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This is barely a probability question, but I needed to check to make sure the solution is as simple as I believe it to be.

What is the the expected number $n$ of independent trials needed to have $x$ success (not necessary to be consecutive) given probability $p$?

I would assume since each trial is independent the solution would be $n = x/p$, but perhaps I am overlooking something here.

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    The answer is correct but the reasoning is a bit wobbly. If this were what you wrote on an exam or on homework, it would likely not be accepted.2012-01-26
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    Your intuition is right. Let $W_1$ be the waiting time (total number of trials) up to first success, $W_2$ the waiting time from first success to second, and so on. Each $W_i$ has geometric distribution. So expected waiting time to $x$-th success is $xE(W_1)$. And $E(W_1)=1/p$. This is intuitively very reasonable, but in probability the intuition is all too often wrong. However, the fact that $E(W_1)=1/p$ is not hard to verify.2012-01-26
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    yes thank you, I was simplifying it.2012-01-26

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