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A probability question.

  1. Apr 12, 2007 #1
    1. The problem statement, all variables and given/known data
    A group of 12 people is going out on the town on Saturday night. The group will take three cars with four people in each car. If they distribute themselves among the cars at random, what is the probability that Rafel and Chantal will be in the same car.

    2. Relevant equations

    P (Rafel&Chantal) = # ways they are in the same car/ total # of outcomes.

    3. The attempt at a solution

    This is what I have so far:

    I place these 2 in a car, so

    (10C2)(8C4)(4C4) / (12C4)(8C4)(4C4)

    Doing this I get 1/11 which is supposed to be the right answer, but however I'm confused and how this works because:

    (12C4)(8C4)(4C4) basically divides these 12 people into 3 groups each using a car, so you could have

    etc, etc

    Where A,B and C are groups, the positions represents a car.

    However the expression at the top (10C2)(8C4)(4C4) forces the 2 ppl to be in the first car ONLY. So I multiplied by 3, doing so I get 3/11 which is the wrong answer. I don't understand why, since the order matters for arranging all of them, how come you can only have the 2 people in the first car?

    Sorry if my I'm not describing the problem well, thanks!
  2. jcsd
  3. Apr 12, 2007 #2


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    Homework Helper

    Doesn't 1/11 feel intuitively like it's too small? It is. Think of the problem another way. There are three people in the car with Chantal and eight people in the other cars. Any one of them is equally likely to be Rafel. So the correct probability is indeed 3/11. And your reasoning is correct, there three different choices for a car for them to be together in.
  4. Apr 12, 2007 #3
    Oh, then it could be a mistake in the solution for our problem sets, I will speak to the teacher about it.
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