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標題:
F.5 math-probability(1)
發問:
最佳解答:
(a) The probability of no congestion at road AXYB = (1 - 0.2)(1 - 0.4) = 0.48 So, the probability of having congestion at road AXYB = 1 - 0.48 = 0.52 (b) The driver chooses AXYB or AZB by random (with probability 1/2). The probability of having congestion at the road chosen = (1/2)(0.52) + (1/2)(0.4) = 0.27 + 0.2 = 0.47 2011-02-27 12:18:02 補充： Sorry, The probability of having congestion at the road chosen = (1/2)(0.52) + (1/2)(0.6) = 0.27 + 0.3 = 0.57
其他解答:
F.5 math-probability(1)
發問:
此文章來自奇摩知識+如有不便請留言告知
圖片參考：http://imgcld.yimg.com/8/n/HA00265776/o/701102270040713873384980.jpg The figure shows two roads AXYB and AZB. X, Y and Z are spots of traffic congestion and their probabilities of congestion are 0.2, 0.4 and 0.6 respectively.(a) What is the probability of having congestion at road AXYB?(b) If a driver... 顯示更多 圖片參考：http://imgcld.yimg.com/8/n/HA00265776/o/701102270040713873384980.jpg The figure shows two roads AXYB and AZB. X, Y and Z are spots of traffic congestion and their probabilities of congestion are 0.2, 0.4 and 0.6 respectively. (a) What is the probability of having congestion at road AXYB? (b) If a driver chooses one of the roads at random, what is the probability of having congestion at the road chosen? *唔該詳解最佳解答:
(a) The probability of no congestion at road AXYB = (1 - 0.2)(1 - 0.4) = 0.48 So, the probability of having congestion at road AXYB = 1 - 0.48 = 0.52 (b) The driver chooses AXYB or AZB by random (with probability 1/2). The probability of having congestion at the road chosen = (1/2)(0.52) + (1/2)(0.4) = 0.27 + 0.2 = 0.47 2011-02-27 12:18:02 補充： Sorry, The probability of having congestion at the road chosen = (1/2)(0.52) + (1/2)(0.6) = 0.27 + 0.3 = 0.57
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