The second patient after an hour

Question

Patients arrive at a clinic as a Poisson process at rate \lambda = 4 per hour. What is the probability that the 2nd patient arrives more than 1 hour into the shift?

Accepted Answer

The 2nd patient arriving after 1 hour is the same as having fewer than 2 patients (that is, 0 or 1) during the first hour: P(	ext{2nd arrival} 1) = P(N(1) \le 1). The count N(1) is Poisson with mean \mu = \lambda t = 4 	imes 1 = 4. So P(N \le 1) = e^{-4}\left(\frac{4^0}{0!} + \frac{4^1}{1!}ight) = e^{-4}(1 + 4) = 5\,e^{-4} \approx 0.0916. About a 9\% chance of such a slow first hour. Takeaway. Translate "the k-th arrival is late" into "too few arrivals so far," then evaluate a short Poisson sum with mean \lambda t.

The second patient after an hour

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