Let there be n balls in a bag. Let k_i be a partition of n such that there are k_1 red balls, k_2 blue balls, etc..., and k_a yellow balls. Give each ball an unique number from 1 to n. What is the probability that a red ball has a number <=j. e.g. consider a bag with 20 balls numbered 1..20, and randomly coloured red (7) and green (13). What is the probability that a red ball is numbered either 1 or 2? Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com
At 01:09 AM 11/30/03 +0000, you wrote:
Let there be n balls in a bag. Let k_i be a partition of n such that there are k_1 red balls, k_2 blue balls, etc..., and k_a yellow balls.
I can't understand why k_2, k_3, and so on, matter to the problem you pose below. Surely the answer is the same regardless of how the balls that are not red are colored. Perhaps I don't understand the problem you are posing.
Give each ball an unique number from 1 to n.
What is the probability that a red ball has a number <=j.
Let k = your k_1. There are (n choose 2
e.g. consider a bag with 20 balls numbered 1..20, and randomly coloured red (7) and green (13).
What is the probability that a red ball is numbered either 1 or 2?
Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com
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My apologies; I sent the previous message prematurely. I was saying: there are n(n-1) ways of picking two balls to be numbered 1 and 2. Of these ways, (n-k)(n-k-1) have neither ball red. So the desired probability is {n(n-1)-(n-k)(n-k-1)}/n(n-1). Simplifying, this is k(2n-k-1)/n(n-1). For the given example, n=20 and k=7, for which the formula gives 56/95. -ACW
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Allan C. Wechsler -
Jon Perry