How would you solve this probability problem?

The question is phrased a little strangely.

First you have to define "at least a B". Does that mean 80% or 16 of the questions right?

Then does "how many combinations are there to consider" mean "how many combinations of T/F get at least 16 right?"

I'll assume both of those are true. Then I'd add up the number "16 right" + "17 right" + "18 right" + "19 right" + "20 right".

Let's look at one of those, 17 right. There are 20C17 = 20C3 ways to choose the 17 right or 3 wrong positions among the 20 questions. Therefore that is the number of T/F choices in which exactly 17 questions are right.

I think that gives you enough to finish the problem.

Well, there are 2 possible answers for each (I did this same exact problem earlier in the year!) You would then realized that the "you need to get at least a B..." is irrelevant. Then you would do 2*20=40. 40/2= 20 possible answers! Lol, anyway. Next, you would do COMBINATION. 2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2... the number two multiplied by itself 20 times is your answer. The answer is 1048576. That sounds a little high though, but I think this might be right because if you only had 2 questions you would do 2*2= 4 possible outcomes which is correct.

Hope this helps!

Can anyone answer my question at http://answers.yahoo.com/question/index;_ylt=AquOp...