Results 1 to 4 of 4

Math Help - I am lost!:(

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    47

    I am lost!:(

    You have 10 balls and 4 boxes. The balls are place din boxes at random.
    a) What is the probability that 2 of the boxes contain four or more balls?
    b) What is the probability that no box contians more than 3 balls?
    c) What is the probability that 2 of the boxes contain exactly one ball and that each of the other 2 contain more than 2 balls?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Given the way you have posted the problem there is no possible answer.
    There are four possible ways to read this: the balls are different and the boxes are different; the balls are identical and the boxes are different; the balls are different and the boxes are identical; both the balls are identical & boxes are identical.
    Which of the four possibilities is the one that fits you question?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2008
    Posts
    47
    Thanks - I will ask my professor this morning.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by wvlilgurl View Post
    You have 10 balls and 4 boxes. The balls are place din boxes at random.
    a) What is the probability that 2 of the boxes contain four or more balls?
    b) What is the probability that no box contians more than 3 balls?
    c) What is the probability that 2 of the boxes contain exactly one ball and that each of the other 2 contain more than 2 balls?
    a) We will assume the boxes are equally likely. If we disregard (temporarily) the order of the boxes, the 10 balls can be placed in the boxes in 4 ways with at least 2 boxes containing at least 4 balls: 5+5+0+0, 5+4+1+0, 4+4+2+0, and 4+4+1+1.

    Let's consider the 5+5+0+0 case first. If, for example, boxes 1 and 2 contain 5 balls each, then the probability of this event comes from a multinomial distribution: \binom{10}{5,5,0,0} (1/4)^{10}. But boxes 1 and 2 aren't the only choice: all together, there are \binom{4}{2,2} choices of the boxes that result in a 5+5+0+0 mix. So the 5+5+0+0 pattern occurs with probability  \binom{4}{2,2}\binom{10}{5,5,0,0} (1/4)^{10}.

    The other cases are similar:

    The pattern 5+4+1+0 occurs with probability \binom{4}{1,1,1,1} \binom{10}{5,4,1,0} (1/4)^{10}.

    The pattern 4+4+2+0 occurs with probability \binom{4}{2,1,1} \binom{10}{4,4,2,0} (1/4)^{10}.

    The pattern 4+4+1+1 occurs with probability \binom{4}{2,2} \binom{10}{4,4,1,1} (1/4)^{10}.

    Adding the 4 probabilities above, the total probability is about 0.1024.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. I am lost, please help
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: February 5th 2010, 01:35 AM
  2. I'm a little lost
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 21st 2010, 07:28 PM
  3. A Bit Lost...
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 29th 2009, 04:44 PM
  4. Lost...
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: December 11th 2008, 01:41 PM
  5. Lost!
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 1st 2005, 12:34 PM

Search Tags


/mathhelpforum @mathhelpforum