概率問題（好蛋和壞蛋）

至少一隻蛋是壞的概率
= 1 - (5/8)(4/7)
= 9/14

所求概率
= (3/8)(2/7)/(9/14)
= 1/6

2011-05-25 20:15:18 補充：
P(壞蛋x2丨沒好蛋) = P(壞蛋x2)/P(沒好蛋)

2011-05-25 20:30:39 補充：
每次都"不把"抽出的蛋放回

若一開始就抽了1隻好蛋(5/8),則會餘下4隻好蛋,連壞蛋計共7隻(4/7)

2011-05-25 20:36:13 補充：
倒是從8隻蛋抽2隻蛋只是一個動作,你指的概率是?

2011-05-25 22:11:47 補充：
8隻蛋抽2隻只是一個動作,並非一件可以用概率描述的事情,我們關心的是抽出來的蛋是怎樣的,這些特徵才是一件概率可描述的事情。3隻蛋抽2隻蛋的情況一樣。

By Law of Conditional Probability,
P(2 eggs are bad given that at least 1 egg is bad)
= P(2 eggs are bad and at least 1 egg is bad)/P(At least 1 egg is bad).
P(2 eggs are bad and at least 1 eggs is bad) = P(2 eggs are bad) = P(1 st egg is bad) x P(2nd egg is bad) = (3/8)(2/7) = 6/56.
P(At least 1 eggs is bad) = 1 - P(Both eggs are good) = 1 - (5/8)(4/7)
= 1 - 20/56 = 36/56.
So answer = (6/56)/(36/56) = 6/36 = 1/6.

我想問002, 你這個4/7是怎樣算出來的？

2011-05-25 21:07:04 補充：
既然該題目說至少有一隻雞蛋是壞蛋,題目要求的是兩隻蛋皆是壞蛋的概率;那我們就要先求8隻雞蛋抽兩隻的概率,再求3隻壞雞蛋中抽出兩隻壞雞蛋的概率。兩個分數相乘就是了。

3/8 * 2/7

6/56