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Looking for confirmation on probability homework question?

  1. Mar 27, 2013 #1
    Looking for confirmation on probability!

    Sorry, posted in wrong forum. It was a probability question which is why I posted it here. Can someone move it to homework section if need be. Thanks!
    -------------------------------------------------------------------------

    Can someone let me know if I am doing this correctly, because the answer I am getting seems to small. I thought the probability would be much higher since the woman out number the men by over 50% and we are just looking for 1.

    Question: Suppose that a programming team has 17 women and 15 men. Three people must be chosen to work on a special project. What is the probability that the group selected has at least 1 women in it?

    My work:

    (17C1) -> To choose 1 women
    (15C2) -> To choose 2 men
    (32C3) -> Total groups of 3 among all people

    (17C1)(15C2)
    --------------
    32C3


    17 * 105
    --------
    4960

    1785
    ----- = .360
    4950


    I've seem some examples where they subtract this answer from 1 which not looks more reasonable, but is it correct and can someone explain why?

    ex.. 1 - .360 = .640

    Thanks for any input!!
     
    Last edited: Mar 27, 2013
  2. jcsd
  3. Mar 27, 2013 #2

    DrClaude

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    Staff: Mentor

    Read the statement carefully: at least one woman.
     
  4. Mar 27, 2013 #3
    Ahh, so would I use:

    (17C1 + 17C2 + 17C3) / 32C3 to account for having 1,2 or 3 women chosen? Or would I have to account for the 8 possibilities of women/men?

    Thanks for feedback. I hope this doesn't get deleted for my lack of reading. I was just getting ready to erase it and repost..

     
  5. Mar 27, 2013 #4

    DrClaude

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    Staff: Mentor

    Not quite. If you have for instance exactly 2 women, is the number of possiblities 17C2? Aren't you forgetting something?

    And this might be a good time to use the other method you mention: 1 - something.
     
  6. Mar 27, 2013 #5
    I'm forgetting the men.. So would it be..

    (17C1)(15C2) + (17C2)(15C1) + 17C3
    -------------------------------------
    32C3

    I was thinking of subtracting the result from 1. Do I have to account for no women..



     
  7. Mar 27, 2013 #6

    DrClaude

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    Staff: Mentor

    Right.

    Why would you do that? Calculate the value numerically and see what makes sense. There is a way to calculate the initial problem as 1-something. Can you figure that one out?

    Account in what way?
     
  8. Mar 27, 2013 #7
    Hmm, I think I got it now.

    1 - ((15C3) / (32C3)) = .908

    So calculate the probability of no women from the total, than subtract it from 1.

    I calculated it my long way above and I get the same answer. Can you confirm this is correct :<)

    Thanks for your help on this.


     
  9. Mar 27, 2013 #8

    DrClaude

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    Staff: Mentor

    That is indeed correct.

    You're welcome!
     
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