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Homework Help: (2) questions on Probability

  1. May 10, 2009 #1
    1. In how many ways could a committee of 3 or 4 people be formed from a staff of 15.
    I got the answer for this which is 1820 ways, but then it asks another question later on.
    For the situation of Exercise 17, what is the probability that Henri, Marcy, and Jorge are all on the committee?

    2. n!/r!(n-r)!

    3. 3!/1820 + 2!/1820 + 1!/1820
    = 9/1820
    but my answer is wrong it should be 13/1820. How do you do this problem?

    2nd question

    1. Two cards are drawn in succession from an ordinary deck. The first is not replaced before the second drawn. Find the probability of each event.
    Both are red hearts

    2. no needed formulas

    3. 26/52 * 25/51
    = 650/7652
    = 325/3826 but my answer is wrong it should be 25/102. How do you do this problem? Thank You!
  2. jcsd
  3. May 10, 2009 #2


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    For the first one, I don't really understand what you are doing with the 3! + 2! + 1!.
    I would simply count how many out of the 1820 possible committees contain all those three persons. Note that in those 1820 you have already discarded the order in which they are elected. So of all the possibilities where the committee is formed of 3 people, there is just 1 which contains those three specific people. How many committees of 4 people are there which contain those three?

    For the second one, you have made a calculation error. The answer is indeed 26/52 * 25/51, but 52 * 51 is 2652, not 7652.
  4. May 14, 2009 #3
    hmmm I don't really get what your saying on the first explanation like if there is only 1 possibility that contains those 3 people then the answer is 1/1820?

    For the second one which is 650/2652 I can't seem to simplify it into getting 25/102. Is it because my answer is wrong or I'm not simplifying right.
    I first divide by 2 getting 325/1326. ---- This seems to be all I can do. ​
  5. May 14, 2009 #4


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    No, I said that there is one possibility in the set of committees with just three people and you need to take into account the committees with 4 people as well.

    For the second one: try a factor 13.
  6. May 15, 2009 #5
    For the second question, 26/52 is the probability of a red card, not of a "red heart".
    All hearts are red, so not sure if they mean "what is the probability of getting two hearts successively". If it does, your numbers are wrong.
  7. May 19, 2009 #6
    o so is it because out of the 15 committee memebers they are 3 that are in a group of 4
    so i multiply 4 *3 = 12 and there is one group of 3
    = 12/1820 + 1/1820
    = 13/1820 Is my reasoning right?

    I get the second one now.
    Thank you for helping me!! I have a test coming up and I want to do good on it. :)
  8. May 22, 2009 #7


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    I'm really bad at these questions but this is my current understanding of the problem.

    If you do not repeat events (which is the case in the 1st question) then you will be looking at getting 15*14*13 places because we can choose the first person in 15 ways the second in 14 ways the third in 13 ways.
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