Question

Events occur according to a Poisson process with rate \(\lambda=2\) per hour. (a) What is the probability that no event occurs between \(8 \mathrm{P} . \mathrm{M} .\) and \(9 \mathrm{P.M.?}\) (b) Starting at noon, what is the expected time at which the fourth event occurs? (c) What is the probability that two or more events occur between \(6 \mathrm{P.M}\). and 8 P.M.?

Short answer

(a) The probability of no events occurring between 8 PM and 9 PM is \(P(0) = e^{-2}\). (b) The expected time for the fourth event to occur is 2 PM. (c) The probability of two or more events occurring between 6 PM and 8 PM is \(P(k \geq 2) = 1 - (P(0) + P(1))\).

Step-by-step solution

a) Probability of no events occurring between 8 PM and 9 PM

We want to find the probability of no events occurring in a 1-hour interval. In this case, the Poisson PMF can be used, which is given by the formula: \(P(k) = \frac{(\lambda t)^k}{k!}\) \(e^{ -\lambda t}\) where \(k\) is the number of events, \(\lambda\) is the event rate, and \(t\) is the time interval. For this question, \(k = 0\), \(\lambda = 2\) events per hour, and \(t = 1\) hour. Plugging in these values, we get: \(P(0) = \frac{(2 \times 1)^0}{0!}\) \(e^{ -2 \times 1}\) \(P(0) = e^{-2}\) Now calculate the value of \(P(0)\) to find the probability.

b) Expected time at which the fourth event occurs

We need to find the expected waiting time for the fourth event to occur starting at noon. For a Poisson process, the waiting times for the events are exponentially distributed with parameter \(\lambda\). The expected waiting time for the \(k\)-th event is given by the formula: \(E[T] = \frac{k}{\lambda}\) In this case, we are interested in the fourth event, so \(k = 4\), and the event rate, \(\lambda = 2\) events per hour. Plugging in these values, we get: \(E[T] = \frac{4}{2}\) \(E[T] = 2\) So, the expected waiting time for the fourth event to occur is 2 hours. As we start at noon, the expected time for the fourth event is 2 hours past noon, which means 2 PM.

c) Probability of two or more events occurring between 6 PM and 8 PM

We need to find the probability that two or more events occur in the 2-hour interval between 6 PM and 8 PM. We can do this by calculating the probability of 0 or 1 event occurring in this interval, and then subtracting this probability from 1. We will use the Poisson PMF again. Since the time interval is 2 hours, the rate during this interval will be \(\lambda = 2 \times 2 = 4\). The probability of \(k\) events occurring in this interval is: \(P(k) = \frac{(\lambda t)^k}{k!}\) \(e^{ -\lambda t}\) The probability of 0 or 1 event is the sum of their probabilities, which can be calculated as follows: \(P(0) + P(1) = \frac{(4)^0}{0!}\) \(e^{-4} + \frac{(4)^1}{1!}\) \(e^{-4}\) Now calculate the probability of two or more events occurring between 6 PM and 8 PM: \(P(k \geq 2) = 1 - (P(0) + P(1))\) Plug in the calculated probabilities of 0 and 1 event and compute the final probability value.