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

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