Results 1 to 5 of 5

Math Help - probability problem counting techniques.

  1. #1
    Junior Member
    Joined
    Aug 2007
    Posts
    26

    probability problem counting techniques.

    can anyone helpe with this problem?:
    Companies A,B,C,D, and E each send three delegates to a conference. A committee of 4 delegates, selected by lot, is formed. Determine the probability that company A has exactly one representative on the commitee.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member F.A.P's Avatar
    Joined
    Dec 2006
    Posts
    26
    Quote Originally Posted by rbenito View Post
    can anyone helpe with this problem?:
    Companies A,B,C,D, and E each send three delegates to a conference. A committee of 4 delegates, selected by lot, is formed. Determine the probability that company A has exactly one representative on the commitee.
    There will be 3 A-delegates and 12 non-A-delegates. Probability of choosing exactly one A-delegate is found through

    \frac{{3\choose 1}{12\choose 3}}{{15\choose 4}}=44/91

    or equivalently through

    \frac{3}{15}\cdot \frac{12}{14}\cdot \frac{11}{13}\cdot \frac{10}{12}\cdot 4=44/91,

    where we have to multiply by 4 at the end because the A-delegate can be either the first, second, third or last person to be chosen ...
    Last edited by F.A.P; September 27th 2007 at 11:43 AM. Reason: Read it wrong the first time... Soroban didn't...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,706
    Thanks
    625
    Hello, rbenito!

    Companies A, B, C, D, and E each send three delegates to a conference.
    A committee of 4 delegates, selected by lot, is formed.
    Determine the probability that company A has exactly one representative on the commitee.

    There are 15 delegates; 4 are chosen for the committee.
    . . There are: . {15\choose4} \:=\:1365 possible committees.

    There are 3 delegates from company A and 12 Others.
    To pick one A and 3 Others, there are: . {3\choose1}\cdot{12\choose3} \:=\:660 ways.

    Therefore, the probability is: . \frac{660}{1365} \;=\;\frac{44}{91}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member F.A.P's Avatar
    Joined
    Dec 2006
    Posts
    26
    You're right Soroban... I read the question too fast and had 4's stuck in my head... there where only three from each company... I've now changed the numbers in my previous post... (where an alternative approach also is presented.....)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,706
    Thanks
    625
    Hello, F.A.P!

    I read the question too fast and had 4's stuck in my head.
    I thought that's what happened.
    (If I had a nickel for every time I've done that . . . )

    And I like your alternate solution.
    It's very neat . . . if everyone understands the "times four".

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Maths Problem Counting Techniques?
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 8th 2009, 09:33 AM
  2. Probability Probelms & Counting Techniques
    Posted in the Statistics Forum
    Replies: 11
    Last Post: June 10th 2008, 08:03 PM
  3. Probability Probelms & Counting Techniques
    Posted in the Statistics Forum
    Replies: 1
    Last Post: June 10th 2008, 05:21 PM
  4. Probability Probelms & Counting Techniques
    Posted in the Statistics Forum
    Replies: 2
    Last Post: June 10th 2008, 04:58 PM
  5. Probability Probelms & Counting Techniques
    Posted in the Statistics Forum
    Replies: 2
    Last Post: June 10th 2008, 04:52 PM

Search Tags


/mathhelpforum @mathhelpforum