急求概率论题解答！x服从泊松分布，p(x=0)=0.4,求p(x>2)！要详细过程！在线等！

p(x=0)=0.4=e^(-λ) λ=-ln0.4 p(x=1)=-0.4ln0.4 p(x=2)=0.4ln²0.4 p(x>2)=1-P(x=0)-P(x=1)-P(x=2)=1-0.4(ln²0.4-ln0.4+1)=0.6-0.4(ln²0.4-ln0.4)