n个球n个盒子,一个扔一个球,会随机落到一个盒子里.求：（要有过程,回答正确会有追加分的）

(1)
x:number of empty boxes
p:probabilty that the box is empty (p1=p2=...pn=p)
n个球n个盒子 => p=(n-1/n)^n
E(x) = sum(pi * 1) = np = n* ((n-1)/n)^n = (n-1)^n / n^(n-1)
(2)
x:number of one-ball boxes
p:probabilty that the box has one ball (p1=p2=...pn=p)
n个球n个盒子 => p=n*(1/n)((n-1)/n)^(n-1)=((n-1)/n)^(n-1)
E(x) = sum(pi * 1) = np =n*((n-1)/n)^(n-1) = (n-1)^(n-1) / n^(n-2)
(3)
x:number of two-ball boxes
p:probability that a box has 2 balls
y:投多少次球before任何一个盒子有两个球之前
y个球n个盒子 => p=[y(y-1)/2]*(1/n)^2*((n-1)/n)^(y-2)
choose y so that E(x) = 1
=>E(x) = sum(pi * 1) = np =1
=> n[y(y-1)/2]*(1/n)^2*((n-1)/n)^(y-2)=1 and solve this equation and this will imply E(y)
(y cannot be written explicitly as n from above equation.If you need to write explicitly,you need to make assumption that n is sufficiently large,and the binomial distribution can be approxmate by POISSON distribution.)

( .空盒子数量的期望) 期望是啥？

什么是期望？？？是概率么？如果是，那我来解。当你有一个盒子一个球时，概率为100%。
当你有2个盒子2个球时，概率为33.3%
三个盒子三个球时，概率为10%
以此类推

题目没看明白；

古典概型，比较麻烦，找一本概率论辅导书，有这种题型，算是比较难的。
考研数学参考书也有这种类型，在概率统计前几章。陈文灯，李永乐的书都有，如果你方便的话自己找一下。