# Expectation of number of broken machines by time t

$$n$$ machines in total (starting working at the same time), working time of each is i.i.d. $$Exp(\lambda)$$. How to calculate the expectation of the number of broken machines at time $$t$$? (If broken, they stay broken forever)

ps: how to calculate the fraction of time when there are at least one broken machines by time t if repeating this process for infinite times? Can I use the expected time until the $$1st$$ broken down here?

• Are you assuming they all start out working, and, once broken, they stay broken forever? – Michael Apr 11 at 2:19
• Yes. I will revise the original question. – cxxu96 Apr 11 at 2:50
• what is the probability of one machine be broken at time $t$? – pointguard0 Apr 11 at 2:53
• Any machine will break down after a random time ($Exp(\lambda)$), so it might be $1-exp(-\lambda t)$? – cxxu96 Apr 11 at 2:59
• So you want to use indicator functions by defining the number of broken machines at time $t$ as $B(t) = B_1(t) + ... + B_n(t)$ where $B_i(t) \in \{0,1\}$ is for machine $i$. – Michael Apr 11 at 3:09

The probability that any given machine has failed by time $$\ t\$$ is $$\ p=1-e^{-\lambda t}\$$. If the machine failures all occur independently of each other, then the number $$\ F_t\$$ which have failed by time $$\ t\$$ follows a binomial distribution: $$\mathbb{P}\left(F_t=j\right) = {n \choose j} p^j\left(1-p\right)^{n-j}\ .$$ So the expected number of machines broken at time $$\ t\$$ is the mean of this distribution, $$\ n p\left(1-p\right) = n\left(1-e^{-\lambda t}\right)e^{-\lambda t}\$$.
As I understand the question in the ps, as now clarified, its answer is $$\ \mathbb{P}\left(T_1\le t\right)\$$, where $$\ T_1\$$ is the time when the first machine breaks down. This can be obtained from the identity $$\mathbb{P}\left(T_1\le t\right)=1-\mathbb{P}\left(F_t=0\right)= 1-e^{-n\lambda t}\ .$$
• Thanks! The question in the ps is about the fraction of time when there are more than one broken machines if repeating this process for $n \rightarrow \infty$ times. – cxxu96 Apr 11 at 14:42