P(abCdef,abcDef,abcdEf,abcdeF,abcdef}=.681^2*.319+.681^3*.319+..

來源: jinjing 於 2011-02-06 17:43:32 [] [舊帖] [給我悄悄話] 閱讀數 : (219 bytes)
回答: why P(abCdef,abcDef,abcdEf,abcdef}=.535? 由 wushuihe 於 2011-02-06 12:31:45

Sorry,I'm outside, a litttle carelessness.Thank you remind me,It should be:

P(notA)=P(notB)=...P(notF)=0.681.P(A)=P(B)=...=.319

So,P(notA*notB)=P(notA)*P(notB),they are independent.

 

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• 回複：P(abCdef,..,f}=.681^2*.319+.681^3*.319+..=.464=.681*.681 -jinjing- ♀ 給 jinjing 發送悄悄話 (0 bytes) () 02/06/2011 postreply 17:47:52

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