Probability of No Events in First Two Hours for Poisson Process with Rate 2

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SUMMARY

The discussion focuses on calculating the probability of no events occurring in the first two hours of a Poisson process with a rate of 2 events per hour. The exponential distribution is utilized, with the probability density function (pdf) defined as pdf = (1/2)e-x/2 for x ≥ 0. The probability that the time until the first event exceeds two hours is equivalent to the probability of zero events occurring in that timeframe. The correct approach involves calculating P(x > 2) using the cumulative distribution function.

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Homework Statement


Consider a Poisson process for which events occur at a rate of 2 per hour.
(a) Give the probability that the time until the first event occurs exceeds 2 hours. Use an exponential distribution to find the probability.

Homework Equations


The Attempt at a Solution


\lambda = 1/2

pdf = (1/2)e-x/2, x >= 0

The probability that the time until the first event occurs exceeds two hours, is equal to the probability that 0 events occur in the first two hours.
I have the pdf, but I'm not sure how to use it to find the probability.
 
Last edited:
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P(x > 2) = 1 - p(x=1) - p(x=2)
 

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