This is a typical preparation worksheet for a graduate school comprehensive examination. Ten possible questions are listed, and the candidate is informed that four of the ten questions will appear on the exam. Out of the four questions, the candidate has his choice of two questions to answer.
What the typical student confronts is a question as to how to maximize his efficiency in studying for the examination. How deeply to study each potential question is a function of his time, which is limited. This is ultimately a probability question, which is underpinned by counting.
There are only a couple of important questions, when approaching this combinatorics problem. The first is order. Does the order of the selection matter? In this case it does not matter if the proctor decides to select question no. 4 before question no. 2. The second question is one of repetition. Is it reasonable to expect the proctor to select any one question more than one time? The implication here is that it is not. If the proctor selects question no. 4, he will consider that question unavailable for his next selection.
5040/24=210
Which is the number of possible arrangements when choosing 4 questions from a pool of 10.
Now our student wants to maximize his time and study as many problems as deeply and efficiently as he can. Let's suppose that he decides to study five questions, and discard the other five. At this point, we're working in the abstract, so it doesn't matter which five he ultimately chooses. What he's really doing is dividing the total pool of problems into two sets.
Set D={discarded questions} and Set S={studied questions}
In his first example, he's going to let the cardinalities of D and S be equal to 5. What is the probability that of the 4 chosen questions, all will be in the set D of discarded questions? In other words, if a student studies 5 out of the ten problems, how likely is it that he will fail his comprehensive examination, and be thrown out of graduate school?
This is, again, a combinatorics problem. We have our number of possible selections, and now we need to know what statisticians call the event space, or the number of ways our event might happen. The first thing we need to know is how many ways can our proctor select questions that we won't be studying?
Guys who are pros with arithmetic call this "5 choose 4" or 5C4. The number of ways in which our proctor can choose only questions our student has not studied is 5. We divide this by the total we came up with earlier, and get an estimate as to our student's chances of failing the exam.
5/210 or 1/42
Of course, our student will also fail the test if three of the four questions include those he hasn't studied. We can add this probability 5C3/210 in our heads if we remember our stats courses, and come up with about a 1/14 sum.
To summarize, we have a student who has been presented with ten questions, of the ten, four will show up on an examination. Of the four, he has his choice to answer two questions. He wants the maximum amount of time to study the smallest number of questions possible, while minimizing his chances of failing the exam. What he doesn't want is sit down and be presented with three or four unstudied questions.
If he chooses to study five of the ten questions, he will have a 1-(1/14) or 13/14 (about 93%) chance of passing his comprehensive exams.
Ideally, of course, our student would study all ten questions deeply, but we don't live in an ideal world. If I were such a student, I would want better odds than 93%, and would choose six of the questions (preferably those with which I had the least familiarity) to study. That would put me at about (41/42) or 98%
Good luck!
