# How to calculate the probability that no defective devices are produced for a given day?

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?

• The random variable $Y$ that you introduce is a confusing one. It is connected with some fixed $k$ and just seems to "fall out of the sky". Is it defined on the same probability space as $X$? If yes, then how did you do that? If no then you cannot compare $X$ and $Y$ so that independence does not even has a meaning here. – drhab Jan 18 '18 at 15:31

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 $p\mu$.
• If we say in our model that the number of produced devices on a sunny day in october is $N$ then $N$ is not a fixed number but a random variable. It can take values $0,1,2,3,\dots$, and if $N$ is assumed to have Poisson distribution then each of the events $\{N=0\},\{N=1\},\{N=2\}\dots$ has positive probability. – drhab Jan 18 '18 at 19:11