(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.)
