Hi So I have this question below. I know I need to model each lane as a separate Poisson distribution. The possible answers are:

a) 11.4%; 22.4%; 33.4%; 44.4%; 55.4%

b) 2.74%; 4.74%; 12.74%; 34.74%; 64.74%

but obviously I need to know how to get to the answer.


Vehicles arrive at a rate of $1200$ vehicles per hour at a traffic signal. $15\%$ of the vehicles turn right while $85\%$ travel straight through. The street has one lane per direction, which is widened at the intersection to one straight through lane, and one right-turn lane. The right-turn lane can accommodate $5$ vehicles but will be blocked if $5$ or more straight-through vehicles are waiting at the intersection. Determine the following:

a) 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 probability that more than $5$ right-turn vehicles will arrive at the intersection during the next red period of $45$ seconds.

For question a) I think I need to find $\Pr(X=5)$ for the straight lane while $\Pr(X=0)$ for the right lane. $$ \Large P_P(x; qt) = \frac{(q\cdot t)^x\cdot e^{-q\cdot t}}{x!} $$

Where I am having trouble though is how do I determine $q$?

  • 1
    $\begingroup$ Can you provide the reference for the text? $\endgroup$ – Antoni Parellada Oct 25 '16 at 16:41
  • $\begingroup$ Unfortunately the source is from a question my lecturer gave in a previous year's assignment. The question was a multiple choice type question. with the answers: a) 11.4%; (22.4%); 33.4%; 44.4%; 55.4% b) 2.74%; 4.74%; 12.74%; (34.74%); 64.74% $\endgroup$ – Kyle Oct 26 '16 at 12:56
  • $\begingroup$ Can we assume that the red-light cycle of 45 sec. applies to question a? $\endgroup$ – Antoni Parellada Oct 26 '16 at 13:36
  • $\begingroup$ @AntoniParellada See answer below. I think I have it. Obviously the "answers" I showed in the question were incorrect. Furthermore, the 45sec must be used in a), surely? $\endgroup$ – Kyle Oct 27 '16 at 7:15

It's tricky to answer homework problems, but one indirect way of looking into part (b) of the problem is to simulate (the closed-form solution still up to you). It seems as though your question elicited quite a bit of interest, and it could possibly get more entries if you erased the actual answers. So here it goes...

In R:

lambda_right = 0.15 * 45 * 1200/60^2        # Rate parameter of 2.25 cars/period (45 sec.)
nsim = 10^6                                 # One million simulations
right_arrivals = rpois(nsim, lambda_right)  # Poisson process rate lambda
mean(right_arrivals > 5)                    # Percentage with more than 5 R turning cars
[1] 0.027423

Here is the plot:

enter image description here

The mean value is clearly consistent with the setup on the simulation ($\lambda =2.25$), and it is clear that the majority of arrivals are below $5$ cars/period, explaining the low percentage for $>5$.


Take 2:

a)Conditional probability of 5 cars going straight, given that there are 5 cars already.

enter image description here




enter image description here

=2.74% Can anyone confirm my answers?

  • $\begingroup$ Part (a) is asking for a conditional probability. Given that there are five vehicles waiting in total, what's the probability that none of them are in the right-turn lane, as opposed to one in the right-turn lane & four in the straight-through lane, or all five in the right-turn lane? (You don't need to make any assumptions about the cycle time of the signal.) $\endgroup$ – Scortchi Oct 27 '16 at 14:23
  • $\begingroup$ @AntoniParellada: You're welcome! But don't you think you're giving away too much too soon in your answer? With regard to (b): there are only 3 seconds on average between cars - can a Poisson process be a sensible model for arrivals? $\endgroup$ – Scortchi Oct 27 '16 at 22:19
  • $\begingroup$ @Scortchi thanks for the help. I think i understand what you mean. See my second attempt. Can you confirm my answers now? $\endgroup$ – Kyle Oct 28 '16 at 7:07
  • $\begingroup$ Furthermore, I should have mentioned that the question was supposed to be answered as either Binomial or Poisson. I used Poisson due to the time relation... Is this a correct assumption? $\endgroup$ – Kyle Oct 28 '16 at 7:08
  • $\begingroup$ @Kyle: Your assumptions need stating & justifying. You seem to have assumed two independent Poisson distributions & calculated the conditional probability from their joint distribution & the marginal distribution of the total. (The LHS notation looks the wrong way round by the way.) Though that's a reasonable approach, you could in fact get the same result from weaker assumptions. $\endgroup$ – Scortchi Oct 28 '16 at 8:47

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