What's the Probability That Two Will Agree?

Date: 11/10/97 at 18:15:46
From: Jessica Hawkins
Subject: Probability

Problem: Of the nine members of the board of trustees of a college, 
five agree with the president on a certain issue. The president 
selects three trustees at random and asks their opinions. What is the 
probability that at least two of them will agree with him?

I've tried working this out but I just can't seem to set it up.

I know that Pr(E) = [number of outcomes in E].
                    -------------------------
                              N

And I know the bottom part of the equation is 9 but after that I'm 
                                              3
stuck so could you please help me?
Thanks so much.

Date: 11/10/97 at 19:17:20
From: Doctor Tom
Subject: Re: Probability

Hi Jessica,

Let's call the result "favorable" if the president happens to pick
a set of three folks at least two of whom agree with him.

How many ways can this happen?

Well, either exactly two agree with him (and one doesn't) or all three 
agree with him.

How many different ways are there to select a team with 2 agreers 
and one disagreer? Well, you can choose 2 from the set of 5 in 
(5 choose 2) ways. For each of those ways, you can choose any one 
of the 4 disagreers, so altogether, there are 4*(5 choose 2) ways 
= 4*10 = 40 ways.

How about 3 agreers (and no disagreers)?  You have to pick 3 from the 
set of 5, and this can be done in (5 choose 3) ways. Once they're 
picked, you're done - you don't get to make any other selections - 
so there are (5 choose 3) = 10 ways to do that.

So altogether, there are 40+10 = 50 ways to choose a favorable team.

Now, how many total ways are there of choosing a team? In this case, 
there are 9 folks, so the answer is just (9 choose 3), which is 
9*8*7/(3*2*1) = 84 ways. So 50 times out of 84 the president will get 
a favorable result. The probability is thus 50/84 = 25/42 = .595238...

Of course if you know anything about real college presidents, it's
pretty unlikely that the selection will be random, and the "true"
probability of getting a favorable team will be very close to 1.0 :^)

-Doctor Tom,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/