F.4 MATHS..............................
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F.4 MATHS
2010-07-16 9:42 pm
回答 (1)
2010-07-16 9:54 pm
✔ 最佳答案
a) (Q1 , Q2 , Q3)(TTT) , (TTF) , (TFT) , (TFF)(FTT) , (FTF) , (FFT) , (FFF)b)i) P(E) = 1 - P(all wrong) = 1 - (1/2)^3 = 7/8ii) P(E) = P(Q2 correct) * P(Q3 wrong) = (1/2)(1/2) = 1/4or by a) , we can see that (TTF) and (FTF) is the required cases ,P(E) = 2/8 = 1/4
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100716000051KK00711