Quick states question.

Quick stats question
If there are ten essay questions to prepare for, and two of them will be selected on my final tomorrow morning, but I only have to answer one of the two, and I choose to not study for 2 of them, there is only a 2% chance that the two that I don't study for will be the two on the final.

[b61476781f1]True/False[/b61476781f1]

As far as I can tell, it is true, and I like those odds, but wanted to double check with others first. On the other hand, I beat the 33% odds that the test I took yesterday wouldn't have one of the essay questions I didn't study for, so maybe I am just an excellent gambler.


Okay here's the states question just to settle the typo in the title
What's the capital of Alaska?

Oh fuck, in my calculations I didn't account for the fact that the same essay question could not be selected twice.

That makes my odds of getting fucked tomorrow much larger but I'm not sure by how much. HALP?

isn't it just (2/10) x (1/9)

[quoted0f8f44daa="akalic"]isn't it just (2/10) x (1/9)[/quoted0f8f44daa]
That's 2.22%.

The capital of Alaska is Juneau. ;)

[ba9cfacb671]edit I am teh dumb, and didn't read the question properly. For your amusement, I leave my calculations for the wrong question below[/ba9cfacb671]

OK, your question was poorly worded, because I don't know if you are asking "What are the odds that BOTH questions I choose to skip are on the exam" or of you are asking "What are the odds that AT LEAST ONE of the questions I choose to skip is on the exam".

Let's assume that the possible questions are numbered 0-9, and that each has an equal chance of being on the final (probably not likely...).

There are 45 possible combinations of questions. Obviously we don't care which question comes first (eg. if the exam has questions 1 & 2, that's the same as if it had 2 & 1).

The calculation for this is 10 choose 2 (can't write it properly here, it's a 10 over a 2 all in brackets). 10 choose 2 = 10!/(8!li2!). The combinations are like this

0&1, 0&2, 0&3, 0&4, 0&5, 0&6, 0&7, 0&8, 0&9, 1&2, 1&3 ... etc

Let's say that you choose to skip studying questions 1 & 2. There is only one possible combination out of the 45 that includes both of those questions, so there is a 1/45 chance of BOTH questions that you skipped being on the exam.

However, what are the odds of AT LEAST ONE of the questions being on the exam?

There are 8 pairs of questions which include question 1, but not question 2 (0&1, 1&3, 1&4, 1&5 ...), and there are 8 pairs of questions which include question 2 (0&2, 2&3, 2&4, 2&5 ...), but not question 1.

So there are 16 possible combinations which include one (but not the other) of your two skipped questions.

So all together there are 17/45 combinations which have AT LEAST ONE of your skipped questions on the exam, which is a 38% chance of getting one of those questions.

Oh crap, you only have to answer 1

Um....recalculating...

Yeah, ok, never mind. 1/45 chance of you getting buggered.

[quote107834c970="TFOAF"][quote107834c970="akalic"]isn't it just (2/10) x (1/9)[/quote107834c970]
That's 2.22%.

The capital of Alaska is Juneau. ;)[/quote107834c970]

Haha i dunno, i didn't bother to punch it in the calculator as I just answered it super quick from what I know about stats lol. the question is true or false anyways P

Yeah, I figured it was about 2%, but that seemed to fail to take into account the fact that the same question can't appear twice. But maybe I just don't remember my stats very well.

[b1ef3f24a9a]For those interested[/b1ef3f24a9a], it turns out the chances are 100% that both questions will appear on the test, because that is exactly what happened. I had a study guide for the ten essay questions, did not even bother reading those two questions because they listed a bunch of countries and dates (the rest were more conceptual), but those were the two on the test.

I did however manage to write a pretty bitching essay despite this setback. Fuck yeah.