数学5
在150个产品中有15件次品,135件正品,任取3件,求恰有有3件次品的概率
恰有有3件次品的概率C(15,3)/C(150,3)=91/149/740≈0.000825
P=C(15,3)/C(150,3)=13*14*15/148/149/150≈0.000825 恭喜曼丽回答正确
P=C(15,3)/C(150,3)=91/149/740≈0.000825
答:恰有有3件次品的概率C(15,3)/C(150,3) 至少有1件正品的概率=1-C(15,3)/C(150,3)详情>>
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