Suppose that we know the count of Aviation accidents in a specific country in a year has a poisson distribution of $\lambda =5$.
Find probability that 4 or more accident happen in a year?
Find probability that no consecutive accidents happen in a 6 month interval
We start to count accidents at time $T$. what is the probability that 5th accident happen before 9th month.
My problem is with 2nd and 3rd part.
For first part we simply have: $P(X = k) = 5^k e^{-5} / k!$
$answer = 1 - P(X=0) - P(X=1) - P(X=2)- P(X=3) = 0.7412$
For the second part I actually don't know what to do. One approach is that we know in 6 month the distribution is a new poisson with $\lambda = 2.5$ maybe no consecutive accident in 6 month means: $P_{new}(X=0) + P_{new}(X=1)$
For the third part, I think we have poisson with $\lambda = 3.75$ and we want to find $P_{new}(X=5) + P_{new}(X = 6) + ... = 1 -P_{new}(X=0) - P_{new}(X=1) - P_{new}(X=2) - P_{new}(X=3) - P_{new}(X=4) $
Is my approaches right? If not, what should I do?
Q2 : What you are looking for is inter-arrival time. For a poisson distributed random variable, the inter-arrival time is exponentially distributed. You may have learnt that exponential distribution has the property of memorylessness. So it doesn't matter where $t$ (time of the first accident) the probability of the next occurence is the same.
Let $T$ denotes the inter-arrival time, as I mentioned it is exponentially distributed with parameter $\lambda$ hence $$P(T\geq0.5)=1-(1-P(T\leq0.5))=e^{5\cdot0.5}=e^{0.25} = 0.082$$ $0.5$ means half a year since our rate $\lambda$ is yearly rate.
Q3 : It is correct