機率問題-這種只有抽球問題，需要分取出放回和取出不放回嗎？

設黑球有 B 顆 , 白球有 W 顆 :
P(抽到二黑球) = [B/(B+W)] * [(B-1)/(B+W-1)]
P(抽到二白球) = [W/(B+W)] * [(W-1)/(B+W-1)]
P(抽到一黑一白球) = P(B , W) + P(W , B)
= B/(B+W) * W/(B+W-1) + W/(B+W) * B/(B+W-1)
= 2BW / [(B+W)(B+W-1)]
得
[B/(B+W)] * [(B-1)/(B+W-1)] = 5 [W/(B+W)] * [(W-1)/(B+W-1)]
B(B-1) = 5W(W-1)......(1)
及
2BW / [(B+W)(B+W-1)] = 6 [W/(B+W)] * [(W-1)/(B+W-1)]
B = 3(W-1)......(2) , 代入 (1) :
3(W-1) [3(W-1) - 1] = 5W(W-1)
3 [3(W-1) - 1] = 5W
9W - 12 = 5W
W = 3
B = 3(3-1) = 6
黑球有六顆


2010-04-29 00:36:40 補充：
連續抽取二球代表取出不放回。

這種題意是一次取兩球，如果硬ㄠ可能先取一球放回再取一次，
那就該像我這麼寫，否則就太奸詐了．
本題前者可求得6黑3白，
以後者算根本無解.