# probability of passing an exam

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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Your chance of passing on the 1st try is $p$. The probability of you succeeding at the second try is the probability that you fail the first time (which is $1-p$) and then pass the second time (which is $p$). The probability of succeeding on the third try is the probability that you fail the first two times, and then pass the third time. Do you see the pattern? – Chris Taylor Jun 12 '12 at 13:39
@ChrisTaylor so P+(1-p)*p+(1-p)(1-p)p+...$\sum_{n=1}^{\infty} p^n(1-p)^{n-1} so he would just need to get the expected value. – yiyi Jun 12 '12 at 14:18 MAoYiyi, could you please explain in more details the link between your comment and the one from ChrisTaylor ? – TimZEI Jun 12 '12 at 15:51 There is, of course, in all this, the unstated assumption that each exam attempt is an independent trial of an experiment on which the probability of success is$p$. One would hope that in real life a student would learn something from each failure and that the probability of failure would be smaller on each successive try. – Dilip Sarwate Jun 12 '12 at 18:13 Yes of course, I assume it to be independent. – TimZEI Jun 12 '12 at 19:46 ## 1 Answer You have to fail the first with probability$(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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So which solution is correct, this one (which seems the correct one to me) or then one described by Chris Taylor and MaoYiyi ? – TimZEI Jun 12 '12 at 15:10
This is identical to the answer that I described (except that I described the process of getting there, rather than giving you the final answer.) – Chris Taylor Jun 12 '12 at 15:15
Sorry my comment was misleading. Indeed your comment is the same as the answer given by Marcos. What I wanted to say rather is that I don't see the link between MaoYiyi's answer and yours (or Marcos'). – TimZEI Jun 12 '12 at 15:19
This is the well-known geometric distribution. It is often defined in terms of this property. – Michael Chernick Jun 12 '12 at 17:22