dcluxx wrote:
此題無解,原題目又沒限定原來袋子裡的球只能為黑色或白色,只說了不知道是白是黑,那就有可能是任何一種顏色,那再取出是白球的機率根本條件不足,無法作答.
這不影響喔:D
只要是服從古典機率的原則
也就是袋內兩顆球拿出來的機會均等的話
原先袋內那顆不管是什麼顏色，只要分成白色與非白色兩種就可以了
因為題目沒有問到拿出黑色或綠色或紅色的機率這類的問題
只要把我剛才的作法中，黑色的部份套用成非白色
答案還是會一樣的
袋內球的組合為：
(原袋內球白,新放入為白) (原袋內球非白,新放入為白)
P(第二顆拿出為白球|第一顆拿出為白球)
P(第二顆拿出為白球且第一顆拿出為白球)
=-------------------------------------
P(第一顆拿出為白球)
(原袋內白,新放入白,先拿到原袋內白再拿新放入白)(原袋內白,新放入白,先拿新放入白,再拿原袋內白)
=--------------------------------------------------------------------------------------------------------
(原袋內白,新放入白,拿到原袋內白)(原袋內白,新放入白,拿到新放入白)(原袋內非白,新放入白,拿到新放入白)
=2/3
結果一樣是不改變的
對機率有興趣的朋友，可以參考Sheldon Ross寫的A First Course in Probability
這本是大多數統計系跟應數系用的機率學入門書
the answer is 2/3 as others have already stated.
we can split this question into two cases:
(1) the first ball is black, and you put in a second ball that is white.
(2) the first ball is white, and you put in a second ball that is white.
your result of your first draw is a white ball and you are looking for the probability of drawing another white ball ===> this is called "conditional probability."
use the formula: P( A|B )=P( A and B )/P( B )
A=the second draw is a white ball
B=the first draw is a white ball
A and B=the first draw and the second draw are both white balls
so for case one, we have P( B1 )=50%=0.5; for case two, we have P( B2 )=0.5*0.5=0.25
P( B )=P( B1 )+P( B2 )=0.75
P( A and B )=0.5
therefore, P(A|B)=2/3
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