Suppose there are 2 defectives among 5

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suppose there are 2 defectives among 5...

suppose there are 2 defectives among 5, test one at a time until you find the 2nd defective, find:

1.you need at most 3 tests
2.given the 2nd defective found at 3rd test, find the p that you find the 1st defective at 1st test.

Thanks.
 
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shinkansenfan said:
suppose there are 2 defectives among 5, test one at a time until you find the 2nd defective, find:

1.you need at most 3 tests
2.given the 2nd defective found at 3rd test, find the p that you find the 1st defective at 1st test.

Thanks.

Forum rules require you to show your work. Tell us what you have done so far, and where you are stuck.

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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