I asked this question earlier, but am hoping for some clarity on how the answer

Question

I asked this question earlier, but am hoping for some clarity on how the answer was arrived at. Please help! A local commuter bus service advertises that buses run every 15 minutes along a certain route. Assume that the bus service follows an exponential distribution. 1. What is the probability of a bus picking up the passengers at a given bus stop in no more than 15 minutes? 2.What is the probability of a bus picking up the passengers at a given bus stop in more than 15 minutes? 3.What is the probability of a bus picking up the passengers at a given bus stop in between 15 and 25 minutes?

Explanation / Answer

Let X buses stop in an hour. Then, X ~ P(m = 3). 1) P(X = 2) = e(-3)*3^2/2! = 0.2241 2) P(no buses stop) = P(Y=0) = e(-0.75)*0.75^0/0! = 0.4724 3) P(at least one) = 1 - P(no buses) = 1 - P(Z=0) = 1 - 0.2231 = 0.7769 4) P(Somebody who arrives at the bus stop will have to wait more than fifteen, but not more than thirty, minutes for a bus) = P(no buses during 15 minutes)*P(at least one during the next 15 minutes) = 0.4724*0.5276 = 0.2492 5) P(There was no bus in the first fifteen minutes, given that there was exactly one bus in a thirty minute period) = P(There was no bus in the first fifteen minutes and there was exactly one bus in a thirty minute period/P(exactly one bus in 30 minutes) = 0.4724*(e^(-01.5)*1.5^1/1!)/(e^(-01.5)*1… = 0.4724

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