Central Limit Theorem and Lightbulbs

Question

"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.

Answer 1

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 $W\sim\mathsf{Gamma}(\mathrm{shape}=3,\mathrm{rate}=1/5).$ 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")