In general terms, this is a renewal process problem. In a renewal process, the times (or distances, etc.) between successive events are independent and identical, but they can take any distribution. In a Poisson process, the times between successive events are described by independent identical Exponential distributions. The Poisson process is thus a particular case of a renewal process. The mathematics of the distributions of number of events in a period (equivalent to the Poisson distribution for the Poisson process) and the time to wait to observe x events (equivalent to the Gamma distribution in the Poisson process) can be quite complicated, depending on the distribution of time between events. However, Monte Carlo simulation lets us bypass the mathematics to arrive at both of these distributions, as we will see in the following examples.

It is known that a certain type of light bulb has a lifetime that is Weibull(0, 4020, 1.3) hours distributed.

**First question**: If I have one light bulb working at all times, replacing each failed light bulb immediately with another, how many light bulbs will have failed in 10 000 hours?

The example model One Lightbulb provides the solution to this question. Note that it takes account of the possibility of 0 failures.

The links to the One Lightbulb software specific models are provided here:

The chart shows the distribution of the outcome:

**Next question**: If I have 10 light bulbs going at all times, how many will fail in 1000 hours assuming that I immediately replace my failed bulbs?

The model Ten Lightbulbs shows a model to provide the solution to this question. It follows exactly the same logic as the model above.

The links to the Ten Lightbulbs software specific models are provided here:

The figure below compares the results for this question and for the previous one. Note that they are significantly different. Had the time between events been Exponentially distributed, the results would have been exactly the same:

**Last question**: If I had one light bulb going constantly, and I had ten light bulbs to use, how long would it take before the last light bulb failed? The answer is simply the sum of 10 independent Weibull distributions.