Probability of n events over a time period

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SUMMARY

The discussion focuses on calculating the probability of n drops from a leaking tap over a one-minute period, where the probability of a droplet occurring each second is 1%. The probability of no drops in a minute is calculated as 0.99^60, leading to the probability of one or more drops being 1 - 0.99^60. The problem is modeled using a binomial distribution with parameters N = 60 trials and p = 0.01. The formula for the probability of exactly n drops is given by P(n) = C(N, n) * p^n * (1-p)^(N-n), where C(N, n) represents the binomial coefficient.

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MrOd67
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How do you calculate the probability of n number of events over a time period given a constant probability p of the event occuring independently at any given time?
Let's say you have a leaking tab, and the probability of a droplet in any given second is 1%, regardless of whether there was a drop previously.

How would you calculate the probability of n drops in a minute?

No drops in a second is 0.99, so no drops over a minute is 0.99^60. Hence one or more drops is 1-0.99^60. But how about exactly n drops?
 
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The way you are modelling it you have a binomial distribution with effectively a trial every second and a independent probability of ##p= 0.01## of success in each and every trial.

https://en.wikipedia.org/wiki/Binomial_distribution

The probability of exactly ##n## drops involves binomial coefficients: $$P(n) = \binom{N}{n}p^n(1-p)^{N-n}$$ where ##N = 60## is the total number of trials.

You could model this sort of thing using a Poisson distribution:

https://en.wikipedia.org/wiki/Poisson_distribution

Although, for a dripping tap, maybe neither of these is quite right.
 
PeroK said:
The way you are modelling it you have a binomial distribution with effectively a trial every second and a independent probability of ##p= 0.01## of success in each and every trial.

https://en.wikipedia.org/wiki/Binomial_distribution

The probability of exactly ##n## drops involves binomial coefficients: $$P(n) = \binom{N}{n}p^n(1-p)^{N-n}$$ where ##N = 60## is the total number of trials.

You could model this sort of thing using a Poisson distribution:

https://en.wikipedia.org/wiki/Poisson_distribution

Although, for a dripping tap, maybe neither of these is quite right.
Perfect, that's exactly what I was looking for. Thank you very much!
 

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