Suppose there are $n=2000$ laptops in the factory, some laptops are alright and some of them are defective. A laptop is defective with probability $p_1 = 0.001$. If a laptop is defective, a controller detects it with probability $p_2=0.9$ (and no chance they can mark an alright laptop as defective).
What is the probability that exactly $k$ defective laptops will be wrongly marked as alright by the controller?
I approached to the problem as follows: let $X$ be the number of defective laptops that were wrongly marked as alright by the controller, and let $Y$ be the number of defective laptops among the $n$ laptops we have. Then $$P(X = k) = \sum\limits_{m=k}^n P(X=k|Y=m)P(Y=m)$$
We can safely assume $Y \sim \text{Poisson}(np_1)$, but can we assume $P(X=k|Y=m) \sim \text{Poisson}(m(1-p_2))$? Anyway, either we set $P(X=k|Y=m) = \dfrac{(m(1-p_2))^k}{k!}e^{-m(1-p_2)}$ or $P(X=k|Y=m) = \binom{m}{k}(1-p_2)^kp_2^{m-k}$, this sum isn't really computable.
Any ideas? I would really appreciate any clue to the solution. Or you can just write a comment "I tried it too and stuck with the sum", so I know it's not just me.
This is not a situation for using a Poisson Distribution.
You have $X$, the count for faulty detection of a defect among $n$ items, when the rate of 'success' for wrong detection is independent and identical for each of the items. This 'success' rate is $p_1\cdot (1-p_2)$; the probability of having the defect and being wrongly declared defect free.
What family of distribution does this describe?