Question

抽检概率题一批280万件产品,有2%的次品.检验方法为,随机抽取20件产品,如未检出次品即判定此批产品合格,如检出次品,

Answer 1

A=﹛一次检验（没有复检）合格﹜

B=﹛复检合格﹜

C=﹛货物合格﹜

∴P(A)=C﹙2800000×98%,20﹚／C﹙2800000,20﹚=C﹙2 744 000,20﹚／C﹙2800000,20﹚

P(B)=[1-C﹙2 744 000,20﹚／C﹙2800000,20﹚][C﹙2 743980,20﹚／C﹙2799980,20﹚+C﹙2 743980,19﹚C﹙56000,1﹚／C﹙2799980,20﹚+C﹙2 743980,18﹚C﹙56000,2﹚／C﹙2799980,20﹚]

P(C)=P(A)+P(B)=

C﹙2 744 000,20﹚／C﹙2800000,20﹚

+[1-C﹙2 744 000,20﹚／C﹙2800000,20﹚][C﹙2 743980,20﹚／C﹙2799980,20﹚+C﹙2 743980,19﹚C﹙56000,1﹚／C﹙2799980,20﹚+C﹙2 743980,18﹚C﹙56000,2﹚／C﹙2799980,20﹚]