# Central Limit Theorem and Lightbulbs

Mathematics Asked by Maksymilian5275 on March 1, 2021

"Suppose I have lightbulbs that have a lifetime which is exponentially distributed with an average lifespan of 5 years. Each time a bulb dies, it is replaced with an another of the same type.

What is the probability that in 10 years, at least three bulbs break?"

So far I have that $$mu=5$$, E$$[X]=frac{1}{5}=0.2$$, Var$$[X]=frac{1}{lambda^2}=frac{1}{25}=0.04$$.

My next step was going to be approaching the problem by plugging in these values into the formula for the central limit theorem, namely:

$$chi=frac{N-0.2}{0.04}$$

But this just returns $$245$$ when I plug in $$N=10$$ years which doesn’t seem to be right. If anyone could explain where my understanding of the topic is failing and where I’m going wrong that would be great.

Multiple Exponential Waiting Time as Poisson or Gamma

Poisson (as in Comments by @MatthewPilling and me): The ten-year mean is $$lambda_{10} = 10(1/5) = 2.$$ Let $$X sim mathsf{Pois}(lambda_{10}=2),$$ then you seek $$P(X ge 3) = 1 - P(X le 2) = 0.3233.$$ In R, where ppois is a Poisson CDF:

1 - ppois(2, 2)
[1] 0.3233236


Gamma (equivalently): The waiting time for three successive exponential failures (mean=5, rate=1/5) is $$Wsimmathsf{Gamma}(mathrm{shape}=3,mathrm{rate}=1/5).$$ You seek $$P(W le 10) = 0.3233.$$ In R, where pgamma is a gamma PDF:

pgamma(10, 3, 1/5)
[1] 0.3233236


Simulation:

set.seed(1112)
w = replicate(10^6, sum(rexp(3,.2)))
mean(w <= 10)
[1] 0.323841

hist(w, prob=T, br=60, col="skyblue2",
main="Wait for 3")
curve(dgamma(x, 3, .2), add=T, lwd=2, col="orange")
abline(v=10, lty="dotted", lwd=3, col="darkgreen")


Correct answer by BruceET on March 1, 2021