It's been 15 years ago I did studies in statistics, so most of the knowledge is gone. I've looked up Poisson's distribution but the tooth of time has taken a too large munch of my skills.
If I know that an alarm occurs 10 times a day, I have the average time between them to 144 minutes, which is my expected value. But what is the likelihood of the following events?
- No alarm for 144 minutes
- Precisely 1 alarm in 144 minutes
- No alarm for 288 minutes
- At least 3 alarms in 288 minutes
Please note that I'm not looking for the answer but rather a pointer on how to approach it. And in case this resembles school assignment and a lazy student, I'm not. However, it's a bit difficult to prove a non-fact. Perhaps a credential consisting of my diploma work with a 10+ years worth of dust on it will do.
Are you assuming alarms are independent events with Poisson distribution? If so, it straightforward. Use $p_n=e^{-\lambda}\frac{\lambda ^n}{n!}$. where $p_n$ is the probability of $n$ alarms in an interval where $\lambda=1$ for the first two questions and $=2$ for the second two.
Note for question 4, the answer requires $\sum_{n=3}^\infty p_n$. It is easier to get $1 - \sum_{n=0}^2 p_n$.