I have this question where i am unsure how to solve it.
X...how often a machine does not work E(X)= 3 per day= 1/8 per hour
X-Poisson distributed
What is the probability that no machine breaks down for more than 5hours.
I know that the time between poisson distributed events is exponentially distributed.
So P(Y>5)? With Y being exp. distributed. That would be the same as 1-P(Y<5)
But what is lambda, i know that if Y($\lambda_1$) is exponentially distributed than X is distributed with ($\lambda_2*(t_2-t_1))$
Can I know say that $\lambda_1/(t_2-t_1)$
$\lambda_2=1/8$ than $\lambda_1=1/40$
And we would get $e^{(-5/40)}$ as solution but this just does not seem correct to me.... I don’t feel like i really get what i am doing.....
It is handsome to define $N_t$ as the number of break downs that take place within $t$ hours.
Then for every $t>0$ random variable $N_t$ has Poisson distribution with parameter $\lambda t=\frac18t$.
(Note that e.g. $\mathbb EN_1=\lambda1=\frac18$ matching with the info in your question.)
So you are asked to find $P(N_5=0)$ where $N_5$ has Poisson distribution with parameter $\lambda 5=\frac58$.
We find:$$P(N_5=0)=e^{-\frac58}$$
For clarity if we define $X_1$ as the time that the first break down takes place then the events $\{N_5=0\}$ and $\{X_1>5\}$ are exactly the same.
So you are indeed justified to find the answer by calculation of $P(X_1>5)$.
Indeed $X_1$ has exponential distribution, and this with parameter $\lambda=\frac18$, so that:$$P(X_1>5)=e^{-\frac58}$$