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The title says pretty much all. The event is fully random, it has the same chance any minute, it can happen multiple times at the same minute. How is it possible to calculate that?

I made a simulation code in c++, which says 78%, is that good answer? https://gist.github.com/bakaiadam/4f4732f4147fc3a5c68f121bf57b919f

Edit: since people say it's not clear what I mean and I'm not really familiar with random distribution types, I tell you the concrete example that I was thinking about: My idea came from seeing group of people passing in a forest path. Let's say i have been there for 2 hours and saw 2 group of people. What is the probability that i will see at least 1 group of people in the next 10 minutes? Obviously there can be more than one group of people passing at the same minute, and obviously their chance to be there is independent from each other.

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  • $\begingroup$ We can't answer this question without knowing what distribution the time between event follows etc $\endgroup$
    – Xiaomi
    Oct 18, 2018 at 16:03
  • $\begingroup$ In the simulation, you have assumed two events in 20 minutes or 120 minutes. IF we assume exponential, with 120 minutes, we get a probabiity of nothing happening in the next 10 minutes is 0.98. $\endgroup$
    – Satish Ramanathan
    Oct 18, 2018 at 16:12
  • $\begingroup$ @SatishRamanathan the thing that confuses me is he says "fully at random" - so presumably independent. If that's the case, why would the condition on past events be of consequence at all $\endgroup$
    – Xiaomi
    Oct 18, 2018 at 16:13
  • $\begingroup$ We call it exponentially distributed random variable. It is still random $\endgroup$
    – Satish Ramanathan
    Oct 18, 2018 at 16:14
  • $\begingroup$ 0.98 is too high. It will be a little over 0.8. $\endgroup$
    – Paul
    Oct 18, 2018 at 16:17

2 Answers 2

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In these cases a Poisson didn’t distribution is typically assumed, although there might be reasons to assume otherwise. In the case, you have a Poisson parameter of $1/(60min)$. From this you can find out you probability of 0 occurrences. Your probability looks too low here.

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I'm not really familiar with random distribution types

You need to get a grasp of probability distributions to understand the problem and solution.

You can think of the time between the events i.e. some one passing through the path that you are looking at as a random variable. It could be 1 minute or 8 minutes or 10 seconds etc. To get a handle on the randomness of this variable you need a mathematical function that gives you the probability that this random variable is less than some value or greater than some value or is in some interval, etc. Exponential distribution is the mathematical function that will give us that facility.

What is the probability that some random event won't happen in the next 10 minutes given it happened exactly twice in the last 120 minutes?

In your example you say that the event occured twice in the last 120 minutes. So, the rate at which it occurs is 120/2 i.e. 1 event per 60 minute. Let $X$ denote the time between the first event the next $10$ minutes.

So, the desired probability is $P[X>x] = \exp^{-(\lambda*x)}$. Here $\lambda$ is the rate parameter which is nothing but $1/60$ i.e. the number of events happening per minute. And $x$ is the time till the event.

So, plugging the numbers you get $P[X>10] = \exp^{-(10/60)}$.

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