A is an average student who has a probability of $1/3$ to pass any exam he gives. He needs to pass $4$ more to finish his studies. Let $X$ be the number of exams he will finally give until he finishes his studies. What is the expected number of exams he will give $ E[X] $?
Obviously, this means I will have to find the PDF and use $ E(x) = \sum_{x=4}^{\infty} \left[xf(x)\right] $. What my intuition says is that $f(x) = \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^{x-4} $. Since he will have to give at least $4$ exams to finish his studies and $(x-4)$ other exams that he didn't pass. But what I found is that $ \sum_{x\in S} f(x) \neq 1 $ , which it should be. I'm missing something and I don't know what. Can someone please clear this out for me? Thanks
We define a successful trial as failing an exam, and a failed trial as passing an exam, so that Wiki's formula for the mean of a negative binomial distribution can be applied.
After each exam (each Bernoulli trial), the student will continue to take exam(s) with probability $2/3$ (success parameter $2/3$), and this will stop upon his graduation with $4$ passed exams ($4$ failed Bernoulli trials). The Negative Binomial distribution models the number of failed exams (successful trials), so $X - 4 \sim NB(4,2/3)$.
$$E[X] = \frac{\frac23 \cdot 4}{1 - \frac23} + 4 = 8 + 4 = 12$$