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Following my previous question, which can be found here: probability of passing an exam, I found out that the probability of passing an exam at the nth try is $p(1-p)^{n-1}$.

If I now assume that taking an exam takes me one hour of work, how many hours on average will I have worked if the maximum number of retries is N (regardless of whether I end up passing the exam or not) ?

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    Right. ${}{}{}{}{}{}{}{}{}$2012-06-12

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The probability of taking the exam $n$ times is then $p(1-p)^{n-1}$ except for the last, which has probability $(1-p)^{N-1}$ as you stop in any case. The average or expected number of hours is then $\sum_{i=1}^N P(i)i=\sum_{i=1}^{N-1} ip(1-p)^{i-1}+(1-p)^{N-1}N$ To sum the series, let $q=1-p$. Then $\sum_{i=1}^{N-1} ip(1-p)^{i-1}=p\sum_{i=1}^{N-1} iq^{i-1}=p\sum_{i=1}^{N-1} \frac d{dq} q^i=p \frac d{dq} \frac {q-q^N}{1-q}$

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    @TimZEI: I was trying to answer the question you asked-what is the expected number of hours of studying-not the chance that you don't pass by N tries. My comments about the chance that you don't pass by N tries were explaining the deletion of $p$ from the last term. So they should agree, as we are working on the same question. If you pass on try i, you study$i$hours. If you don't pass on N you study N hours.2012-06-13
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A person can give 3 trial exams for the clearing test. In 1st attempt probability of passing is 40 , Those who have failed have 60 probability of passing the exam in 2nd attempt. Those who have failed in 2nd attempt have 20 probability of passing the exam