Say you're taking a multiple choice test and each question has 4 choices and you're going to randomly select all your answers.
If you know that you can ignore one of the answers in each question because you know it is incorrect, is that relevant other than just making it a 1/3 chance of being right per question instead of 1/4? Or if disregarding two answers, a 1/2 chance? I guess I'm not certain whether it actually is just as simple as making it 1/3 and 1/2, or if there's something else I have to do.
If I wanted to find the probability of getting a certain amount of questions on the test right, say 6 right out of 10 questions, is there some other way I have to take the ignoring an answer part into account or can I just use 1/3 and 1/2 instead of 1/4 in my binomial distribution formula?
It is as simple as you predicted, if you can throw out one answer which you know for sure is incorrect, the probability would change to 1/3. But if you have a question of the type "probability of getting x right out of n", remember that to use 1/3 you must be able to throw out one answer from EVERY question. Otherwise, it becomes a little more difficult.