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Homework Help: Simple Probability Question

  1. Apr 7, 2008 #1
    This one seems simple, but I want to be sure I'm using the correct thinking.

    1. The problem statement, all variables and given/known data

    If there are 30 students in a class and 5 of them are to be assigned A's, 20 to be assigned B's, and the remainder C's, how many different sets of students receiving A's are possible?

    2nd Part

    Determine the probability of a group of 5 students being selected at random and none of the 5 being assigned an A.

    2. Relevant equations

    3. The attempt at a solution
    Would this be just a 30 choose 5 question since there are 5 students who can make an A out of 30?
    Last edited: Apr 7, 2008
  2. jcsd
  3. Apr 7, 2008 #2


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    For the first part.
    The second part is a bit more involved.
  4. Apr 7, 2008 #3
    I hate to ask, but any idea of how I should start with the second part?
  5. Apr 7, 2008 #4
    Now you need to find out how many sets of 5 students not receiving an A are possible. This is not the final solution but it gets you started...
  6. Apr 7, 2008 #5
    Maybe it well help to give you a similar problem. Say you have ten pens and two are broken. How many sets of one member are there of broken pens? Well, two, because there are two broken pens. How many sets of one member are there of functioning pens? Eight. Now...how many sets of two are there of broken pens? One. How many sets of two are there of working pens?
    Last edited: Apr 7, 2008
  7. Apr 7, 2008 #6
    Wouldn't it be nCr=28?
  8. Apr 7, 2008 #7
    Would the prob be 30 choose 25 for the prob of not choosing a student with an A, 142506?
  9. Apr 7, 2008 #8
    No, a probability will never be greater than 1. You need to calculate the number of ways to choose 5 students from the group of other than A students. Then divide that number by the number of ways to choose 5 students from the entire group.

    You have 25 students that did not get an A. Choose 5.

    You have 30 students total. Choose 5.

    This is close to the complete solution...
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