Poisson Question - number of signals received at station A

Question

the question:

 The number of signals received at station "A" during an hour is poissonian distribution with parameter 5.
It is decided to close station "A" after the first 2 consecutive hours in which it will NOT receive signals,
What is the probability that station A will close after 5 hours?

my approach here is very simple:
$P(x=0) = e^-5$ for not receiving any signal in hour
$1-e^-5$ to receive at least one signal in hour
Im looking for $P(will close > 5 hours)$ so its easier to calculate $1 - P(will close <= 5hours)$ what i did was write all options and sum them where "0" = no signal, "1" = at least 1 signal, the options are: "00", "100", "1100", "0100", "11100", "01100", "10100"

but ive been wondering if there is an easier/simpler solution? because what if it was more complex situation and i was trying to find $P(will close>15)$ for example

Answer 1

Let $a_n$ be the probability of being open after the $n$th hour, and $p=e^{-5}$. You can be open after hour $n+1$ either because you got a signal in hour $n+1$, or because you didn't get a signal but you got one the previous hour. Therefore the possible strings corresponding to being open after $n+1$ are:

• a string of length $n$ with a $1$ added to it

• a string of length $n-1$ with a $10$ added to it

and these cases are disjoint. So $$a_{n+1}=(1-p)a_n+(1-p)pa_{n-1}$$ This gives an easy to compute or solve recursion relation, and may strike you as reminiscent of the Fibonacci sequence.