I found this problem on http://poissonlabs.com/blog/how-insurance-works. I'm not quite sure how to solve it.
Suppose that $1000$ factories belong to an industry, and that each factory faces an independent 10% chance that they will be closed down (because of a machine failure they can't pay for). To protect themselves, each factory buys an insurance policy, which is capitalized in such a way that it will pay for $128$ machine failures before ruin.
So, under the policy, what is the probability that a factory will be closed down because of this peril?
The article makes some observations. For "your" factory to close down,
- your machine would have to break.
- 128 other machines would have to break first.
I do understand all the concepts, but I was just never very good at combinatorial probability.
It does look like the number of failures is binomial. But how do we account for the ordering?
We want the probability that at least $k=129$ out of $n=1000$ factories have a broken machine. The probability that $i$ chosen factories have a breakdown is (with $p=0.1$) $$ p^i(1-p)^{n-i} $$ Now, there are $\binom ni$ ways to choose the $i$ factories. This gives a probability of $$ \binom ni p^i(1-p)^{n-i} $$ Now we have to sum from $k$ to $n$ to get the probability that at least one factory closes down: $$ P\{\text{at least one factory closes down}\} = \sum_{i=k}^n\binom ni p^i(1-p)^{n-i} $$