Probability of passing a multiple choice exam, general formula? You are presented with a multiple choice exam where each question has one correct answer. 
 You firstly solve all questions you know for sure, and then pick the rest at random.
What are your odds of passing it for a given passing grade?
After a qucik peak on the Binomial distribution, I have came up with the following expression:
(Correct me if I made a mistake somewhere, but I believe that won't be necessary)

$$ P = \sum_{t = \lceil p  q \rceil-n}^q \binom{q}{t}\frac{(a-1)^{q-t}}{a^q}$$
$p$ - passing grade from $0$% to $100$% $(0\le p \le 1)$
   $q$ - number of unknown questions 
   $n$ - number of known (correct) questions 
   $a$ - number of answers (choices) per question
If $t \le 0$ then use $P = 1$, you passed for sure.

But this seems like too much to calculate if you are going to do it by hand and are dealing with large or "ugly" values even thought its as simple as I believe it can be.
I'm looking for easiest and quickest ways & formulas to do this calculation, or at least a good and calculable by hand approximation. 
I guess you can always round up a bit the "ugly" values, and for the large values looking up in a table of binomial coefficients? But then again is there a better way or why not?
 A: Let me define some variables of my own:
$n$=number of questions
$k$=number of questions where the answers are known.
$u$=number of questions where the answers are unknown.
$a$= number of answers (choices) per question.
$c\leq u$ is the minimum number of questions where the answers are unknown and you have to choose the right answer to pass the exam.
Thus $q=\frac1a$ is the probability that the correct answer is selected if the answer is unkown.
$\frac{k}{n}\leq P \leq 1$ ist the treshold to pass the exam.
The inequality is 
$\frac{k}n+\sum_{i=0}^{c} \binom{u}{i} q^i\cdot (1-q)^{u-i}\geq P$
$\sum_{i=0}^{c} \binom{u}{i} q^i\cdot (1-q)^{u-i}\geq P-\frac{k}n$
If $X$ is distributed as $X\sim Bin(u,q)$ then the equation can be written like
$P(X\leq c)\geq P-\frac{k}n$
If $u$ is large enough you can apply the central limit theorem (Approximation of the binomial distribution by the normal distribution). Otherwise you have to calculate $\sum_{i=0}^{c} \binom{u}{i} q^i\cdot (1-q)^{u-i}$ for some $c.$
