Thread: Poisson Process and Two Exponentials

I understand how a Poisson process works if you have many exponentials (like buses coming to a bus stop), but what about a finite number? For example two light bulbs fail according to the exponential distribution and I'm asked the probability one failing. How do I work the fact that their are two into the equation? Or does it matter? The parameter in the gamma distribution should be the number that fail, correct?

The number of random events occurring within a defined time interval follow a Poisson distribution, so the number of light bulbs that fail each month would be an example of a Poisson Random Variable. While the time between random events follows an exponential distribution, the time until failure of each light bulb for instance. So if you want the probability of both light bulbs failing in a particular time interval you must use the Poisson distribution, the parameter in this distribution will be the average number of light bulb failures per unit of time being considered and this information can be determined from the average time for 1 light bulb to fail.

So when using the gamma distribution to find the probability of say 2 failing in a given time, I use k=2 (using the pdf from wikipedia), and don't care about the fact that there are only 2?

Use the PDF for the Poisson distribution,



For x = 2 and lambda being the rate parameter i mentioned previously.