Risk Faced With Insurance I'm studying a simple models of insurance.  Suppose that a machine breaks down with probability $p$, and suppose that an insurance company collects enough in premiums to pay for $k$ breakdowns.  What is the probability that there are at least $k$ break downs before yours breaks down?  Note that some number $n$ machines are insured.  Since it is insurance, uninsured machines don't "count".
It's pretty straightforward to put these in a summation form.  If there are $n$ machines getting insured, then the number of machines that break down is binomial (or normal, under CLT approximation), so that the probability that $k'$ machines break down is:
$$
   f(k') = \binom{n}{k'} p^{k'} (1 - p)^{n-k'}
$$
Now, if your machine is one of the $k'$ machines that breaks down, for $k' > k$, there are $(k' - k)!k!$ orders of breakdowns that preserve the condition (of course, I'm assuming each order is equally likely).  So the probability that my machine breaks down after the $k$th machine, given that there are $k'$ breakdowns, is $\binom{k'}{k}^{-1}$.  So using the law of total probability, we get:
$$ \sum_{k'=k}^{k'=n} \binom{n}{k'} \binom{k'}{k}^{-1} p^{k'} (1-p)^{n-k'}  \\
$$
Is there a way to evaluate that sum?
 A: The binomial distribution does not apply under the given statement of the problem.  In fact, there is insufficient information in the problem to determine the desired probability, because the question asks for a probability related to observing one event (at least $k$ failures) before observing another event (your machine fails).  Without a probability distribution for the failure time of a machine, it is not possible to answer the question.
To help you see why, think of it this way.  Suppose I have two fair coins, called $C_1$, and $C_2$.  I select one and flip it and observe the result.  I repeat the process, each time, selecting one of the two coins, flipping it, and observing the result.  The probability any coin flip lands heads is always $1/2$--this is your "$p$".  Now suppose I am interested in the probability that $C_1$ lands heads before $C_2$ does.  But I did not tell you how I chose the coins.  I did not say I chose them randomly.  I did not say I chose them alternately.  You cannot tell me the answer to the question, because I could have decided to always flip $C_1$, thereby giving me a probability of $1$; or I could have decided to always flip $C_2$, thereby giving me a probability of $0$.
