NO.PZ2017092702000094
问题如下：
A stock is priced at $100.00 and follows a one-period binomial process with an up move that equals 1.05 and a down move that equals 0.97. If 1 million Bernoulli trials are conducted, and the average terminal stock price is $102.00, the probability of an up move (p) is closest to:
选项：
A.0.375.
B.0.500.
C.0.625.
解释：
C is correct.
The probability of an up move (p) can be found by solving the equation: (p)uS + (1 – p)dS = (p)105 + (1 – p)97 = 102. Solving for p gives 8p = 5, so that p = 0.625.
100到105是move up，概率为p；100到97是move down，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
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？
