Probability Analysis of Arrival Times in a Poisson Process within a Single-Server Queueing Model at a Theme Park

Question

I'm working on a problem regarding a Poisson process and I could use some assistance. The question is as follows:

A theme park ride follows a Poisson process, where people arrive at a rate of 3 people per minute. The ride can accommodate only one person at any given moment, and the duration for each ride is 2 minutes. As soon as one person's ride ends, the next person (if present) boards the ride instantly. If there's no one in the queue, the next arrival can get on the ride right away.

The problem states that exactly three people were in queue at the end of the fourth minute and at that precise moment, the first of them began their ride. In the following 7 minutes, only one additional person arrived. The question asks us to calculate the probability that this new person will have to wait.

Here's my attempt at a solution:

The initial three people in the queue will board the ride one by one, taking a total of 6 minutes. If the new arrival occurs during this 6-minute window $(0 \leq t < 6)$, they'll have to wait. If they arrive in the last minute $(6 \leq t < 7)$, they'll be able to board immediately.

Hence, I computed the probability of zero arrivals in the first 6 minutes, followed by exactly one arrival in the seventh minute. The Poisson probability of zero arrivals in a minute is given by:

$P(0) = \frac{e^{-3} \cdot 3^{0}}{0!} = 0.0498$

And for one arrival in a minute, it is:

$P(1) = \frac{e^{-3} \cdot 3^{1}}{1!} = 0.1493$

Multiplying these probabilities, the likelihood of no arrivals in the first 6 minutes and then one arrival in the seventh minute is:

$P(\text{0 in first 6 minutes, 1 in seventh}) = 0.0498^{6} \cdot 0.1493 = 2.278 \times 10^{-9}$

Therefore, the probability that the new arrival will have to wait is:

$P(\text{waiting}) = 1 - 2.278 \times 10^{-9} = 0.9999$

I would appreciate if someone could review my solution and confirm if my approach and calculations are correct, or provide a better way to solve the problem, if there is one. Thank you in advance.

Disclaimer: While the original problem and my attempted solution are my own, I used ChatGPT, for assistance with phrasing the content more clearly and coherently.

Answer 1

That is not what I would have guessed. If exactly one person arrives in the next $7$ minutes, then I would have thought the probability they arrive in the first $6$ minutes of those would be $\frac67$.

Let's try it your way:

• The probability $0$ arrive in $6$ minutes and $1$ in the seventh is $e^{-18}\frac1{0!} \times e^{-3}\frac3{1!}=3e^{-21}$
• The probability $1$ arrives in $6$ minutes and $0$ in the seventh is $e^{-18}\frac{18}{1!} \times e^{-3}\frac1{0!} =18e^{-21}$

So, given that one arrived in $7$ minutes, the conditional probability they arrived in the first $6$ is $$\dfrac{18e^{-21}}{18e^{-21}+3e^{-21}}=\dfrac67.$$