How many exam topics should be covered?

April 25, 2006 4:03 AM

There are 10 possible exam topics. 6 will appear on the paper. 2 questions must be attempted. How well is an ass covered based on the number of topics covered?

Assume the 6 are selected from the 10 at random.
From covering 10 topics to just 2, what are the answers to the following questions?

What level of topic choice is ensured in the exam?
What levels of topic choice are likely?
What is the risk of not having covered two topics which appear?

I feel that there is a lot of extra useful data which could come out of this question. If you feel there is another angle which I have not requested please feel free to show me the data.

posted by kenaman to education (13 comments total)

You might assume that each topic is selected at random, with the condition that no topic appear twice in the same paper. In which case someone here will without a doubt do the mathematics for you.

In my experience of exam papers however, topic selection is not random. It's a function of which if any topics are related and which have come up in recent years. The best way to get a handle of it is to look at past papers.

You need at least six topics to be sure that you have two questions you've revised for, as the 4 you didn't can't then make up all 6 that are on the paper if topics cannot repeat.
posted by edd at 4:12 AM on April 25, 2006

I'm not sure exactly what you're looking for, whether it's some kind of probability question or you actually have this exam coming up.

If the latter, then to be safe for the exam you need to be comfortable with 6 topics - this allows for four of the topics not to appear on it, and you answer the remaining two.
posted by jacalata at 4:14 AM on April 25, 2006

on post: what he said
posted by jacalata at 4:14 AM on April 25, 2006

Great question. I'm doing a similar set of maths for my exams.

I think that if they are giving you 10 questions, study for 6. If they are giving you 8 or 6, study for 4?
posted by k8t at 4:23 AM on April 25, 2006

Past exam papers will be consulted and patterns searched for, however I really cant ask metafilter to factor that in.

It is a real exam being prepared for by a friend.

I am looking for the data describing risks/guarantees, best worse case senarios associate with the various different levels of coverage.

e.g.
10 ensures choice from any 6
0% risk of not having 2 covered.
.
.
.
5 ensures 1 will be present
x% risk of not having 2 covered
Best case senario all five covered.
Worst case senario 1 covered....
posted by kenaman at 4:41 AM on April 25, 2006

1 Studied Questions
--Odds 0 of your studied questions will appear: 40.0%
--Odds 1 of your studied questions will appear: 60.0%
2 Studied Questions
--Odds 0 of your studied questions will appear: 13.3%
--Odds 1 of your studied questions will appear: 53.3%
--Odds 2 of your studied questions will appear: 33.3%
3 Studied Questions
--Odds 0 of your studied questions will appear: 3.3%
--Odds 1 of your studied questions will appear: 30.0%
--Odds 2 of your studied questions will appear: 50.0%
--Odds 3 of your studied questions will appear: 16.7%
4 Studied Questions
--Odds 0 of your studied questions will appear: 0.5%
--Odds 1 of your studied questions will appear: 11.4%
--Odds 2 of your studied questions will appear: 42.9%
--Odds 3 of your studied questions will appear: 38.1%
--Odds 4 of your studied questions will appear: 7.1%
5 Studied Questions
--Odds 0 of your studied questions will appear: 0.0%
--Odds 1 of your studied questions will appear: 2.4%
--Odds 2 of your studied questions will appear: 23.8%
--Odds 3 of your studied questions will appear: 47.6%
--Odds 4 of your studied questions will appear: 23.8%
--Odds 5 of your studied questions will appear: 2.4%
posted by onalark at 6:23 AM on April 25, 2006

All sorts of unnecessary discourse on combinatorics follows.

This assumes all sorts of things like independence and no prior knowledge like previous tests, but this was assumed in the original question (I hope).

I used Matlab to crunch and output the numbers. The basis of my logic for conducting the likelihood tests was this:

I used C the choose function as defined in probability C(n,k) = (n!)/(n-k!*k!). It defines the number of ways I can pick k objects from a set of n.

For any i studied questions and j of the studied questions appearing on the test you want to divide the total number of ways the professor can make j of the i questions appear on the test by the number of possible ways to choose 6 questions from a set of 10.

The denominator, 6 questions from a set of 10, is hopefully straightforward: it's C(10,6). Up top is a little harder, but it comes out to being equal to the number of ways I can avoid the j of the i studied questions [C(10-i,6-j) * the number of ways I can arrange the i studied questions j times [C(i,j)].

To clarify, I have two separate equations in my numerator. These correspond to the two groups of possible questions, those that you have studied for and those you haven't. If you studied 3 questions and the professor put 2 of your studied questions on the exam (i=3, j=2), then there are C(3,2) = 3 different sets of 2 questions you studied he could have asked {Q1 Q2}, {Q2 Q3}, {Q1 Q3}. This is where the C(i,j) comes from.

Remember the second term corresponds to questions you haven't studied for. If you studied a total of 3 questions, there are 10-i=10-3=7 questions you haven't studied for to choose from. On a 6 question exam, if the professor has chosen 2 questions you studied for on the test, there are 6-j=6-2=4 questions you didn't study for on it. Thus he chooses 4 questions from a set of 7 or C(7,4), which leads to C(10-i,j-6).

Lemme know if this looks ugly, TeX makes math so much more readable...
posted by onalark at 6:48 AM on April 25, 2006

And of course, some Matlab code (sans comments)

C = @(n,k) (factorial(n))/(factorial(n-k)*factorial(k))

for i=1:5
fprintf('%d Studied Questions\n', i);
for j=0:i
if 10 - i < 6 - jbr> jProb = 0;
else
jProb = C(i,j)*C(10-i,6-j)/C(10,6);
end
fprintf('--Odds %d of your studied questions will appear: %.1f%%\n',j,jProb*100);
end
end

posted by onalark at 6:49 AM on April 25, 2006

onalark - thats exactly what I was looking for. Now I can see that:

Studying 5 topics gives a 97.6% chance of covering at least 2 topics.
Studying 4 gives an 88.1% chance of minimum coverage.

Your data coupled with an examination of past exam papers will allow my friend to make an informed decision regarding their exam preparation. Thank you

on preview - thanks for the explanations also....they look interesting!
posted by kenaman at 6:51 AM on April 25, 2006

Hmnn, only thing that came out funky is the line that begins:

if 10 i- i < 6...br>
That should end with 6 - j and the next line should begin with

jProb = 0;

I'm a muck when it comes to putting stuff in comments without them breaking.
posted by onalark at 6:52 AM on April 25, 2006

Let my official advice to your friend be to study 6 topics. That's an absolute guarantee. My numbers are certainly suspect until andrew cooke shows up and verifies my maths.
posted by onalark at 6:56 AM on April 25, 2006

well, of course one should cover all six, but sometimes you have to make choices - depends on the kind of exam, but sometimes it's better to have depth in fewer topics than breadth across the spectrum. This way s/he can prioritize the topics so that there are say four areas that s/he has really well, and two where the knowledge is more superficial - doable if necessary, but not optimum. That would probably give an overall greater likelihood of a higher grade if the quality of work is going to be a major factor than trying to cover all topics equally.
posted by mdn at 8:39 AM on April 25, 2006

I'm a big fan of question spotting in these circumstances.
posted by unSane at 12:22 PM on April 25, 2006