(1)P(4)=(5^4/4I)*e^(-5)=0.175
(2) P(0)=e^(-5)=0.007
P(1)=5*e^(-5)=0.034
P(2)=(5^2/2I)*e^(-5)=0.084
故一只面包所含葡萄干粒数至少为3的概率=1-P(0)-P(1)-P(2)=0.875
(1)P(4)=(5^4/4I)*e^(-5)=0.175
(2) P(0)=e^(-5)=0.007
P(1)=5*e^(-5)=0.034
P(2)=(5^2/2I)*e^(-5)=0.084
故一只面包所含葡萄干粒数至少为3的概率=1-P(0)-P(1)-P(2)=0.875