Hypergeometric Distribution Problem Involving Defective Parts The question is

A manufaturer receives a lot of 100 parts from a vendor.  The lot will be deemed unacceptable if more than five parts are defective.  So the manufacturer is going to randomly select K parts from the lot for inspection and the lot will be accepted if no defective parts are found in the sample.  How large does K have to be to ensure that the probability that the manufacturer accepts an unacceptable lot is less than .10?

So I figure this is hypergeometric with $N=100$, and $M=5$ defectives in a lot.  So I know that if i let $X$ be the number of defectives in the sample of size $K$, then $X$~Hypergeometric$(100,5,K)$. If there are no defectives in the sample, then this is
$$P(X=0)=\frac{\binom{5}{0}\binom{95}{K-x}}{\binom{100}{K}}$$
But I think the probability of the question at hand is a conditional probability.  But what would be the condition?  An unacceptable lot has nothing to do with the sample, i think, it only has to do with more than 5 parts being defective in the whole of the lot.  But the hypergeometric distribution is only on he sample of $K$ items.  Logically I don't know how to proceed.
 A: 
But I think the probability of the question at hand is a conditional probabilty. 

A lot is deemed "unacceptable" if more than 5 defective parts are found in the lot.   A lot will be "accepted" if none from any of those defective parts are found in the sample.
You have (almost) determined the probability of accepting a lot with exactly 5 defective parts.   You actually want the conditional probability of being accepted when unacceptable.   Where $N$ is the count of defectives actually in the lot this is:
$$\mathsf P(A\mid U) = \sum_{n=6}^{100-K} \dfrac{\binom{n}{0}\binom{100-n}{K}}{\binom{100}K}\mathsf P(N=n\mid N\geq 6)$$

An unacceptable lot has nothing to do with the sample

Indeed, you are missing the rate of defective parts in a lot.   Let $p$ be the rate of defectives. Then $$\mathsf P(N=n\mid N\geq 6)=\dfrac{\binom{100}{n}p^n(1-p)^{100-n}}{\sum_{j=6}^{100} \binom{100}{j}p^j(1-p)^{100-j}}$$

Perhaps you should consider a way to approximate this if the rate of defectives in the sample is sufficiently small? 
