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.
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, $p_4=1-p_2-p_3$.
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?
