概率论:某批产品中有10%的次品,进行重复抽样检查,...... 2009-08-12 | 分享

2025-03-22 18:53:52
推荐回答(1个)
回答1:

次品数X
P(X=0)=0.9^10
P(X=1)=(C10 1)0.9^9*0.1
...
P(X=10)=0.1^10

P(X=n)=(C10 n)(0.1^n)0.9^(10-n)
0<=n<=10

2)
P(X=0)+P(X=1)
=0.9^10+10*0.9^9*0.1
=0.9^10+0.9^9
=1.9*0.9^9
=73.61%