1. Fairly Basic/Easy Probablity Problem?

So my professor gave 6 possible essay topics for the final exam. Of these 6 he will choose 2 that will appear on the exam. I am to choose 1 of the 2 essay topics he presents on the exam and respond to it.

So I am strapped for time and wish to study as little as possible. So, naturally my first thought was to only study for 5/6 possible topics--ensuring that I would at least be able to answer one of the topics no matter what the professor chooses.

But what are the odds that if I only study for 4/6 possible topics that I will be left completely in the dark? (read: the 2 topics he choses are the 2 that I didn't prepare for.)

So, is this right?

For his first choice: 2/6 chance that he chooses one that I didn't prepare for.

For his second choice: 1/5 chance that he chooses one that I didn't prepare for.

Total probability = (2/6) * (1/5) = 6.67%

Also if, the above is right, then is it safe to assume that the odds that I know at least one of the 2 essay topics on the exam is = 100% - 6.67% = 93.33%?

Discuss.

2. Originally Posted by eysikal
So my professor gave 6 possible essay topics for the final exam. Of these 6 he will choose 2 that will appear on the exam. I am to choose 1 of the 2 essay topics he presents on the exam and respond to it.

So I am strapped for time and wish to study as little as possible. So, naturally my first thought was to only study for 5/6 possible topics--ensuring that I would at least be able to answer one of the topics no matter what the professor chooses.

But what are the odds that if I only study for 4/6 possible topics that I will be left completely in the dark? (read: the 2 topics he choses are the 2 that I didn't prepare for.)

So, is this right?

For his first choice: 2/6 chance that he chooses one that I didn't prepare for.

For his second choice: 1/5 chance that he chooses one that I didn't prepare for.

Total probability = (2/6) * (1/5) = 6.67%

Also if, the above is right, then is it safe to assume that the odds that I know at least one of the 2 essay topics on the exam is = 100% - 6.67% = 93.33%?

Discuss.
You have a sample consiting of 4 questions you can answer and 2 that you can't.

Let X be the random variable number of questions chosen that you can answer.

$\displaystyle \Pr(X = 0) = \frac{{4 \choose 0} \cdot {2 \choose 2}}{{6 \choose 2}} = \frac{1}{15}$.

(This is an example of a hypergeometric distribution problem: Hypergeometric distribution - Wikipedia, the free encyclopedia)

3. Wow, thank you.

So, in essence, I was right....but now you have given me this model to better conceptualize the problem.