# randomly picking balls

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• Feb 10th 2010, 11:59 PM
billy
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) ?
• Feb 11th 2010, 05:36 AM
awkward
Quote:

Originally Posted by billy
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.