某人独立射击,直到命中目标为止,已知每次射击的命中率为0.75.则射击次数X的数学期望与方差分别是?3/4和4/9

Question

某人独立射击,直到命中目标为止,已知每次射击的命中率为0.75.则射击次数X的数学期望与方差分别是?
3/4 和 4/9

Answer 1

P{X=x}=F(x)=(1-0.75)^(x-1)0.75=0.75*(0.25)^(x-1)
E(X)=∑(1,n)xF(x)=0.75+2*0.75*0.25+3*0.75*0.25^2+4*0.75*0.25^3+...+n*0.75*(0.25)^(n-1)
E(X^2)=∑(1,n)x^2F(x)
D(X)=E(X^2)-(E(X))^2
数列忘了,你再看看怎么做,∑(1,n)中1在∑下面,n在∑上面