Results 1 to 2 of 2

Math Help - randomly picking balls

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    8

    randomly picking balls

    There are m red balls labeled as r1,r2,.,rm and n blue balls labeled as b1,b2,.,bn in a bag. We draw them out of the bag without replacement until the last ball is picked out. Suppose at each drawing, all balls remaining in the bag always have equal probabilities of being picked out. What is the probability that both the red balls and the blue balls are drawn out in the numerical order (that is, ri is the i-th red ball to be picked out and bi is the i-th blue ball to be picked out) ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by billy View Post
    There are m red balls labeled as r1,r2,.,rm and n blue balls labeled as b1,b2,.,bn in a bag. We draw them out of the bag without replacement until the last ball is picked out. Suppose at each drawing, all balls remaining in the bag always have equal probabilities of being picked out. What is the probability that both the red balls and the blue balls are drawn out in the numerical order (that is, ri is the i-th red ball to be picked out and bi is the i-th blue ball to be picked out) ?
    Hi Billy,

    I think the easiest way to work this problem is to observe that the draws of the red and blue balls are independent in the sense that drawing a blue ball does not influence the order of draws of red balls, and vise versa. So

    P(red balls in order and blue balls in order) = P(blue balls in order) * P(red balls in order)

    The probability that the blue balls are in order is 1/n! and the probability that the red balls are in order is 1/m!.

    If that doesn't satisfy you, here is a longer method that leads to the same result. The total number of orderings of the balls (both red and blue) is (m+n)!, and each of the orderings is equally likely to occur. Let's see if we can count the number of these arrangements in which the blue balls and red balls are drawn in order. If we disregard the numbers on the balls, there are \binom{m+n}{m} to select the red and blue balls out of the sequence of m+n balls. Once the places for the red and blue balls are marked, there is only one way to place the balls in order according to there numbers. So there are \binom{m+n}{m} arrangements in which the balls are in order according to their numbers, and the probability that this occurs is

    \frac{\binom{m+n}{m}}{(m+n)!} = \frac{(m+n)!}{m! n!} \cdot \frac{1}{(m+n)!} = \frac{1}{m! n!}

    which is the same answer we got before.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Drawing balls randomly with replacement
    Posted in the Statistics Forum
    Replies: 2
    Last Post: November 14th 2011, 12:59 PM
  2. Randomly chosen family
    Posted in the Advanced Statistics Forum
    Replies: 13
    Last Post: October 12th 2011, 10:06 AM
  3. Replies: 3
    Last Post: October 7th 2011, 04:58 PM
  4. a basket contains 5 red balls, 3 blue balls, 1 green balls
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: May 28th 2010, 02:39 AM
  5. sequential picking 1080 balls from 21 barrels
    Posted in the Statistics Forum
    Replies: 1
    Last Post: June 10th 2007, 07:11 AM

Search Tags


/mathhelpforum @mathhelpforum