Three parts within the first minute

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Parts drop onto a conveyor belt as a Poisson process at rate \lambda = 2 per minute. What is the probability that the 3rd part arrives within the first minute?

Accepted Answer

The 3rd part arriving within the first minute is the same as seeing at least 3 parts in that minute: P(	ext{3rd arrival} \le 1) = P(N(1) \ge 3) = 1 - P(N(1) \le 2). The count N(1) is Poisson with mean \mu = \lambda t = 2 	imes 1 = 2. The small tail is easier to subtract: P(N \le 2) = e^{-2}\left(1 + 2 + \frac{2^2}{2!}ight) = e^{-2}(1 + 2 + 2) = 5\,e^{-2} \approx 0.6767. So P(N \ge 3) = 1 - 0.6767 = 0.323. About a 32\% chance of three parts within the opening minute. Takeaway. "The k-th event arrives by time t" equals "at least k events by t"; compute it as 1 minus the short Poisson tail b…

Three parts within the first minute

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