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NO.PZ2017092702000094 问题如下 A stois price$100.00 anfollows a one-periobinomiprocess with up move thequals 1.05 ana wn move thequals 0.97. If 1 million Bernoulli trials are concte anthe average terminstopriis $102.00, the probability of up move (p) is closest to: A.0.375. B.0.500. C.0.625. C is correct.The probability of up move (p) cfounsolving the equation: (p)uS + (1 – p) = (p)105 + (1 – p)97 = 102. Solving for p gives 8p = 5, so thp = 0.625.100到105是move up，概率为p；100到97是move wn，概率为1-p，这是一期二叉树的情况，此时这个一期二叉树的均值就是105×p+97×（1-p）。根据题干，这个均值为102105×p+97×（1-p）=102，可以直接解得p=0.625 没看答案是 我用的二项分布算的，E(X) =np算呢，p= 102/1million 。。。。没答案，，，然后就不知道知识点了这咋能看出来是二叉树呢，我记得课上讲的时候咋的还有个tree的单词。。。咋理解呀老师
NO.PZ2017092702000094问题如下A stois price$100.00 anfollows a one-periobinomiprocess with up move thequals 1.05 ana wn move thequals 0.97. If 1 million Bernoulli trials are concte anthe average terminstopriis $102.00, the probability of up move (p) is closest to:A.0.375.B.0.500.C.0.625. C is correct.The probability of up move (p) cfounsolving the equation: (p)uS + (1 – p) = (p)105 + (1 – p)97 = 102. Solving for p gives 8p = 5, so thp = 0.625.100到105是move up，概率为p；100到97是move wn，概率为1-p，这是一期二叉树的情况，此时这个一期二叉树的均值就是105×p+97×（1-p）。根据题干，这个均值为102105×p+97×（1-p）=102，可以直接解得p=0.625 请问为什么上涨是涨到105，下跌是跌到97？ 题目不是说的是up move equals 1.05吗？
NO.PZ2017092702000094 问题如下 A stois price$100.00 anfollows a one-periobinomiprocess with up move thequals 1.05 ana wn move thequals 0.97. If 1 million Bernoulli trials are concte anthe average terminstopriis $102.00, the probability of up move (p) is closest to: A.0.375. B.0.500. C.0.625. C is correct.The probability of up move (p) cfounsolving the equation: (p)uS + (1 – p) = (p)105 + (1 – p)97 = 102. Solving for p gives 8p = 5, so thp = 0.625.100到105是move up，概率为p；100到97是move wn，概率为1-p，这是一期二叉树的情况，此时这个一期二叉树的均值就是105×p+97×（1-p）。根据题干，这个均值为102105×p+97×（1-p）=102，可以直接解得p=0.625 老师好，看到这道题时尝试用二项分布求概率来解答，解不出来，感觉跟二叉树有些混淆了，请问怎么区分题目是考哪个知识点？
NO.PZ2017092702000094问题如下A stois price$100.00 anfollows a one-periobinomiprocess with up move thequals 1.05 ana wn move thequals 0.97. If 1 million Bernoulli trials are concte anthe average terminstopriis $102.00, the probability of up move (p) is closest to: A.0.375. B.0.500. C.0.625. C is correct.The probability of up move (p) cfounsolving the equation: (p)uS + (1 – p) = (p)105 + (1 – p)97 = 102. Solving for p gives 8p = 5, so thp = 0.625.100到105是move up，概率为p；100到97是move wn，概率为1-p，这是一期二叉树的情况，此时这个一期二叉树的均值就是105×p+97×（1-p）。根据题干，这个均值为102105×p+97×（1-p）=102，可以直接解得p=0.625 另外，1 million Bernoulli trials are concte题干里面的这句话如何理解呢，二叉树列跟伯努利试验有啥关联吗
NO.PZ2017092702000094 0.500. 0.625. C is correct. The probability of up move (p) cfounsolving the equation: (p)uS + (1 – p) = (p)105 + (1 – p)97 = 102. Solving for p gives 8p = 5, so thp = 0.625. 100到105是move up，概率为p；100到97是move wn，概率为1-p， 这是一期二叉树的情况，此时这个一期二叉树的均值就是105×p+97×（1-p）。 根据题干，这个均值为102 105×p+97×（1-p）=102，可以直接解得p=0.625 我知道要算概率，但是不知道怎样用二叉树计算，麻烦老师讲解一下
