Question about probability and poisson process

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The discussion centers on calculating the probabilities of a single customer arriving at two different systems, B and C, modeled as independent Poisson processes with rates Lambda1 and Lambda2. The user seeks to understand how to compute the probability that the arrival time at system B is less than that at system C. It is suggested to condition on one of the events and apply the law of total probability for the calculation. The user confirms that they found the solution following the provided guidance. This highlights the application of probability theory in analyzing independent Poisson processes.
quacam09
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Hi all, I have a question about probability. Can you help me?

There are 2 events:
- Customer A arrives the system B in accordance with a Poisson process with rate Lambda1
- Customer A arrives the system C in accordance with a Poisson process with rate Lambda2.

Given that Poisson processes are mutually independent. Computing the probability of the event that customer A arrivers the system B and the probability of the event that customer A arrivers the system C?

Thank you!
 
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I'm guessing there is only one custumer. The holding times are exponential with parameter lambda. The question then is \mathbb{P}(\mathbf{e}_{\lambda_1}< \mathbf{e}_{\lambda_2}). Condition on one of them and use the law of total probability.
 
Thank you. As your suggestion, I found the solution.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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