I have some doubts about the following problem:

I have 100 bulbs with a lifetime represented by an exponential distribution, with an expected value of 1000 hours. Find the probability that, at least one bulb, blown down after at most 500 hours.

I have calculated the probability about one bulb with this method:

$$P(X \leq 500)=\int_{0}^{500}\lambda e^{-\lambda x}dx = 1-e^{\frac{1}{2}} = 0.394$$

now, how can I extend this method for all the 100 bulbs? A step-by-step solution is really appreciated, I'm really newbie about statistics/probability arguments.

Thank you so much and best regards.

EDIT: $$\frac{1}{\lambda}=1000$$ hours so $$\lambda = \frac{1}{1000}$$

• Yes, it was my mistake sorry... Have you any idea to extend the solution for $n$ bulbs? Now I have the probability about one bulb May 5, 2020 at 18:28
• Use the binomial distribution and the converse probability: $P(X\geq 1)=1-P(X=0)$, where $X\sim Bin(100, 0.39347)$. Here $n=100$ as given in the exercise. The solution is not very surprising. May 5, 2020 at 18:39
• As I understand it we have $1-p=0.607$. The probability that one bulb blow down after 500 hours. Then $P(X=0)=0.607^{100}\Rightarrow P(X\geq 1)=1-0.607^{100}\approx 1$ May 5, 2020 at 19:08
• The probability that one bulb blow down is 0.39347 like in the integral in my request? 1-p is the probability that 99 works well or not? May 5, 2020 at 19:17
• I will sum these comments up: The probability that at least one bulb dies is $p=0.394$. So the probability that none of the $n=100$ bulbs dies ist $P(X=0)=(1-p)^{100}=0.607^{100}$. What you want is the probability that at least one dies, i.e. $P(X\geq1) = 1 - P(X=0) = 1 - 0.607^{100}$. May 5, 2020 at 20:45

I assume that it means "after at most 500 hours" right? In that case your computation makes sense for one bulb. What is $$\lambda$$ btw?

For the second part, we may assume that the bulbs are all independent and blow down within $$500$$ hours with a probability of $$p=0.394$$. You have $$100$$ bulbs. What is the chance that none of these blows down?

• the statement is not so clear (is one of the problems of this), however I have assumed $\lambda = 1/1000$ and yes, "after at most 500 hours " is a perfect correction About your question I can see that like the complement of 100% blows down? But I don't know how write this May 5, 2020 at 15:35
• Sorry, I overread the part where you say that the expected value is $1000$. I think the second part is simply a Bernoulli experiment with $n=100$ and $p=0.394$ May 5, 2020 at 15:51
• Which quantity I must use for $q$? Or better to write $1-p$? May 5, 2020 at 16:12
• By $q$ you mean $q = 1-p$? Well then this. May 5, 2020 at 16:19
• No not quite. As I mentioned before, it is a Bernoulli experiment with $n=100$. And you want the probability for $k=100$ fails ($q=0.606$ is the probability of $k=1$ fail). The answer is given by callculus above May 5, 2020 at 20:38