I need help please with probability assessment

Well guys, the question is this: We'll will have 8 essay questions on the exam, and we know what the questions are, and on the exam date the teacher will randomly pick out two among those eight, and among those two we will choose one and write and essay for it. Now I don't have time to study all 8 questions so my question is: What is the probability of me being screwed if I study for 5 questions only? I mean I know no math or anything but it seems a pretty low chance for both of the questions to be among the three ones for which I would not have studied. Thanks..

The Finite John schrieb:

There are 8*7 ways to select two questions, and 3*2 of these are unfavorable to you. Your chance of being screwed is 6/56=10.7%.

ObPuzzle: Why is he turning to a puzzle group to ask this?

No and Yes. There are C(8,2) = 8*7/2 ways to select two questions and C(3,2) = 3 unfavorable ways. But the chance of being screwed is indeed 6/56 = 3/28. In mathematics at least, two wrongs sometimes do equal a right.

'Glenn C. Rhoads' schrieb:

Your formula is correct if order doesn't matter; mine is correct if order does; given the problem as stated, it doesn't matter either way as long as you're consistent, and it saves me from explaining to someone who doesn't know elementary statistics (if he did, he wouldn't have had to ask) what a binomial is or why you have to divide by two.

In a nutshell, I was aware of that, thank you.