Little help with some math?

[ozymandias]

Can someone help me with this simple sounding but confusing calculation?

I have a midterm in a few days, prof put out 27 terms, 9 of which will be on the exam. We will have to answer any 3 of those 9.

How many terms can I get away with studying with no risk of having to write a term I did not particularly study?

Thanks!

 

[BubblyMasquerade]

is this an actual math problem or are you just being lazy?

 

[binge]

27

 

[CDFS]

21

 

[aKula]

21 is correct.

Consider if you study 20. Then the prof could choose the 7 you didn't study and 2 you did. So for no risk you must study 21. Then the worst case choice would be 6 you didn't study and 3 you did.

 

[rdawson808]

Originally Posted by ozymandias

How many terms can I get away with studying with no risk of having to write a term I did not particularly study?


You must study all 27 to have no risk. The actual questions you are given will be a random sample, so each is as likely to be chosen as the others.

Besides, you're being lazy. Go study them all.


b

 

[gnatty8]

Originally Posted by rdawson808

You must study all 27 to have no risk. The actual questions you are given will be a random sample, so each is as likely to be chosen as the others.

Besides, you're being lazy. Go study them all.


b

But he only has to answer 3 of the 9, meaning at worst, if he studied all but the 6 that were actually placed on the exam, he would still know enough (3) to pass, no?

 

[Thomas]

wait - what kind of class is this where there are 27 relevant terms, the class will be given 9 (1/3 of the material) to answer, and then the student need answer only 3 (1/9 of the material).?

 

[Douglas]

This is a trick question. He will actually have to study 42.

 

[edinatlanta]

Originally Posted by Thomas

wait - what kind of class is this where there are 27 relevant terms, the class will be given 9 (1/3 of the material) to answer, and then the student need answer only 3 (1/9 of the material).?

I had similar scenarios. We would be given a study guide with six questions that would be used for essays, only four of those would be on the exam and of those four we would have to answer two.

If that makes sense.

 

[scarphe]

Originally Posted by gnatty8

But he only has to answer 3 of the 9, meaning at worst, if he studied all but the 6 that were actually placed on the exam, he would still know enough (3) to pass, no?

if you only need 3 to pass that is ****** grading policy assuuming all the problems are given equal value.

 

[rach2jlc]

Originally Posted by Douglas

This is a trick question. He will actually have to study 42.
Well, that's the answer to EVERYTHING, Doug-san. Anyway, according to a delightful Hamburglar Adventure I once heard about from teh Comic Book Store Guy, the answer is "Fries."

 

[MrG]

Originally Posted by edinatlanta

I had similar scenarios. We would be given a study guide with six questions that would be used for essays, only four of those would be on the exam and of those four we would have to answer two.

If that makes sense.

I had the same experience. There was usually some overlap in the questions, so there weren't as many unique concepts as there were questions.

I always assumed it was a way to force us to study everything without actually having to test us on all of it.

 

[rdawson808]

Originally Posted by gnatty8

But he only has to answer 3 of the 9, meaning at worst, if he studied all but the 6 that were actually placed on the exam, he would still know enough (3) to pass, no?


Unless I'm missing something, he can't know this. This hinges on him wanting no risk.

As I read it, another way to state his question is "how many of the terms do I need to study to guarantee that I have studied the ones that will be asked?" (I'm ignoring the word "particularly" because it is undefined.) The inherent problem is that he has zero knowledge of which ones the prof will choose. So each has equal probability of showing up. And that probability is positive. Therefore he must study each.

But I'm no probability expert. I could be missing something.

 

[Thomas]

Originally Posted by rdawson808

Unless I'm missing something, he can't know this. This hinges on him wanting no risk.

As I read it, another way to state his question is "how many of the terms do I need to study to guarantee that I have studied the ones that will be asked?" (I'm ignoring the word "particularly" because it is undefined.) The inherent problem is that he has zero knowledge of which ones the prof will choose. So each has equal probability of showing up. And that probability is positive. Therefore he must study each.

But I'm no probability expert. I could be missing something.

Actually, if it's probability you want, here's a few good statistics relative to this issue:

So, you will have to answer three questions out of 27 concepts - that's 1 in 27, or roughly 4% chance of being tested on each concept. However, reality tends to work its magic so that:

if you study 13 concepts, you ought to have a 47% chance of your chosen questions being on the exam. however - in practice this runs closer to 12% due to game theory and quantium statistics.

If you study 14 concepts *(commonly known as the "tipping Point" because you've just gone over 50%), you ought to have a 52% chance of your studied questions making the exam, but due to probabilistic tendencies and random chance - you're more likely to experience a 15% exam proficiency rate.

If you study 21 concepts, though - this is 50% more than the tipping point of 14, and therefore...well, I'll cut through the jargon and theoretical aspects of probabililstic calculus and Murphy's Law and tell you that your exam proficiency rate will skyrocket to 30%.

So, if you want to answer three questions well, I think that studying all 27, plus a few spares, should give you a good shot.