one of the five out of a 
total of 15, times picking one of the remaining four out of the 
remaining total of 14, etc. This is (5 * 4 * 3 * 2 * 1)/(15 * 14 * 
13 * 12 * 11), an equation of the form: 1/(n!/k!(n-k)!) (where a! 
= a * a-1 * a-2 * .... * 3 * 2 * 1), which comes up a lot in 
probability and other areas of math.

One name that mathematicians use to refer to the number n!/k!(n-
k)! is n choose k, meaning that this represents the number of 
different combinations of n elements taken k at a time. Here's an 
example:If we let n = 5 and k = 3, we can think of how many 
different ways can we take 3 elements from the set {1,2,3,4,5}.  
We have(1,2,3), (1,2,4), (1,2,5) ... if we write all of them out, 
we'll have

5!/3!*2! = 10 different combinations.

So, had the principal chosen 5 rooms out of 15, (s)he would have 
had 15 choose 5 choices, from which (s)he chose 1. Then the 
probability of choosing that 1 set of five rooms would be 
1/(15 choose 5) = (5 * 4 * 3 * 2 * 1)/(15 * 14 * 13 * 12 * 11).

I hope that this is helpful.

-Doctor Aaron,  The Math Forum