Question

某生产线出次品的概率为0.01,检查员每次随即抽取10个检查,如果出现次品数多于1个,就要调整设备,

Answer 1

设次品数为X,则X服从二项分布B(10,0.01)

本题要求P{2≤x≤10}=1-P{X=0}-P{X=1}=1-0.99^10-10*0.01*0.99^9≈0.004