Question

设随机变量X~b(3,0.4),求Y=X^2-2X的分布律

Answer 1

P(X=0)=0.6^3=0.216,此时Y=0

P(X=1)=3*0.4*0.6^2=0.432,此时Y=-1

P(X=2)=3*0.4^2*0.6=0.288,此时Y=0

P(X=3)=0.4^3=0.064,此时Y=3

上述概率可转化为：

P(Y=0)=0.504

P(Y=-1)=0.432

P(Y=3)=0.064

故其分布为

y 0 -1 3

P 0.504 0.432 0.064