共取5件样品,求其中次品数等于0,1,2,3,4,5的概率 解:P(O)=C3(0.2)08)3≈0.3277 ()=C(0.2)(0.8)4≈0.4096 3(2)=C3(0.2)(0.8)≈0.2048 3(3)=C3(0.2)(0.8)2≈0.0512 13(4)=C(0.2)(0.87)≈0.0064 r3(5)=G3(0.2)(0.8)≈0.0003例1 一批产品中有20%的次品，进行放回抽样检查， 共取5件样品，求其中次品数等于0,1,2,3,4,5的概率 5 P (2) = 解： 2 2 3 5 C (0.2) (0.8) 0.2048.  5 P (1) = 1 1 4 5 C (0.2) (0.8) 0.4096.  5 P (3) = 3 3 2 5 C (0.2) (0.8) 0.0512.  5 P (0) = 0 0 5 5 C (0.2) (0.8) 0.3277  5 P (4) = 4 4 1 5 C (0.2) (0.8) 0.0064.  5 P (5) = 5 5 0 5 C (0.2) (0.8) 0.0003. 