Expected Grade on Strange Testing Scheme A teacher wants to give a test for a class. She gives the students 12 problems to study, 8 of which will be on the test. Students can pick 4 questions to answer and only those will be graded. If a student memorizes the answer to k problems, what is his expected grade?
My teacher did this for one of our tests and I got curious. I can work out the expected grade for k < 4 with a simple binomial distribution, but I'm not sure how to do this for the other possibilities. 
 A: To answer the question, we need to make some assumptions. The least plausible is that if the student has not memorized the answer, she will get $0$ on the question. We also assume that the questions to be memorized, and the questions on the test, are all equally likely to be chosen, and independent.
If $k\ge 8$, the student will get a perfect score, which we call $4$. So now we need to deal with $0\le k\le 7$. The arguments for $k\le 4$ and $5\le k\le 7$ are a little different.
Case $k\le 4$: For any $i$ from $0$ to $4$, we compute the probability $p_i$ that exactly $i$ of the memorized questions are on the test. There are $\binom{12}{8}$ equally likely ways to choose the $8$ questions. 
There are $\binom{k}{i}$ ways to choose $i$ memorized questions and for each of these ways there are $\binom{12-k}{8-i}$ ways to choose the rest of the $8$ questions from the $12-k$ unmemorized. Thus
$$p_i=\frac{\binom{k}{i}\binom{12-k}{8-i}}{\binom{12}{8}}.$$
We are using the convention that if $a\lt b$ then $\binom{a}{b}=0$.
Finally, the expected score is $\sum_{1}^4 ip_i$.
Cases $5\le k\le 7$: We do a sample case, say $k=6$. Then $p_0=p_1=0$. 
The probabilities $p_2$ and $p_3$ are calculated just as in the case $k\le 4$. But $p_4$ is different, since it is possible for the teacher to choose more than $4$ questions the student has memorized. The fix is easy, $p_4=1-p_2-p_3$.
A: Here are some hints.
Memorizing $8$ of the $12$ questions will ensure that you ace the test.  Worst case there will be $4$ on the test you did memorize, and $4$ you didn't, in which case you answer the $4$ you did.
There are a total of $_{12}C_8 = 495$ possible tests.
Let's say you memorized questions $1$ through $7$.  Then, you can either get a test that has at least four of the questions you memorized, or only three (questions $8$ through $12$ showed up on the exam).  Let's count the number of tests that you won't ace.  We know questions $8$ through $12$ are on there, so we pick three from $1$ through $7$ ($_7C_3 = 35$).
So, if you memorized $7$ questions, $35$ tests will result in a $3/4$, and $460$ will result in a grade of $4/4$.  Then your expected grade is
$$G(7) = \frac{35 \cdot 3 + 460 \cdot 4}{495}  \approx 3.93.$$
Can you take it from here?
