Probability and time. The chance of failure under warranty Let's say the chance of a car failing in its 5 year warranty is p=0.005 based on warranty return statistics.
Let's say I sell $n$ cars, it is a fairly simple calculation (I think) to determine the probability that $x$ or more cars fail in that 5 year period:
$$\sum_{r=x}^{n} {p^r(1-p)^{n-r} \binom{n}{r}} \ where \ x \le n$$
I am slightly unsure how I extend this to a calculation that $x$ or more cars fail on the same day, same hour, same month etc if we assume for example that over the 5 year warranty the probability of failure is equally likely.
Can anyone provide a way to proceed?
EDIT: Clarification
If you want to calculate the probability that exactly x  cars will fail in the same year in a 5 year period without specifying the year, this is more work, and you need to be more precise about what exactly you want to compute
Yes I would like to calculate that x or more cars will fail in the same period in a 5 year period without specifying the period.
Thanks
 A: If you're assuming the probability of failure is equally likely on any day or hour (hopefully it's not for new cars), then the probability of failure only depends on the length of the time interval.  Since it is $.005$ for any 5-year period, it is, for instance $.005/5 = .001$ for any 1-year period.  So you just need to divide this again by 365 (or 365.25 if you use simplified leap year rules) for the probability of failure on any day.  Similarly you can compute the probability of failure for a given month or hour or whatever. 
Now you can similarly compute the probability that exactly $x$ cars fail on a, say given, year.  
(I'm not sure if this is what you meant.  If you want to calculate the probability that exactly $x$ cars will fail in the same year in a 5 year period without specifying the year, this is more work, and you need to be more precise about what exactly you want to compute.)
Edit below:
Here's one approach to calculating the probability that at least $x$ cars fail in the same (say calendar) year in a 5 calendar year period.  First, calculate $q_m = $ the probability that exactly $m$ cars fail in the 5-year period.  Now, number these failing cars $1, 2, \ldots, m$.  There are $5^m$ possible configurations for which cars fail in which year, and all configurations are equiprobable, i.e., they each have probability $q_m/5^m$.  So what you need to do is count how many configurations there are at least $x$ cars in one year.  It's probably easier to actually count the complement of this set.  If you're fine with computational evaluation, it's not hard to write a program to calculate this.  If you want to do a more mathematical analysis, this is somewhat involved--you might want to do a little research on the balls-into-bins problem.
