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 - Reinstate Monica 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 - Reinstate Monica 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 - Reinstate Monica Oct 28 '16 at 8:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.