Im trying to prepare for an university exam in Probability, but I got stuck on this practice sheet question:
A company produces lots of highly complex devices. As a consequence, these devices may not always work properly. Suppose, individual devices are defective with a probability p, and these problems occur independently. Also the number X of devices produced on a certain day is random. It follows a Poisson probability law: $$ P[X = k] = \frac{\mu^k}{k!}e^{-\mu} ; k\ge0, \mu>0 $$ On a sunny day in October, how large is the probability to have no defective device at all?
My take on it was to define a new random variable Y as the number of deffective devices out of $k$ produced devices.Then the probability to get $n$ deffective devices out of $k$ produced would follow the binomial law, thus: $$P[Y = n] ={ k \choose n}p^n(1-p)^{k-n} $$ and for zero defective devices this gives us: $$P[Y = 0] =(1-p)^{k} $$ If X and Y are independent then the probability to get $0$ defective devices on a day is just going to be the product of those probabilities yielding: $$P[Y=0|X=k] = \frac{\mu^k}{k!}e^{-\mu}(1-p)^{k}$$ The problem is that now I am not sure whether X and Y are actually independent and also I couldn't come up with any way of checking whether my answer makes sense. Could someone please clarify and if I am wrong with my approach explain how to actually calculate the desired probability?
Let $Z$ denote the number of defective devices on a day. Then actually you found that: $$P(Z=0\mid X=k)=(1-p)^k$$ and can go on with:$$P(Z=0)=\sum_{k=0}^{\infty}P(Z=0\mid X=k)P(X=k)=\sum_{k=0}^{\infty}(1-p)^ke^{-\mu}\frac{\mu^k}{k!}=e^{-\mu}e^{(1-p)\mu}=e^{-p\mu}$$
It can be shown that $Z$ has Poisson distribution with parameter $p\mu$.