So we have seen before (here: probability of passing an exam (continued)) that the average number of hours I would have worked for an exam, assuming I work one hour for an exam that I can pass with a probability p, and assuming I can try to pass it at most N times, is $\sum_{k=0}^{N-1}q^k$. (q=1-p)
I know have one last question. What if this exam is actually composed of r tests, which follow the rules described above (so all tests can be taken at most N times), and the test i can only be passed if i-1 is taken (so if I reach N once, the whole exam is stopped, and thus my number of hours spent into this exam stops growing).
What would then be the average number of hours spent into the exam ?
If the question is unclear please state so and I'll try to reformulate.
edit: maybe a little more explanation is needed as to why the exam requires on average $\sum_{k=0}^{N-1}q^k$ hours of work: $E(X)=\sum_{i=1}^{N-1}iP(X=i)+P(X=N)$, and $P(X=N)=pq^N$. The sum can be calculated by noticing that it's a sum of derivative, and it comes $E(X)=\frac{p}{(1-q)^2} \left( (N-1)q^{N}-Nq^{N-1}+1 \right)$+$pq^N$=(1-q^N)/(1-q)
My question has still not been answered.