第1个回答 2012-03-06
设Xi是第i个坛子里取到的球的编号,
P(max(Xi,X2,...,Xk)<=m )=P(X1<=m,X2<=m,...,Xk<=m)
=P(X1<=m)*P(X2<=m)*...*P(Xk<=m)
=(m/n)^k
同理
P(max(Xi,X2,...,Xk)<=m-1 )=P(X1<=m-1,X2<=m-1,...,Xk<=m-1)
=[(m-1)/n]^k
于是,
P(max(Xi,X2,...,Xk)=m )
=P(max(Xi,X2,...,Xk)<=m )-P(max(Xi,X2,...,Xk)<=m-1 )
=(m/n)^k-[(m-1)/n]^k