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I have an exam that I can pass with a probability p.

If I fail the exam, I can retry as many times as I want until I do (the chances to succeed at the 3rd try is still p).

What's the probability that I succeed at the n th retry (which is to say that I took n-1 exams, failed them, and then succeeded at the nth) ?

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    Yes of course, I assume it to be independent.2012-06-12

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You have to fail the first with probability $(1-p)$, in the second ..., in (n-1)th and pass in the nth. Then the probability is $(1-p)^{n-1}p$.

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    This is the well-known geometric distribution. It is often defined in terms of this property.2012-06-12