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If you have a given probability for a specific outcome in a specific event or period of time, how can you calculate the probability of it happening over a length time or number of events?

I've seen answers that might explain what I'm talking about, but I don't understand the notation.

If I have a 10% chance of something happening in a given day, it is NOT a 100% chance over 10 days, but how do you calculate the actual odds?

I thought the way to do it was subtracting the chance it didn't happen from one, for example:

If I have a .1 chance of an outcome on a given day, I think the likelihood of it happening at some point over 10 days is 1-.9^10 which is approximately .65

Did I do this right? Does a 10 percent chance of an outcome per event only have a 65% chance of happening over the course of 10 events?

For a 1% probability over 100 events I tried 1-.99^100 which approximated to 63.4%.

If this is wrong, how do you accrue the likelihood of an event over a number of periods when only the likelihood in a single period is given?

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    $\begingroup$ You are correct. If the probability of the event occurring on a given day is $p$, and different days are independent, then the probability that the event occurs at least once in $n$ days is $1-(1-p)^n$. $\endgroup$ – angryavian Jan 10 at 21:11
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Yes, I agree to you and the comment of angyavian. Maybe a simplification of the term is useful. We have $1-\left(0.99\right)^{100}=1-\left(1-\frac{1}{100} \right)^{100}$

This term can be approximated. For large $n$

$\left(1-\frac{x}{n} \right)^{n}\approx e^{-x}$

In your case it is

$$1-\left(1-\frac{1}{100} \right)^{100}\approx 1-e^{-1}\approx63.2\%$$

In both cases the probability is approximately $63\%$.

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