# two independent Poisson Arrivals

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.

• It's easy to solve this problem via Monte Carlo simulation. – Digio Feb 8 at 22:45
• Do you mean the probability that both events $A$ and $B$ occur? – Michael Hardy Feb 10 at 1:20
• @Digio : This should admit an easy closed-form solution. (Provided we can be sure exactly what question is intended.) – Michael Hardy Feb 10 at 1:21
• 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. – Michael Hardy Feb 10 at 1:22
• 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. – Michael Hardy Feb 10 at 3:02