Let $n$ items be drawn in order without replacement from a shipment of $N$ items of which $N\theta$ are bad. Let $X_{i} = 1$ if the $i$-th drawn is bad and $X_{i} = 0$ otherwise. Show that $\sum X_{i}$ is sufficient for $\theta$ directly and by the factorization theorem.
The problem is I do not know which probability model should we use this context, which makes difficult to assert the given proposition. Can someone help me out? Thanks in advance!