题目内容
某批产品中有次等品,进行重复抽样检查,共取五个样品,求其次品数等于0,1,2,3,4,5的概率.
思路解析:直接使用独立重复试验的概率公式![]()
解:由题意,n=5,p=0.2,1-p=0.8,则
P5(0)=0.85=0.327 68.
P5(1)=
×(0.2)×(0.8)4=0.409 6.
P5(2)=
×(0.2)2×(0.8)3=0.204 8.
P5(3)=
×(0.2)3×(0.8)2=0.051 2.
P5(4)=
×(0.2)4×(0.8)1=0.006 4.
P5(5)=(0.2)5=0.000 32.
所以次品数等于0,1,2,3,4,5的概率分别为0.327 68,0.409 6,0.204 8,0.051 2,0.006 4,0.000 32.
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