【答案】应助回帖

2楼: Originally posted by Edstrayer at 2015-05-19 03:24:49
随机变量X的取值为m,m+1,m+2,\cdots,m+k,\cdots,n
随机变量X取值m+k(k=0,1,2,\cdots,n-m)的概率为：
P(X)=\frac{C_{n-m}^kC_m^m}{C_n^{m+k}}
所以，x的期望就等于：
E(X)=\sum\limits_{k=0}^{n-m}(m+k)P(X)
因 ...