# Homework Help: Probability: me vs. textbook

1. May 26, 2010

### Shawn Garsed

1. The problem statement, all variables and given/known data
A. Mrs. Moyer’s class has to choose 4 out of 12 people to serve on an activity
committee.
B. The four chosen students are then selected for the positions of chairperson, activities

What is the probability that any one of the students is chosen to be the
chairperson?

2. Relevant equations
Permutations and combinations

3. The attempt at a solution
I first calculated the number of outcomes for part A: 12!/((12-4)!4!)=495. Then I calculated the number of outcomes for part B: 4!=24. Which gives the total number of outcomes: 495*24=11880.

After that, I calculated the favorable number of outcomes for A and B:

A: (1*11*10*9)/3!=165

B: 1*3*2*1=6

Which gives the total number of outcomes: 165*6=990.

Therefore, the probality is 990/11880 or 1/12.
The problem is that the book says it's 1/2970.

Who's right?

Shawn

2. May 26, 2010

### sebb1e

Are you quite sure that's the question being asked?

As you've phrased the question, it has to be 1/12 because part A is irrelevant, all students have equal chance of being the chairperson.

3. May 26, 2010

### Shawn Garsed

That's exactly how the question is phrased. I even calculated it in different ways, the simplest being: 4 out of 12 means you have a 4/12 or 1/3 chance of being chosen and the chance of being chosen as chairperson is 1/4. (1/3)*(1/4)=1/12.

Last edited: May 26, 2010
4. May 26, 2010

### sebb1e

Well you are right about there being 1180 different combinations of 4 students. 2970 is a quarter of 1180 so it's not a totally random number but it definitely isn't the answer to your question.