I have two types of customers (type 1 and type 2) enter a shop. Their arrival processes are independent and follow Poisson process with the arrival rates of $\lambda_1$ and $\lambda_2.$

Consider two events where $A = \{\text{customer } q+1 \text{ is type 1}\}$ and $B = \{\text{more than } q \text{ customers from all types arrive} \\ \text{ within a specified time}\}.$

What is the probability of $P(A,B)$? I want mathematical formulation for this problem.

Thanks in advance for considering my question.

  • $\begingroup$ It's easy to solve this problem via Monte Carlo simulation. $\endgroup$ – Digio Feb 8 '19 at 22:45
  • $\begingroup$ Do you mean the probability that both events $A$ and $B$ occur? $\endgroup$ – Michael Hardy Feb 10 '19 at 1:20
  • $\begingroup$ @Digio : This should admit an easy closed-form solution. (Provided we can be sure exactly what question is intended.) $\endgroup$ – Michael Hardy Feb 10 '19 at 1:21
  • $\begingroup$ I suspect you actually meant $\Pr(A\mid B),$ the conditional probability of $A$ given $B,$ since otherwise you'd probably have said something about how much time has passed. $\endgroup$ – Michael Hardy Feb 10 '19 at 1:22
  • $\begingroup$ Some things are unclear in your question. If customers arrive at rate $\lambda,$ then the average time until the next customer arrives is $1/\lambda.$ The event $B$ seems to be that at least $q$ customers have arrived. Does that mean by some particular time that many have arrived? Or before some particular event happens? Are you asking about the conditional probability $\Pr(A\mid B)$? Or about $\Pr(A\ \&\ B)$? Or something else? That is not clear. However the question about the probability that the $(q+1)$th customer is of type $1$ is clear. $\endgroup$ – Michael Hardy Feb 10 '19 at 3:02

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