If this switch is installed in a circuit, what is the probability that it will perform $35$ cycles before it fails?

Question

On average a very low quality electrical switch can be operated $20$ times before it fails to operate the circuit it is installed in.
a.) If this switch is installed in a circuit, what is the probability that it will perform $35$ cycles before it fails? ($36$ total)

I'm trying to solve this using negative binomial distribution:

$$P(X=x) = ({_{x-1} \mathbb C}_{r-1})P(s)^r[1-P(s)]^{x-r}$$

Right now I have $x = 35$, $P(s) = 19/20$ but I'm not sure if $r$ is correct or if my $P(s)$ is correct. I thought $r$ might be $35$ but that gives too high a probability $(.95)$ and then I tried $20$ but that seemed too low. I'm confused where to get these variables even though I know $r$ should be the $r$th success that happens on the $x$th trial.

Answer 1

I believe this is a geometric distribution. They tell you that on average it works 20 times before failure, hence we are counting the number failures until success, where I am defining failure are "the item works", and success as "the item fails". Thus, this is a geometric with mean $(1-p)/p = 20$ over the set $\{0,1,2,3,\dotsc\}$. Thus, if we call this variable $X$, then the question asks $P(X = 35)$.

Notice this is not a negative binomial since this would be counting something like the $r$th success, where success means "item fails", in this case. We are only counting until it fails once (which if you really want, is a negative binomial with $r=1$ over $\{0,1,2,\dotsc\}$).

Answer 2

You really can't answer this question unless you know how the switch fails. The switch may fail randomly with a fixed probability p every time you try it, so you would get a binomial distribution.

On the other hand, every time you operate the switch, some damage might occur. In an extreme case, you might have a switch that always breaks down after 18 to 22 operations due to accumulated damage.