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Let N(t) be a Poisson process with intensity λ=5, and let T1, T2, ... be the corresponding inter-...

Let N(t) be a Poisson process with intensity λ=5, and let T1, T2, ... be the corresponding inter-arrival times. Find the probability that the first arrival occurs after 2 time units. Round answer to 6 decimals.

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Answer #1

The distribution of arrivals is poisson, so distribution of the waiting time would be exponential with the same parameter 5. Therefore the distribution here is given as:

T \sim exp(5)

a) The probability here is computed as:

P(T > 2) = \int_{2}^{\infty } 5e^{-5t}dt = e^{-5*2} = 0.000045

Therefore 0.000045 is the required probability here.

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