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I am trying to solve the following question.

A pedestrian needs at least 8 seconds to cross a dual-lane carriage-way. Vehicles in the near-side lane arrive at the rate of 10 per minute and vehicles in the far-side lane at the rate of 8 per minute. Making suitable assumptions:What is the probability that the nearest vehicle (in time) will be in the near-side lane?

Thoughts towards solution

We have been given two Poisson processes the first with mean $\lambda_1 = 10 $ cars per minute and the second with $\lambda_2 = 8 $ cars per minute. The interval between two successive arrivals has a negative exponential distribution in each case with means $\dfrac {1}{\lambda_1}$ and $\dfrac {1}{\lambda_2}$. I think if i understand the problem correctly we are interested in computing

Let $T_1$ be the arrival time of the next vehicle in the near-side lane. Let $T_2$ be the arrival time of the next vehicle in the far-side lane.

$$P(T_1 <T_2) = \int _{0}^{8}P(T_{1} < t)P(T_{2}>t)dt $$ $$= \int _{0}^{8}(1-e^{-10t})(e^{-8t})dt $$

This evaluated to 0.0694 which seems intuitively kind of low. I suspect that i have made a mistake. Any help would be much appreciated.

Thanks in advance.

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A pedestrian needs at least 8 seconds to cross a dual-lane carriage-way: not useful? – Did Aug 28 '12 at 19:25
@did There is actually two more sub parts to this question this information was useful for one of them. Since i knew how to solve that part myself and in the interest of reducing the noise i skipped mentioning that sub part, but forgot to remove the 8 secs comment. – Hardy Aug 29 '12 at 19:13
up vote 3 down vote accepted

You correct that you need to evaluate: $$ \mathsf{Pr}\left(T_1 < T_2 \right) = \mathsf{E}\left(\mathsf{Pr}\left(T_1 < T_2 | T_1\right)\right) = \mathsf{E}\left( \mathrm{e}^{-\lambda_2 T_1} \right) = \left.\frac{\lambda_1}{\lambda_1-t } \right|_{t = -\lambda_2} =\frac{\lambda_1}{\lambda_1+\lambda_2 } $$ where $\mathsf{Pr}\left(T > x\right) = \mathrm{e}^{-\lambda x}$ was used, as well as the moment-generating function $\mathsf{E}\left(\exp(t T)\right) = \frac{\lambda}{\lambda-t}$.

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You have two independent Poisson processes $X_t$ and $Y_t$ with rates $\lambda_1$ and $\lambda_2$. Now here's a way to construct such processes. Start with a Poisson process $Z_t$ of rate $\lambda_1 + \lambda_2$, and independently assign each occurrence to $X_t$ or $Y_t$ with probabilities $\lambda_1/(\lambda_1 + \lambda_2)$ and $\lambda_2/(\lambda_1 + \lambda_2)$ respectively. Check that this does give you independent Poisson processes with the given rates. By construction, the probability of the next occurrence being from $X_t$ is $\lambda_1/(\lambda_1 + \lambda_2)$. So this must also be the case for your independent Poisson processes.

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