 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2012/m/m5/b/solution

N_{5} - N_{3} \sim Poisson (2 λ)

. Therefore

P (N_{5} - N_{3} = 4) = e^{- 2 λ} \frac{( 2 λ ) ^{4}}{4 !} .

There is zero probability of an arrival at either deterministic endpoint, so endpoint inclusion does not alter the result.

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