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Suppose that we know the count of Aviation accidents in a specific country in a year has a poisson distribution of $\lambda =5$.

  1. Find probability that 4 or more accident happen in a year?

  2. Find probability that no consecutive accidents happen in a 6 month interval

  3. 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?

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  • $\begingroup$ All look good to me. $\endgroup$
    – antkam
    Commented Nov 7, 2019 at 20:23
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    $\begingroup$ You are making the slightly stronger assumption that the accidents follow a homogeneous Poisson process, but that seems expected if you are to be able to answer the question $\endgroup$
    – Henry
    Commented Nov 7, 2019 at 20:51

1 Answer 1

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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

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  • $\begingroup$ I found your answer for Q2 a little odd. because 0.082 is also equal to $P_{poisson}(\lambda= 2.5 , X=0)$ should we consider $P(X=1)$ somehow? $\endgroup$
    – amir na
    Commented Nov 8, 2019 at 10:32
  • $\begingroup$ It is not mere coincidence. My slightly longer answer describes the connection between poisson distribution and exponential distribution. Think of it like this if there is an accident on march what is the probability that no event occur in between march and september. This quantity happens to be $P(X=0;\lambda=2.5)$, this should holds True for any $t$ where $t$ denotes time of first occurence or for this case if the first accident occurs on February or even December. tl;dr we are interested in probability of no occurence between 6 months time frame, hence $P(X=1)$ is irrelevant. $\endgroup$
    – Yohanes Alfredo
    Commented Nov 8, 2019 at 10:57

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