There is 100 boxes of electronic components.
Number of defective components 0 1 2 3 4 5Number of boxes 76 15 4 3 1 1
3boxes are chosen . In how many ways canthe total number of defective components in these 3 boxes be fewer than 3 ?
Maths and sta question 求救 20點
2011-06-16 3:31 am
回答 (1)
2011-06-16 8:17 am
✔ 最佳答案
Denote n(a,b,c) as the number of ways that the numbers of defectivecomponents in these 3 boxes are a, b and c respectively.Number of ways that the total defective components is fewer than 3
= n(0,0,0) + n(0,0,1) + n(0,1,1) + n(0,0,2)
= 76C3 + (76C2x15C1)+ (76C1x15C2) + (76C2x4C1)
= (76!/3!73!) + [(76!/2!74!)x(15!/1!14!)] + [(76!/1!75)x(15!/2!13!)] + [(76!/2!74!)x(4!/1!3!)]
= 70300 + (2850x15) + (76x105) + (2850x4)
= 132430
參考: andrew
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