I just did the first part of this problem:
You have a lot of $50$ items and are taking a sample size of $15$. In the lot $3$ items are defective. The lot is accepted if the number of defective items, $X$, is less than or equal to $1$. Find the probability that the lot is accepted.
$$\frac{\large{47 \choose 14}{3 \choose 1}}{\large{50 \choose 15}}+\frac{\large{47 \choose 15}{3 \choose 0}}{\large{50 \choose 15}}=.7892...$$
I was then asked to find the variance. Up until this point, I've always had some pdf or CDF from which to work with. Am I suppose to recognize this as a type of binomial distribution and thus, $V(X)=npq$. Where $n=50, p=\frac{47}{50},$ and $q=\frac{3}{50}$? Or, is the probability the lot is accepted suppose to be seen as some type of mean? If so, what would $E(x^2)$ be then?
I'm not certain if this is appropriate or not, but for the sake of completeness, I thought I'd add my solution based off of georg's comment above. This does appear to be the correct answer.
We have a hypergeometric probability distribution. Thus,
$$V(X)=n\left(\frac{M}{N}\right)\left(1-\frac{M}{N} \right)\left(\frac{N-n}{N-1}\right)$$ $N=50$ $M=47$ $n=15$ $$V(X)=15\left(\frac{47}{50}\right)\left(1-\frac{47}{50}\right)\left(\frac{50-15}{49}\right)$$ $$=.6042...$$