Question

某人有n把钥匙,其中只有一把能打开门,现从中任取一把试开,试过的不再重复,直至把门打开为止,求试开次数的数学期望和方差.

Answer 1

设随机变量代表到打开为止时的开门次数,则

P(X=m) = (n-1)/n .(n-2)/(n-1) ...1/(n-m+1) = 1/n

其中 m=1,2...,n

因此

E(X) = 1*(1/n) + 2*(1/n) + ...+ n*(1/n) = (1+2+...+n) * (1/n) = (1+n)/2

E(X^2) = (1^2)*(1/n) + (2^2)*(1/n) + ...+ (n^2)*(1/n)

= (1^2+2^2+...+n^2) * (1/n)

= n(n+1)(2n+1)/6 * (1/n) = (n+1)(2n+1)/6

V(X) = E(X^2) - [E(X)]^2 = (n+1)(2n+1)/6 - (1+n)^2/4 = （这个式子的化简请自己算 :P）