I have trouble with the following exercize, expecially with the use of the “OR” conjunction between the intervals (perhaps suggesting intersection?)
Let {N(t),t∈[0,∞)}be a Poisson process with rate λ. Find the probability that there are two arrivals in (0,2]or three arrivals in (4,7].
The solution is not to multiply the prob (2 arrivals in (0,2])*prob (3 arrivals in (4,7], is it?
I suppose the solution is much more complex than this but I could not understand how to solve it correctly. Please, I really need help! Thanks in advance
Edit: the given solution suggests $\frac{(e^{-\lambda*2})*(λ*2)^2)}{2!} + \frac{(e^{-\lambda*3})*(λ*3)^2)}{3!} + \frac{(e^{-\lambda*5})*(λ*36)^5)}{2!3!}$
if it is Prob A + Prob B - Prob A and B (intersection) I still don't get why the second addend $\frac{(e^{-\lambda*3})*(λ*3)^2)}{3!}$ is raised to the square and not to the third degree.
Is the third element the result of the multiplication of the first and the second? Sorry but I am very confused by this topic.
In probability, "$A \mbox { or } B$" represents the union of the two events. If the events are mutually exclusive, then the probability of their union is $P(A \mbox{ or } B) = P(A) + P(B)$.
If the events are not mutually exclusive, then $P(A \mbox{ or } B) = P(A) + P(B) - P(A \mbox{ and } B)$, where $A \mbox{ and } B$ is the intersection of the two.
If the events are independent (and because Poisson processes are "memoryless", event occurring in non-overlapping intervals are independent of each other), then it is often easier to note that the event "$A \mbox{ or } B$" is the conjugate of the event "$\neg A \mbox{ and } \neg B$" - i.e., the event "either A or B or both happen" and the event "neither A nor B happens" cover the whole probability space with no overlap. So you can work out $P(\neg A \mbox{ and } \neg B) = P(\neg A)P(\neg B)$ (because of independence), and hence $P(A \mbox{ or } B) = 1 - P(\neg A \mbox{ and } \neg B)$.