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For a Bernoulli random variable with success probability $p$, how does one compute the probability of having a ''success'' at the $k$-th trial out of $N$ repeated and independent trials where $k \le N$ ?

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    $\begingroup$ Isn't it just $p$? Maybe I'm misunderstanding the question. $\endgroup$ – mark999 Mar 6 '16 at 8:30
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As described in more general case here, it follows geometric distribution. You have $k$ independent trials with the same probability of success $p$, so you expect $k-1$ failures and then a success, what translates to

$$ \underbrace{(1-p)\times(1-p)\times\dots\times(1-p)}_{k-1 ~ \text{failures}} \times \underbrace{p}_{\text{success}} = (1-p)^{k-1}p $$

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