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The following couldn't be farther removed from any homework problem, although it arose from getting stuck on a question posted here with too much homework flavor as to probably remain unanswered. To avoid having to click on the hyperlink:

It's a queuing problem with traffic: $1200$ cars/hour go through a traffic signal at an intersection. $85\%$ is thru traffic; and $15\%$ is right turning traffic. There is a right-turning lane before the intersection with capacity for $5$ cars. This lane becomes inaccessible if there are more than $5$ cars in the straight-traffic lane. Questions:

(a) [posted just because the answer makes reference to it] The probability that no vehicles are waiting in the right-turn lane while there is a total of $5$ vehicles waiting at the intersection.

(b) [the part I'm interested in] The probability that more than $5$ right-turn cars will be at the intersection during the next $45$ seconds.

The expected rate per second (Poisson process) is $\small \lambda= 1200/60^2=0.\overline{3}\text{ cars/sec.}$; and the expected interarrival time (exponential) is $\small 1/\lambda=3 \text{ sec./car}$ (3 seconds of separation between cars).

My specific question: How do you calculate that $\small \color{red}{28\%}$ of the cars will be just a second behind the one in front? This is one of the comments (and already too much of a follow-up thread to keep on asking).

I see that $28\%$ of the expected number of cars in $45$ sec is $.28\times\lambda\times 45=4.2$. So the expected arrival time of, rounding, $4$ cars should be $4/\lambda=12$ sec. A result that seemed immediate given the interarrival time of $3$ sec. Not $1$ second. What am I missing?

It would be great to also get some guidance regarding the part of the comment indicating that a Poisson process may not be optimal, and the advantage of "assuming a constant flow of traffic with right-turners randomly interspersed".

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If the inter-arrival times follow an exponential distribution then 28% comes straight from the cumulative distribution function: $1-\exp\left(\frac{-1\,\mathrm{s}}{3\,\mathrm{s}}\right)\approx 28.3\%$. It's perhaps reasonable to use a Poisson process to model traffic on a quiet road with plenty of opportunity for drivers to over-take cars in front; but not to model heavy traffic, when cars impede each other.

For the first part of the question, asking for the probability that there are no right turners among the first five to arrive at the junction after the light change, it would be better therefore to avoid relying on an unnecessary & implausible assumption—a binomial distribution can be justified directly without being derived as a conditional distribution.

For the second part, though with a mean inter-arrival time of 20 seconds between right-turners alone, the Poisson model seems less inappropriate, one might assume that all cars arrive like clockwork at fixed intervals, 15 of them in 45 seconds; & again use the binomial distribution for the no. right-turners. Under this binomial model the variance of the count over any time period is 85% of the variance under the Poisson model:

$$0.15 \cdot \frac{t}{3\,\mathrm{s}} = 0.05 t \,\mathrm{s}^{-1}$$

compared with

$$\frac{t}{3\,\mathrm{s}} \cdot 0.15 \cdot 0.85 = 0.0425 t \,\mathrm{s}^{-1}$$

You'd expect them to give similar answers insofar as the time period is long & the proportion of right turners is small, by the Law of Rare Events.

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  • $\begingroup$ Thank you very much for answering. What would be the exact lambda parameter and number of counts to get the $28 \%$? $\endgroup$ – Antoni Parellada Oct 30 '16 at 17:24
  • $\begingroup$ @Toni: The rate parameter's given in the question, 1200 cars per hour, 1/3 per second. Not sure what you mean by the no. counts. $\endgroup$ – Scortchi Oct 30 '16 at 22:44
  • $\begingroup$ OK... I am done now... Here's the bottom line... I was entering $1-\exp(-3)$ with the very simple misunderstanding being that the $\lambda$ is the Poisson rate parameter (as well as mean and variance), and not double-checking the tricky part: in the exponential $\lambda$ (same as Poisson) is the rate parameter, but the mean is $1/\lambda$, and it was this $1/\lambda = 3$ that I was entering as $1-\exp(-3)$, or pexp(1, 3) as opposed to pexp(1, 1/3). $\endgroup$ – Antoni Parellada Oct 31 '16 at 1:03
  • $\begingroup$ For the first part: $\Pr(5 \text{ straight}\vert 5 \text{ waiting}) =\Pr_{\text{binomial}}(5 \text{ successes}, 5 \text{ trials}, 0.85) = 44.4\%$, or dbinom(5,5,.85) 0.4437053, as already posted in the now erased answer. For the second part, I see a huge discrepancy in the actual numerical result of pbinom(q = 4, size = 15, prob = .15, lower.tail = F) [1] 0.06170539 ($6.2\%$), and ppois(q = 5, lambda = 0.15 *45 * 1200/60^2, lower.tail = F) [1] 0.02736535 ($2.7\%$)! I was expecting to get identical values... $\endgroup$ – Antoni Parellada Oct 31 '16 at 2:55
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    $\begingroup$ Right, that would exclude ($=5$): pbinom(q = 5, size = 15, prob = .15, lower.tail = F) [1] 0.01681009. $\endgroup$ – Antoni Parellada Oct 31 '16 at 10:59

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