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I am not an mathematician so forgive me if my terminology is incorrect. I believe this question is similar to this one but I am not sure.

Imagine there is a process that happens once per second and takes 17 ms to finish. That is at t = 0 the process starts and runs for 17 ms, then the process is stopped for 0.983 s, then at t = 1 the process starts again, and so on.

Now I randomly select an instant of time, i.e. 0.23 s, 0.550203 s, 0.9999 s, from every second and do so for, say, 8 hours (28800 selections).

What is the probability that I would have selected, say 1000, instants that coincide with the running process?

I have a feeling that I am making this more complicated than I have to.

The probability that a single selection occurs during the running process is 17 ms / 1 s = 0.017.

That's about as far as I've got.

Edit: really what I want to know is the probability that I select at least 1000 instants that coincide.

cholz
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1 Answers1

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If you assume that each time is chosen independently of the others, then the random variable that describes the number of selections that coincide with the running process follows a binomial distribution.

In your example, if we want exactly $1000$ coincidences, the probability will be: $$P= {28800 \choose 1000}(0.017)^{1000}(1-0.017)^{27800}.$$

Why? We perform $28800$ selections, and we must choose $1000$ to be our "successes" trials, what can be made of $28800 \choose 1000$ ways. If we want exactly $1000$ coincidences, those $1000$ selections must coincide, and this happens with probability $(0.017)^{1000}$ AND the other selections must NOT coincide, and this happens with probability $(1-0.017)^{27800}$.

Of course, calculate this probability exaclty is difficult. In this case, since the probability of success is small and the number of trials is large, we can approximate the Binomial Distribution by a Normal one.

Célio Augusto
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  • Thanks. This helped a lot. I guess the difficulty in computing the answer lies in the difficulty of the factorial calculation? – cholz Sep 30 '20 at 18:55
  • Yes. Note that, since the number trials is large, the probability of exactly $1000$ successes is pratically $0$. It's most interesting to compute things like "at most" $1000$ successes and the like. You can use this online calculator to help you https://stattrek.com/online-calculator/binomial.aspx – Célio Augusto Sep 30 '20 at 19:23