Inference problem, having issues translating problem from words into probabilistic terms This is the question:
There are particular readers for swipe cards, Under normal operating conditions, there is a small probability of 0.02 that swiping a card fails to open
the entrance - some of these readers have a particular
fault. If this fault occurs, then the probability of swipe failure rises to 0.1.
an estimate says that 22% of all the readers it has sold
so far have this fault. It decides to send technicians around to test all the readers it has sold,
and replace the faulty ones. You are tasked with designing a rule to help the technicians decide
whether or not a card reader is faulty. They are to test each reader by swiping a card through it
repeatedly in successive independent trials, until a failure occurs on the N-th trial. Based on N,
the technician must decide whether or not the tested reader is faulty.
My question and answer:
Translating problems like this from words to math has always been a problem for me; I'm trying to work out the two main random variables as per a question, my answer so far has been that it is two Bernoulli random variables where one describes the probablility of a faulty card reader and the other one describes the probability of swipe failure, where you can use the law of total probability to determine the values.
P(Fail|Fault) = 0.1, P(Fail|Fault') = 0.02, P(Pass|Fault) = 0.9, P(Pass|Fault') = 0.98.
The bit that confuses me is when they ask for the two main random variables, as far as I know N is a random variable too correct? So am I interpreting the conditional variables incorrectly in the above paragraph?
 A: Welcome to CV. Nice question. Here's my attempt:
Let $X$ be random variable such that $X=k$ denotes, first swipe failure occurs in $k^{th}$ swipe. Let the probability of swipe failure in any random swipe be $p_x$. So we have:
$$X \sim Geometric(p_x)$$
Now define another random variable $Y$ that takes value $1$ if machine is faulty and $0$ otherwise. So  we have:
$$Y \sim Bern(p_y)$$
Also,
$$Pr(X=x)=\sum_y Pr(X=x|Y=y)Pr(Y=y)$$
Now, $$Pr(X=N|Y=1)Pr(Y=1) = (1-p_{x|y=1})^{N-1}p_{x|y=1}p_y = (1-0.1)^{N-1}*0.1*0.22$$
Similarly, $$Pr(X=N|Y=0)Pr(Y=0) = (1-p_{x|y=0})^{N-1}p_{x|y=0}(1-p_y) = (1-0.02)^{N-1}*0.02*0.78$$
Therefore, $$Pr(X=N)=0.1*0.22*0.9^{N-1}+0.02*0.78*0.98^{N-1}$$
Now the technician only observes $N$ and must decide whether machine is faulty. So we are interested in:
$$Pr(Y=1|X=N)$$
Using Bayes:
$$Pr(Y=1|X=N)=\frac{Pr(X=N|Y=1)Pr(Y=1)}{Pr(X=N)}$$
Let us assume that technician decides that the machine is faulty if $Pr(Y=1|X=N)\geq0.95$
So, we finally have:
$$\frac{0.1*0.22*0.9^{N-1}}{0.1*0.22*0.9^{N-1}+0.02*0.78*0.98^{N-1}}\geq0.95$$
From here you can calculate an upper bound on $N$. Hope there aren't any mistakes.
