Statistics to estimate the student's knowledge Short Story
I come from here, Calculate probability of passing in an exam
Im tring to calculte the probability to pass in the exam, knowing that the stundent knows $K$ out of $N$ questions, the exam as $m$ questions and the user needs to get at least $i$ questions right to pass.
$$
P(Y \geq k) = \sum_{i=k}^{m} \frac{\binom{K}{i}\binom{N-K}{m-i}}{\binom{N}{m}}
$$
I undertand how to do this, but i can't rely on this 100%, i need to estimate the user knowledge
My Problem
I been trying to figure this out but with no luck, i don't understand how to get the aproximate value of $K$ 
$$
P(Y \geq y) = \sum_{i=y}^{m} \frac{\binom{K}{i}\binom{N-K}{m-i}}{\binom{N}{m}} = 0.05
$$
I know i need to use numerical methods to find $K$ but i don't understand how, i know that there is software outhere that does this, but that's not an option for me
Any help is appreciated
 A: It's not too clear from what you wrote, but I think the situation is this.
There are a total of $N$ possible questions, and the student knows the answers to $K$ of them.  A random subset of $m$ questions will be chosen to form an exam, and the student needs to know the answers to at least $y$ of them to pass.  You know $N$, $m$ and $y$; given that the student passes with probability $0.05$, you want to find $K$.
As you said, to do this exactly would require solving 
$$ \sum_{i=y}^m \dfrac{{K \choose i}{N-K \choose m-i}}{N \choose m} = 0.05 $$
However, if $N$ and $K$ are large, there is a useful approximation.  If we ignore the requirement to choose distinct questions and sample with repetition rather than without, it will make very little difference ($N$ being so large, the probability of choosing the same question more than once will be very small anyway).  The student has probability $p = K/N$ of getting any particular question right, and now the random variables for the different questions are independent, so the distribution of the number of correct answers is binomial with parameters $p$ and $m$.  You want to find $p$ such that $P(X \ge y) = 0.05$ for a binomial random variable with parameters $p$ and $m$.
This is still not easy to calculate by hand, so let's make a further approximation: if $m$ is large, by the De Moivre-Laplace theorem, $X$ is approximately normal with mean $\mu = mp$ and standard deviation $\sigma = \sqrt{mp(1-p)}$.  For
a normal random variable with that mean and standard deviation, 
$P(X \ge y) = 0.05$ if $y = \mu + 1.64485 \sigma$ (approximately).
Thus you want $p$ such that $$y = mp - 1.64485 \sqrt{mp(1-p)} \tag{1}$$
Subtract $mp$ from both sides and square, and you get a quadratic equation to solve for $p$:
$$ (m^2 + 2.70553 m) p^2 - (2 m y + 2.70553 m) p + y^2 = 0 $$
There will be two solutions to the quadratic: you want to check which one is between $0$ and $1$ and works in (1).
