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?

  • 1
    $\begingroup$ I'm having a little trouble thinking through it clearly at the moment (it's late), but I suspect these probabilities are given by the Hypergeometric distribution. $\endgroup$ – Jonathan Christensen Jan 2 '13 at 6:02

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.

  • $\begingroup$ I just noticed a semantic ambiguity in my question. Did you take L as the number of questions the lazy student studied where L<(Z+X-Y) or did you take L as the number of question below the ideal Z+X-Y that the lazy student studied, ie he studied Z+X-Y-L questions? (Thank you for responding, btw.) $\endgroup$ – Double AA Jan 2 '13 at 7:39
  • $\begingroup$ The $L$ is the number of questions actually studied. There is no explicit restriction on $L$, that is taken care of by the convention mentioned about $\binom{a}{b}$. $\endgroup$ – André Nicolas Jan 2 '13 at 7:54
  • $\begingroup$ Ok I have edited the Q to clarify. +1 (I like to hold off on checkmarks for a day or so just to see if anything else happens. Don't worry; I won't forget.) $\endgroup$ – Double AA Jan 2 '13 at 8:00
  • $\begingroup$ The expression you've derived is exactly the Hypergeometric distribution. In Wikipedia's notation, your $L$ is $n$, your $X$ is $N$, and your $Y$ is $K$. There's not a useful closed form for the cdf, so you just have to do the sum. $\endgroup$ – Jonathan Christensen Jan 2 '13 at 18:20

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