How to find probability in repeated trials?

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Info: I made the mistake of starting engineering grad school years after finishing my undergrad, and forgot my basic prob/stat. This is related to reliability engineering.

Homework Statement



How many arbitrarily selected units do I have to test so that my probability is 0.85 of finding one at least one defective product if I have a 0.05 defect rate for this particular product.

Homework Equations



Binomial Probability Formula

binomial%20probability%20formula.gif


The Attempt at a Solution



I tried setting the formula to be equal to 0.85, k=1 (because I only need one to fail), and left number of trials as "n" to solve for. I have tried to solve for this and was having a tough time, and I am not even sure if this is the correct approach.

Edit: p=0.05 and q=1-p=0.95
 
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TiredEngineer said:
Info: I made the mistake of starting engineering grad school years after finishing my undergrad, and forgot my basic prob/stat. This is related to reliability engineering.

Homework Statement



How many arbitrarily selected units do I have to test so that my probability is 0.85 of finding one at least one defective product if I have a 0.05 defect rate for this particular product.

Homework Equations



Binomial Probability Formula

binomial%20probability%20formula.gif


The Attempt at a Solution



I tried setting the formula to be equal to 0.85, k=1 (because I only need one to fail), and left number of trials as "n" to solve for. I have tried to solve for this and was having a tough time, and I am not even sure if this is the correct approach.

Edit: p=0.05 and q=1-p=0.95

The probability of getting at least one failure after k trials is one minus the probability of getting no failures after k trials. You should find the 'no failures' case easier to solve.
 
What Dick said. To elaborate, setting k=1 gives the probability of exactly one failure - you want at least one.
 
What Dick said. To elaborate, setting k=1 gives the probability of exactly one failure - you want at least one.
 
Thanks for the help guys. That was a herp-derp moment for me.
 
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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