Results 1 to 4 of 4

Math Help - Joint Probability Question

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    3

    Joint Probability Question

    Hi, I am having trouble with the following question:
    Five balls are drawn without replacement from an urn containing 7 blue balls, 4 green balls, and 8 yellow balls. Let B, G, and Y be the total number of blue, green and yellow balls drawn, respectively. Find
    a) The joint probability mass function of B, G, and Y.
    b) The joint probability mass function of Y and B
    c) THe marginal probability mass function for B, G, and Y separately. Do these marginal probabilities look familiar? ID them.

    For a) I tried modelling the joint probability mass function using a combinatorial approach.
    c(7 choose B)c(4 choose G)c(8 choose Y)/c(19 choose 5).
    And B+G+Y=5 (because you can only choose 5 balls). However, after this I'm stuck.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    these are all hypergeometrics

    joint is P(B=b,G=g,Y=y)={ {7 \choose b} {4 \choose g} {8 \choose y}\over  {19 \choose 5}}.

    b is really the same...

    P(B=b,G=g)={ {7 \choose b} {4 \choose g} {8 \choose 5-b-g}\over  {19 \choose 5}}.

    I'll only do B's distribution. It's B vs the world so...

    P(B=b)={ {7 \choose b} {12 \choose 5-b} \over  {19 \choose 5}}.

    you do the other two.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    3
    I understand how you got part c, but
    I don't really understand how you got part b. Is it because you are focusing on B and G, so Y is just what's left over?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    I was in a rush.
    I wrote the joint distribution of B and G.
    Here's the joint of Y and B...

    P(B=b,Y=y)={ {7 \choose b} {8 \choose y} {4 \choose 5-b-y}\over {19 \choose 5}}.

    This is because b+y+g=5.
    So, if you know two of them, you know the third one.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Joint Probability question
    Posted in the Statistics Forum
    Replies: 2
    Last Post: July 21st 2011, 10:33 PM
  2. joint probability question
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: March 25th 2011, 01:01 PM
  3. X and Y joint pdf probability question
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: December 16th 2010, 10:34 PM
  4. Joint Probability question
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: July 28th 2010, 05:11 PM
  5. Joint probability question???
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: October 17th 2008, 12:15 AM

Search Tags


/mathhelpforum @mathhelpforum