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Customers arrive at a service center according to a Poisson process with a mean interarrival time of 15 minutes. 1) If two customers were observed to have arrived in the first hour, what is the probability that both arrived in the last 10 minutes of that hour? 2) If two customers were observed to have arrived in the first hour, what is the probability that at least one arrived in the last 10 minutes of that hour?


My understanding: The arrivals are uniformly distributed. Let X=customer arrives in the last 10 minutes. Then $P(X)= \frac 16$. The arrivals are independent, so the arrival of 2 customers would have the probability of $\frac 16*\frac 16=\frac 1{36}$ (answer to 1).

Regarding the 2). Let Y=number of customers arriving in the last 10 minutes. Then $P(Y>=1)=P(Y=1)+P(Y=2)=\frac 16 + \frac 1{36}=0.19$.

Is it correct ?

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  • $\begingroup$ Why do you think that the arrivals are uniformly distributed? $\endgroup$ – Alfredo Jul 30 at 7:18
  • $\begingroup$ @Alfredo this is a basic property of the Poisson process. It can be shown using the independence of arrivals in disjoint intervals. Note that this does not contradict the fact that the time to the next arrival is exponential. $\endgroup$ – pre-kidney Jul 30 at 8:03
  • $\begingroup$ @pre-kidney What I want to say is that the arrivals are not uniformly distributed. That is not the same that saying that the arrivals are uniformly distributed in the interval $[0,U]$ knowing they came before the time $U$. $\endgroup$ – Alfredo Jul 30 at 8:25
  • $\begingroup$ @Alfredo the logic used in the question statement is correct. Here is a precise formulation. Let $S$ be the (random) set of arrivals in a Poisson process on an interval, with arbitrary intensity. Fix a deterministic integer $k\geq 1$ and let $X_1,\ldots,X_k$ be uniform random elements of the same interval. Then the law of the random set $\{X_1,\ldots,X_k\}$ is equal to the conditional law of $S$ given $|S|=k$. Does that clear up your confusion? $\endgroup$ – pre-kidney Jul 31 at 1:30
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Your reasoning is correct as well as your answer for 1) but not for 2), due to an incorrect value for $\mathbb P(Y=1)$. Perhaps a simpler way of solving 2) is to calculate the complementary probability: $$ \mathbb P(Y=0)=\frac{5}{6}\cdot\frac{5}{6}=\frac{25}{36},\qquad \mathbb P(Y\geq 1)=1-\mathbb P(Y=0)=\frac{11}{36}. $$ Now comparing with your method, we can see that $\mathbb P(Y=1)$ does not equal $\frac16$ as you wrote, but rather $\frac{10}{36}$. How to see it directly? Call the arrival times $T_1$ and $T_2$. Then either $T_1$ falls in the last 10 minutes and $T_2$ falls in the first 50 minutes, or vice versa. Thus $$ \mathbb P(Y=1)=\frac{1}{6}\cdot \frac56 + \frac56\cdot\frac16=\frac{10}{36}. $$

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$\lambda = \frac{1}{15}$

Probability that two customers arrive in the first hour

$$=\frac{e^{-\frac{60}{15}}4^2}{2!}\tag 1$$

Probability that two customers arrive in the last 10 minutes and that 0 customers arrive in the first 50 minutes. Use $\lambda t$ for the mean of the poisson probability.

$$=\frac{e^{-\frac{10}{15}}(\frac{2}{3})^2}{2!}.\frac{e^{-\frac{50}{15}}(\frac{10}{3})^0}{0!}\tag 2$$

The required probability in part I $$=\frac{(2)}{(1)} =0.0278 $$

Part II Probability that atleast one customer arrive in the last 10 minutes has two cases. Case 1 is one arrive in the first 50 minutes and 1 arrive in the last 10 minutes and Case 2 is none arrive in the first 50 minutes and 2 arrive in the last 10 minutes.

$$\frac{e^{-\frac{50}{15}}(\frac{10}{3})^0}{0!}.\frac{e^{-\frac{10}{15}}(\frac{2}{3})^2}{2!}+\frac{e^{-\frac{50}{15}}(\frac{10}{3})^1}{1!}.\frac{e^{-\frac{10}{15}}(\frac{2}{3})^1}{1!}\tag 3$$

The required probability in Part II $$ = \frac{(3)}{(1)} =0.30556$$

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  • $\begingroup$ I have downvoted your answer, even though it is technically correct, because it does not answer the question that was asked, and presents an inferior method than was used in the question, leaving the answers in an unsimplified form that obscures the cancellations showing they are simple fractions! ($1/36$ and $11/36$, respectively). You will notice that all the exponential factors you have written cancel out when taking ratios, and this cancellation is one way of seeing why the conditional distribution is uniform (as asserted in the question). $\endgroup$ – pre-kidney Jul 30 at 8:19
  • $\begingroup$ Using Bayes' rule is part of the problem and hence made this argument which comes out to be the same as your method. $\endgroup$ – Satish Ramanathan Jul 30 at 8:25

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