Test-question elimination probablity No, this question is not on a test. It is about tests.
A common strategy among test givers is to provide students beforehand with X problems, of which Y will appear on the test and of which Z will need to be answered. (X,Y,Z are natural numbers such that X>Y>Z)
So, to have a 100% success rate at knowing the answer, a student must prepare Z+X-Y problems (lest the X-Y problems that are left off are all ones he studied).
My question is, as a function of X Y Z and another natural number L, what are the chances that the student will be OK on the test if he studies some number of problems, L, which is less than Z+X-Y?
 A: One needs to assign meaning to the phrase "will be OK on the test." We will assume that the student is a perfectionist, and that her definition of OK means being able to answer at least $Z$ of the questions on the test.   
There are $\binom{X}{L}$ ways to select $L$ questions from the $X$ available.
We count the number of such choices that contain at least $Z$ questions from the ones chosen by the instructor. If this count turns out to be $K$, then the required probability is
$$\frac{K}{\binom{X}{L}}.$$
The number of ways that we can choose $Z$ questions from the $Y$ questions the instructor chose, and $L-Z$ questions from the $X-Y$ the instructor did not choose, is $\binom{Y}{Z}\binom{X-Y}{L-Z}$.
The number of ways that we can choose $Z+1$ questions from $Y$, and $L-Z-1$ questions from $X-Y$ is $\binom{Y}{Z+1}\binom{X-Y}{L-Z-1}$.
The number of ways that we can choose $Z+2$ questions from $Y$, and $L-Z-2$ questions from $X-Y$ is $\binom{Y}{Z+2}\binom{X-Y}{L-Z-2}$.
And so on. Add up to find $K$. So a correct formula for $K$ is
$$K=\sum_{i=Z}^L \binom{Y}{i}\binom{X-Y}{L-i}.\tag{$1$}$$
We are using the convention that if $a\lt b$, then $\binom{a}{b}=0$. There does not appear to be any obvious way to obtain a closed-form expression for $K$.
If the student considers it "OK" to be able to answer a number of questions smaller than $Z$, it is easy to modify the expression $(1)$ suitably: the sum just starts earlier. 
