If two events occur at a rate of 1.8 per hour on average, and this occurrence follows a poisson process, what is the probability that there is at least 1 hour between two events?
My approach for this question is the following:
$$\lambda = \frac{1}{E(x)} = \frac{1}{1.8} = 0.556$$
The probability of the time being under an hour (using the cumulative probability density for the poisson process):
$$P(X<1) = 1-e^{-\lambda x} = 1-e^{-0.556(1)} = 0.4265$$
Therefore the probability of the time being above an hour would be:
$$P(X\ge1) = 1-P(X<1) = 1-(1-e^{-0.556}) = e^{-0.556} = 0.573$$
The logic here seems obvious: The probability of a given wait time for independent events following a poisson process is determined by the exponential probability distribution $\lambda e^{-\lambda x}$ with $\lambda = 0.556$ (determined above), so the area under this density curve (the cumulative probability) is 1. Therefore, the probability of any time range would be the integral of this distribution function over the given time range. The integral from 0 to 1 is the probability of the time between events being under 1 hour, which is $0.4265$, and so the probability of the time between events being above 1 hour is $1-0.4265 = 0.573$.
…however, this is apparently not correct. The answer to this in the back of the book I am using is $0.1653$, and by the logic I've outlined above I am not seeing how. I am very much hoping my approach is proper and that the book is wrong, and am hoping someone can help me correct or confirm my logic.
Best Answer
The Poisson distribution $\lambda$ is $1.8$. The probability of no event in an hour is $e^{-\lambda}$.
Another way: The waiting time has exponential distribution with parameter $\lambda$. Now calculate as you did, with parameter $1.8$.