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I have a homework question that I can't seem to figure out. Any help is appreciated!

The lifetime of light bulbs (in days) is modeled as a Poisson Process with expected lifetime of beta = 200 days. A certain building has 1,800 bulbs. What is the probability that at most 10 bulbs fail on any given day?

I thought I would use a gamma distribution (alpha, beta = 200). My notes say the gamma distribution gives the waiting time until the nth occurrence, so alpha = 10 for the nth failure.. but that doesn't seem right since I believe the 10 goes with X (Prob(X <= 10)) and not alpha. And then I'm not sure what to do with the 1,800 bulbs. That seems like a binomial distribution with n = 1,800. I just don't know if I'm on the right path..and if I am how do I incorporate the two distributions?

user87654321
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1 Answers1

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It depends how you want to model this. You're saying the failure is a poisson process, which technically goes on forever, meaning that you're not taking into account the fact that once a lightbulb fails, it can't fail again. So what happens if your poisson process reports 2000 failures, when you have only 1800 lightbulbs? Perhaps you want to model each lightbulb individually:

Suppose you have $n$ lightbulbs and each fails as an independent poisson process with $\beta=200$. In other words the waiting time for failure is exponential: $f_\beta(t)=\beta e^{-\beta t}$ is the density of failure at time $t$. Suppose we want exactly $k$ lightbulbs to fail before time $t$ and $n-k$ lightbulbs to not fail. The former is equal to $P_\beta(t)^k$, where $P_\beta(t)=\int_0^t \beta e^{-\beta x}dx=1-e^{-\beta t}$. The latter is just $(1-P_\beta(t))^{n-k}$. Putting it all together and realizing we don't care which $k$ lightbulbs fail gives:

$$\binom{n}{k}P_\beta(t)^k(1-P_\beta(t))^{n-k}=\binom{n}{k}(1-e^{-\beta t})^ke^{-\beta(n-k)t}$$

The answer is then the sum from $k=0$ to $k$.

Alex R.
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