Calculating Probability of Waiting 2 Seconds for Event A | P(rate x) = 76.9%

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Homework Help Overview

The discussion revolves around calculating the probability of waiting 2 seconds for event A, given a measured average wait time of 2.6 seconds. Participants are exploring the implications of this average wait time and its relation to probability calculations.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are questioning the validity of using the measured average wait time in probability calculations and whether a change in the average wait time can be assumed. There are discussions about the nature of the process (e.g., Poisson process) and the implications of the average wait time on the calculations. Some participants suggest different formulas and approaches to determine the probability.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have offered insights into the nature of the probability distribution and the need for additional context or information to proceed effectively. There is no explicit consensus on the approach to take.

Contextual Notes

Participants note the lack of information regarding the nature of the process (e.g., whether it is a Poisson process) and the absence of details about the variance of the sample, which limits the ability to draw conclusions.

moonman239
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Problem
I want to find the probability that I will have to wait 2 seconds on average for event A to happen, given that the measured average wait time is 2.6 seconds.

Attempt at a solution:

The probability that event A will occur at any given second is 1/2.6. Then the probability of waiting 2 seconds on the average is (1/2.6 * 2) = 76.9%.
 
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Maybe I'm missing something. If the MEASURED average wait time is 2.6 seconds, then one should probably concede that 2.6 second is the actual average wait time, unless of course the measurements are biased or plain wrong.

What is the probability that the average wait time will change to 2 seconds?

That would require a fundamental change in the underlying system, which cannot be guaged by the information provided.

Then again maybe I'm missing something here?
 
Yes. I want to find the probability that say the next 6 occurances will happen on average 2 seconds apart.
 
Could the formula be P(waiting 2 seconds)^6?
 
To do this problem, we need more information. Is it a Poisson process? That's what I first thought of when I read the problem, but you don't mention that anywhere. Then you said that the probability of it occurring in any particular second was 1/2.6, which further makes it sound like a Poisson process. But then you say the "measured" average wait time, which sounds like a sampling question, where you're looking for error in the measurement. But there's no discussion of the variance of the sample, so we can't do much with that.

Anyway, if it's a Poisson process with mean interarrival time 2.6s, then the underlying PDF for interarrival times is:

[tex]f(x) = 2.6e^{-2.6x} (x > 0)[/tex]
 
Thank you. So that would give me the probability that the wait time is greater than 0, right? So then maybe I can use the binomial dist formula to answer my question?
 
Well, no. That's the probability density function for the interarrival wait times if the process is Poisson. A particular wait time is always greater than zero, right?

It might be helpful if you gave us some context for this problem (ie, more of the problem if there is any, what class it's for).
 
I think I found the answer to that question: e^-mean interval * desired times for event x to happen^mean interval / desired times for event A to happen!
 

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