mathematics!!

First of all, before approaching the centre of this question, we have to consider the events "What is the probability of getting 100 non-defective fuse from 1000 fuses in which n is/are defective" (where n = 0,1,2,3,4,5) and their corresponding probabilities Pn.
So for P0, it is surely that all 100 selected fuses are non-defective since all are non-defective, hence P0 = 1.
For P1, we can think as, out of 1000 fuses, there are a total of 1000C100 possibilities in selecting 100 fuses and, in which there are al total of 999C100 possibilities of selecting 100 non-defective fuses (since 999 are non-defective in this case). Therefore, P1 = 999C100 /1000C100. And the rest can be obtained through such analysis.
Summarizing, we have the following probabilities:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Bayes.jpg

Now, let's back to the centre of the question:
Since the fact that "100 fuses are non-defective" has been confirmed, so the possibility to be considered must be one of the 6 mentioned above and hence, by Baye's theorem, the probability that all 1000 fuses are non-defective is:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Bayes1.jpg

which gives a numerical value of 0.213 (correct to 3 d.p.).

因為none was defective,所以
of no defective fuses係100%